point-set-topology

[rancid umbra]

not an element

so C is a finite collection of F's

[prime elbow]

Okay I see

I got it

@rancid umbra great

[rancid umbra]

nice 😎

[rancid umbra]

oh, btw, you want C to also be non-empty when defining \cal D

[rancid umbra]

why do i feel like there is a proof of this using bundles

[prime elbow]

But they are because they are finite intersection of elements of \cal C

Any hint?

[rancid umbra]

empty set is finite

[prime elbow]

What are bundles?

[rancid umbra]

so f(C) = empty causes a problem if C = emptyset: \bigcap \cal D would be empty!

[prime elbow]

Take C \subset \cal C, such that C is non-empty

[rancid umbra]

yes, exactly

[prime elbow]

C is non-empty, finite set

[rancid umbra]

they are like generalizations of product spaces and covering spaces. in the most general form, a bundle can be just a continuous surjection

[balmy nexus]

Maybe it will help to think that
"connectedness" is purely a topological notion (expressed purely with the language of topology) but

"path connected" uses the reals to describe what's going on. It is saying "not only is this connected, but it is connected in a close way to how R is connected" which is quite a lot because R is a pretty ideal space. Another problem is that path connectedness charts a 1d path through your space, which might be huge. So it's a magnifying glass for discontinuities in your space.

For example R is complete, so it has no jumps or skips in it. When you take a 1d path in 2d space, you may encounter these jumps, while an open set (which may be 2d) will not be able to use those jumps to seperate because infitisimally close there is a section that is denser.

So If you take a space like the infinite broom, where everything is path connected but most of the paths are forced to go through a specific point, you can delete that point and no longer be path-connected.

But the connectedness persists because there is still "density" around the point you remove, in a sense that every two open sets containing the point has more than that point as intersection. There might be a term for this, I just made it up.

K. Conrad has an expository paper on connectedness vs. path connectedness which might also be of help, look for it here: https://kconrad.math.uconn.edu/blurbs/ .

interesting, maybe that's what I was thinking about.

[balmy nexus]

I didn't test this analogy(ies) out in practice, so it might not be really useful... hope it helps

[pliant elk]

The Stone-Čech compactification takes X a Tychonoff space as homeomorphic to a dense subset of a compact Hausdorff space which induces unique commuting extensions to continuous maps X into K compact Hausdorff. If we relax the conditions on this map to just be a continuous bijection into a compact Hausdorff space or into a dense subset of one, do we still get something interesting (also would it work for non-Tychonoff spaces)?

I'm pretty sure the SC compactification still would satisfy the universal property in the second case, so necessarily the closure of its image would be homeomorphic to the SC compactification, but would you get anything about the map itself? I also can't really think of any examples of continuous bijections between (non-compact) Tychonoff spaces and CH spaces (pretty sure they should exist, though maybe not generally?)

(not entirely sure how much this is pointset or functional lol)

[merry wharf]

hey, I am pretty new to toplogy and in general I am not the most adv in abstract math but I am making very solid progres imo(im studying basic toplogy for the soul mostly, and to adv my mathematical skill)

I wanted to ask a pretty creative question which I feel like can clarify the core of the definition of toplogy

basically I imagined a field which changes. imagine an xy graph and you stick a rotating circle in the middle, which will change rotate inside it. I wonder if it fits into the axioms of a topology

if I imagine the sets as connected, i.e not dispersed no holes, then this will break the topology open set intersection and union axioms. but I wonder, can it possibly work in some topology? actually if we are on that can an open set in the axioms of a topology be dispersed or get dispersed i.e get holes?

I assume it fits into discrete topology, but looking more deeply at the wikipedia definitions I might be wrong. I wonder if you guys can add some depth to this question

[balmy nexus]

Can you try to be clearer? What do you mean by "field which changes"?

[merry wharf]

I mean it in the literal sense. similar to how a linear transformation changes a field, I imagine it being localized, for example just pick a circle in a euclidean space and rotate it while the rest of the field is static

or any general topology

actually now that I reflect a bit, that isnt too special, functions can do it hypothetically with a bit of tweaking

so I assume it will fall under the study of another field not topology which is concerned with shapes that dont have holes or sharp corners

my question came from thinking of the definition of a topology in the broadest sense as a field you can define a geometry in, and I wondered what if I rotate the field itself, and now I wonder if there is any mathematical difference between the field itself rotating and simply a function causing an equivalent change in a shape

there must be some difference, but I wonder on the specifics

[unreal stratus]

What do you mean by field here

[prime elbow]

Part e, I am taking X = (0,1) and closed set {1/n | n in N{1} }

And, U_n such that U_n = B(1/n, r_n), where r_n = 1/(6×2^n).
U = \bigcup U_n, n runs over N\ {1}.

I am claiming that there is no such eps>0 such that B(1/n, eps ) \subset U for all n in N \ {1}.

I tried with some eps, but I don't how do I write it ?

[alpine nest]

It is in (0,1)

[random sinew]

oh lmfao i didn’t read

ignore me

[balmy nexus]

Do you mean that the circle has the property of "being rotated", or do you mean the the points on the circle are rotated in regards to how they were before?

The first option is something you'll have to cook your own definition for, the second one depends on how you want to remember the the circle was rotated.

[azure harbor]

Hey guys, im quite confused by this answer to the difference between Sequential space and Frechet-urysohn space, how could we prove that cl(A) will be reached after some iteration?( my own attempt go like: if the space is sequential , first suppose A is not closed,( otherwise the result hold directly) , then there exist a sequence in A that converge to a point not in A ( in boundary of A tho), then suppose after first iteration cl1(A ) is still not closed, we may continue to reach additional boundary points of A, but how could we say that we could reach all boundary points and iteration ends. does this require some sort of transfinite induction ? (https://math.stackexchange.com/questions/269032/understanding-two-similar-definitions-fréchet-urysohn-space-and-sequential-spac)

[tender halo]

yea it requires transfinite induction

Frechet-urysohn spaces are the ones where this induction will end after one step

we will reach all points in cl(A) / A due to cardinality, we add at least new point at each step

so after less than |X|^+ (the successor cardinal of |X|, where |X| is the cardinality of the space) steps we will be done

or after \chi(|X|) (hartogs of |X|) if you want a choiceless argument

[merry wharf]

heres my general idea, and I kinda figured it out on my own in a way that I feel was educational for the ideas in topology

my topology I meant a geometry, in the simplest example a euclidean space. now, imagine a circle in the space independently rotates. the analogy would be a table that is the xy graph, with a circle that rotates for whatever reason, maybe a fancy cup holder

then I thought how this would apply to the logic of topology. I realized that you can assign some value for the circles rotation, lets say t for time, and you can represent it in the power set. so lets say a shape is partially in the cup holder, and it spins independently so it just spins without any effect on the overall shape

I wondered how it would be considered a geometry, but I realized it can be considered as part of a power set of the geometry(if you represent the geometry as a set of course which shouldnt be an issue)

[umbral hamlet]

why is it that the product topology is generated by the basis of products of open sets where only finitely many are not the whole space?

i dont see why this is equivalent to the canonical projections being continuous

[rancid umbra]

you want p_i^{-1}(U_i) = \prod_{j != i} X_j \times U_i open in the product for each i and U_i open in X_i.

the product topology is generated by the subbasis of all of these, so unions of finite intersections of things that look like the above

this is the smallest topology on the product that makes all the projections continuous

[umbral hamlet]

i see

[rancid umbra]

cool

[timber pike]

you can probably pass it to a tychonoff space using a tychonoff reflection and work there

[quartz horizon]

Well it’s not quite equivalent since any finer topology also makes the projections continuous

[quartz horizon]

for me the universal property gives a good motivation for the def of the product topology

[prime elbow]

Is it true that if X is locally compact and Hausdorff then for each x in X has compact open set?

[crisp lintel]

what do you mean by each x in X has compact open set

every compact open subset of a hausdorff space is also closed, so if you take X=R then the only compact open set is the emptyset

[prime elbow]

I mean each x, has open compact set U such that x in U

I got it

[sweet arrow]

Could someone check this proof pls

[quick crane]

what is your definition of $\partial A$?

[gentle osprey]

L

[quick crane]

this result is what I would take as the definition of \partial A, i.e. $\partial A = A \setminus (int(A) \cup ext(A))$

[gentle osprey]

L
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[sweet arrow]

the def is what u wrote \partial A = A \setminus (int(A) \cup ext(A))

[umbral hamlet]

I keep hearing about universal properties but idk what they are

[tender halo]

untrue unless your space is like very disconnected?

or X is compact i guess

[rancid umbra]

everything looks fine as written. however, you can simplify things a lot.
by (the negation of the) definition of Int A, for every nbhd U of p, U is not a subset of A, and so contains a point of X \ A.
similarly, by (the negation of the) definition of Ext A, for every nbhd of a point U of p, U is not a subset of X \ A, and so contains a point of A.

since you are defining bd A = X \ (Int A U Ext A), the reverse direction trivial: any open nbhd of p can't be a subset of A or X \ A, so p can't be in Int A or Ext A.

this proof could be done in one line if you just expand definitions

p in bd A <=>
p in X \ (Int A U Ext A) <=>
∀ U open nbhd of p, U ⊈ A and U ⊈ X \ A <=>
∀ U open nbhd of p, ∃ q in U ∩ A and ∃ q' in U ∩ (X \ A)

[quartz horizon]

universal properties tell you what something "does", how it relates to other things, what it can be used for

the universal property of the product says that continuous maps $Z \to \prod_i X_i$ correspond to families of continuous maps $Z \to X_i$, in that you can interconvert between the two

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[plush folio]

I recommend checking out https://topology.mitpress.mit.edu/, in particular chapter 1.4 to learn about how the product topology is characterized by the universal property

[sweet arrow]

wdym negation of the definition of Int A
I got X \ Int A = ∩ {B superset X \ A : B closed} and how does it. follow

[plush folio]

Lee's ITM also mentions the universal property of various constructions (but he calls them characteristic properties)

[alpine nest]

Yeah, I prefer that term

At least in the contexts in which I've seen it used.

(I've mostly seen them used for uniquely characterizing a construction in terms of how it behaves under some operation)

[quartz horizon]

i don't quite understand the term "characteristic property"

all the ones i've seen are instances of universal properties anyway

and for me i find having a consistent name easier

[rancid umbra]

Int A = {x in X : \exists U open nbhd of x such that U subset A} (this is the same as the union of all open nbhds contained in A), so its complement is {x in X : \forall U open nbhd of x, U is not a subset of A}

[alpine nest]

Yeah, I wasn't proposing "characteristic property" as a distinct term with a different definition, I was just saying that I've only ever seen universal properties used to characterize certain objects, so "characteristic property" is a synonym that better conveys their role for me

[quartz horizon]

Hm what do you mean by characterising certain objects

[rancid umbra]

i think that the way it is being used here is this:

• the characteristic property of the product topology says that functions into the product are continuous if and only if their component projections are
• the universal property says that given continuous maps into each factor, there is a unique set theoretic function into the product that is continuous because of the characteristic property of the product topology

here is an mse post which explains in more detail. i was also curious about the distinction and usually kind of lumped them into being the same thing.

[quartz horizon]

I see

[queen prism]

I thought it was called the characteristic property because the product topology is the only topology on the cartesian product with that property

[quartz horizon]

At least with riehl’s definition these are both universal properties

[rancid umbra]

its not the only one, its the coarsest one; any finer topology will also satisfy the characteristic property.
so in the poset category of topologies on the product (arrows going from finer to coarser), the product topology is terminal, or universal, with respect to this property

i wonder if you can separate this more formally, characteristic prop vs universal prop. i guess like, a characteristic property is a universal property in the right category

[quick crane]

For the product topology, the best definition for me is that it is the smallest topology making all the projection maps continuous. This is to ensure that a sequence (or net) converges in the product space if and only if each component converges.

[queen prism]

yea ok you're right

but then I wonder what lee is talking about here

[rancid umbra]

where specifically?

oh, uniqueness?

[queen prism]

yea

[rancid umbra]

okay, maybe this is wrong then...

the "any finer part"

[queen prism]

I think it's the coarsest topology such that each projection pi_i : X -> X_i is continuous
but that's different from saying that a function f : Z -> X is continuous if and only if its projections f_i : Z -> X_i are

[rancid umbra]

okay yes

lot of subtle properties here that i usually of identify as the same, but they are not

like this one

[alpine nest]

Yeah, the uniqueness is why I referred to it as characterizing the structure: the product topology is the unique topology on the product space satisfying that composition thing.

[prime elbow]

I don't know much, but it is an initial topology with X = Π X_i, where functions are projections, right?

And initial topology has this given characterization

[sly geyser]

ok I have a question

let's define two topologies on X x Y based on the topologies on X and Y

one is just the usual product topology

the other is a topology of all subsets S of X x Y so that all "slices" along X and Y ({(x,y) in S: y = y_1| and {(x,y) in S: x = x_1} are open when respectively projected to X and Y

but it's just taking (x,y_1) to x with a constant y_1 and (x_1, y) to y with a constant x

so it's whatever

what I said was a mouthful but what I basically mean is that they are open if all the "sections" along the X x {y_1} and {x_1} x Y planes are themselves respectively open in X and Y

my question is whether the two are the same

I know that the second is neccessarily finer or equal to the first

it's relatively simple, of course the products of the open sets satisfy the rule for the second, and the finite intersections and unrestricted unions transfer over to the planes

so all the members of the first topology are members of the second

[alpine nest]

It's possible for all the sections to be open but for the set not to be open in the product topology

You can find a counterexample even when X = Y = [0,1] with the standard topology.

[rustic cove]

I've been using Gemini AI to check my work and it's been doing an okay job (I think?), but it's giving me an answer for Munkres Topology 2nd edition that does not seem correct. For Chapter 1, Section 3, Exercise 5 c) it's saying that T = S' whereas I concluded that T contains just pairs from 4 line segments and all pairs from y=x and is not equal to S'. Any insight into this exercise would be appreciated. Anyone else using AI to check proofs? I've found that sometimes it changes its answer when I push it when something seems off.

[gaunt linden]

Whenever something produced by generative "AI" looks like it's mistaken, it probably is.

[alpine nest]

Whenever it looks right, it might well also be mistaken

Considering LLMs are trained to produce output that appears correct.

[rustic cove]

I gotta give it credit though, it does a great job reading my handwritting!

[cosmic mirage]

whenever something produced by generative "AI" it mistaken

anti robot blackout poetry

[worn mortar]

Does this proof make sense

If F is a topology on a set X consisting of open sets I denote the dual family consisting of all the complements of the open sets (the closed sets) by F*

[sly geyser]

bro this is great handwriting wdym "bad handwriting"

[lucid ocean]

it looks almost typeset

[sly geyser]

can you give me an example.... or I guess a counterexample

[alpine nest]

Consider $([0,1] \times [0,1]) \setminus {(t,t): t>0}$, i.e. the square with almost the entire diagonal removed, but with the bottom-left corner kept. The sections of this are either all of $[0,1]$, or sets of the form $[0,t)\cup (t,1]$, or $[0,1)$; in each case they're open in $[0,1]$. However, the set as a whole isn't open in the product topology, because $(0,0)$ has no neighborhood that would be entirely in the set (we are of course considering $[0,1]$ with the Euclidean topology).

[gentle osprey]

Outsider

[alpine nest]

This put me in mind of a harder problem; suppose A is a subset of [0,1]x[0,1] such that the intersection of A with every line on the plane is open (in the relative topology on said line). Does A need to be open? My gut tells me the answer is negative, but the counterexample will probably be trickier to construct.

Unless something like just messing up along a spiral or a parabola would suffice.

Yeah, never mind, the counterexample will be very similar in nature, just remove a circle somewhere within the square, but keep one point of that circle.

What if the intersection with every rectifiable curve is open? Intersection with every curve without qualification wouldn't work because of the space-filling curves.

[sly geyser]

sorry for the ghost ping

I said something, who's main promise might or might not be correct but whose argument was bullshit

(the product topology on two copies of the cofinite topology of Z being different from the section topology, it's true still I'm pretty sure but my previous argument was bullshit)

[alpine nest]

Cofinite topology is not Hausdorff so I have no intuitions

Although having so few open sets probably makes it fairly reasonable to analyze in this context

[prime elbow]

Let X and Y be two linearly ordered sets. Any strictly monotone surjection f: X -> Y is homeomorphism.

Let f is increasing map.
First I will show f is an open map, so take the basis set (a,b), yes there can be a basis set of type [a,b), but first consider (a,b).

Then f( (a,b) ) = (f(a), f(b) ) because of strictly monotone of f. Similarly we can show for other basis open sets, hence open sets maps to open sets.

Similarly f^-1 is strictly monotone shows that f^-1 maps open sets to open sets.

Is it correct?

[rancid umbra]

yes. note that showing f(a,b) = (f(a),f(b)) also uses the fact that X and Y are linearly ordered and that f is surjective

[prime elbow]

Yes

And we can find strictly monotone surjection from Q to Q_≥ 0( positive)

So Q and Q_≥0 are homeomorphic

[sly geyser]

yeah in this case you can just take the codiagonal

[sly geyser]

????

how?

that would mean there's an element q such that f(q) = 0

but that would imply that for every r < q, f(r) < 0 while also beng inside Q>=0?

I'm pretty sure for total orders atleast strictly monotone surjections are just bijections, or more specifically, isomorphisms

[alpine nest]

Yeah, there is a general theorem that says the two spaces in question are homemorphic, but I don't think the monotonicity argument works.

In general I'm not sure how to find an explicit homeomorphism, although it probably can be done.

[sly geyser]

because they're isomorphic any structure created out of their order properties should also naturally be isomorphic

including the topologies

[alpine nest]

What do you mean by "isomorphic" here?

[tender halo]

c squared made an explicit one

it was a little bit lower

oh wait

uhh

[sly geyser]

it's kind of a vague notion but what I kind of mean is that there's a bijection f:A->B, that preserves the properties of the structures of A and B (which are the same kind of structure here btw) across f

for orders it's the strictly monotone thing (a < b <-> f(a) < f(b))

[tender halo]

well is it positive or nonnegative

[alpine nest]

I don't get it, in particular the set of rationals and the set of nonnegative rationals aren't order-isomorphic

[sly geyser]

for topologies it's just homeomorphism

[alpine nest]

So you can't carry over the order topology

In general both of them have preexisting topologies, and you want to show that these two topologies are homeomorphic

[sly geyser]

look the set of non-negative rationals is equivalent to the rational + {-infinity}

basically the rationals plus a "minimum point" corresponding to 0

[alpine nest]

Sure, why not

To be clear, the thing I'm pondering is finding an explicit homeomorphism between the rationals and the nonnegative rationals (with their standard topologies)

Any two countable metric spaces without isolated points are homeomorphic, so we know homeomorphisms exist

But how about an explicit example?

[tender halo]

there is one above

[alpine nest]

Where?

I probably can't read but I don't see it

[tender halo]

fuck no

[sly geyser]

ok sorry I'm not the most experienced myself what is an explicit homeomorphism and how is it different to a normal homeomorphism

[tender halo]

i keep reading nonnegative as positive

[alpine nest]

I mean actually constructing a function in a way that tells you what its value is at a point

The same way arctan is an explicit homeomorphism between the reals and (-pi/2, pi/2)

Explicit because you have a formula: f(x) = arctan x

[sly geyser]

oh ok look the two aren't homeomorphic

[alpine nest]

They are

[sly geyser]

Q is homemorphic to Q>0 but not Q>=0

[alpine nest]

This is a general theorem

Both Q and Q >=0 are countable metric spaces without isolated points.

There is no doubt at all that they are homeomorphic.

[worn mortar]

Can someone clarify this passage for me? I can’t quite understand how the family of open rays given here satisfy the completeness definition given in Definition 1.12.

Note that in the second picture, X is any ordered set

[tender halo]

yea hmm i think you just need to repeat the construction from that proof, because you need to turn a "one sided" nbhd of 0 into a "two sided" nbhd of f(0)

which requires nudging the disconnected structure of Q around

[alpine nest]

Yeah, something is weird here because the intersection of two rays doesn't have to be a ray

If you threw in intervals, you would have a complete family

[alpine nest]

Yeah, that's what I was thinking, map 0 to 0, and then partition Q using some intervals with irrational endpoints, and "shuffle" those around

The endpoints probably have to get dense around 0 but not elsewhere

[alpine nest]

Reference to the theorem I mentioned: https://en.wikipedia.org/wiki/Sierpiński's_theorem_on_metric_spaces

[rancid umbra]

yes, i gave an example of such a map a while ago

[alpine nest]

Oh, you did?

[tender halo]

no its nonnegative this time

[rancid umbra]

yea

oh

[alpine nest]

Ah

Yeah, an explicit homeomorphism between Q and positive rationals is much easier

[rancid umbra]

they specify that they meant positive

or are we just considering

[alpine nest]

They're probably French

[rancid umbra]

the case when its non-negative

[alpine nest]

I'll admit I just read the symbol and didn't read the "positive"

Either way, a homeomorphism exists in both scenarios.

It's just that in one case it's given by a simple formula.

[tender halo]

yea, separate (-inf, 0) into omega clopen copies of Q, (-inf, -r1), (-r1, -r2), ..., do the same thing on the right, and you insert the left bits inbetween the right bits

[rancid umbra]

okay, yea, you can't find an order preserving one when 0 is included in the codomain

[alpine nest]

Would be mildly fiddly to write down in full precision but ultimately I was worried it would be worse

[tender halo]

yea you would need to find nice r_i so that the stretching and bijections are nice

[sly geyser]

r_(n+1) is put right after (r_n, r_(n+1)) right?

[tender halo]

Not sure what you mean

[sly geyser]

where do the r_n and l_n go

I assume 0 just goes to 0

[alpine nest]

Take r_n = sqrt(2)/n

[sly geyser]

my point is

f(r_(n+1)) goes to where (f(r_n), f(r_(n+1))) ends right?

[tender halo]

r_i are irrational

they dont go anywhere

[sly geyser]

oh hmmm

[tender halo]

we shuffle whole segments around

[alpine nest]

Meanwhile I do go to the shops, so good luck if you intend to continue

[sly geyser]

I don't think you need an infinite amount in that case

can't you just biject (-inf, 0) to (irr, inf) then biject (0, inf) to (0, irr) and get 0 to 0 of course

[tender halo]

no because then continuity breaks

(-inf, 0) used to have 0 in its closure

and now it doesnt

yea they need to continue to tend to 0

but they do it from the "other side" now

[sly geyser]

ok wait I don't understand

you basically split an infinite amount of adjacent open sets from both sides and make them adjacent to each other one one side

right?

so

ok

yeah that works wtf

[tender halo]

yea and also importantly they are clopen

so you can just shuffle them around

only point you need to care about the continuity is 0, because we fixed its image separately

[sly geyser]

ok I understand why the clopenness of the interval is important, it makes it so that any open set can be separated into an open set completely inside the interval and one completely outside it

which in this case means it can be that, when not touching 0 at least, any open set can first be cleanly divided into parts inside the intervals, then it can be shown that for every part f[part] is an open set, then you can unite the f[part]'s back together to get that f[theWholeThing] is open

[tender halo]

you could make a nice animation of this construction, first you separate the Q line into these parts, then you stretch the line so gaps appear and you fold it in half at 0 so it fits back together

[sly geyser]

Q ~ 2Q but also Q + {a} + Q, incredibly weird

[tender halo]

first one is not true, the way you write it is Q * 2

instead of 2 * Q

[sly geyser]

ok fair enough

[tender halo]

2 * Q is inserting 2 instead of every point of Q

and you would have successors and predecessors

[sly geyser]

I know I was just being dumb

[tender halo]

but yeah very weird

that Q is homeomorphic to 1 + Q + 1

[sly geyser]

is the general proof of sierpenski's theorem something similar to this

actually nah probably not

[alpine nest]

Sort of, here's the relevant section from Engelking (where it's an exercise):

Here's the referenced 4.3.H:

I quite like Engelking, but it's such an incredibly dense book (no pun intended), for better and worse.

[alpine nest]

By the way, "dense in itself" means "without isolated points"

[alpine nest]

Interesting that the argument goes by way of exploiting the irrational numbers

Or maybe not so surprising since the irrationals are a Baire space whereas the rationals are not

[tender halo]

there was a ludicrous multidimensional analog

or was it a plane one?

lemme find it

[tender halo]

the main observation the proof makes if you fix a point x, then there are only countable amount of possible values of d(x,y)

so there are lot of radii for which a closed ball around x is equal to an open ball

so the space is very disconnected (i.e. zero-dimensional)

then its just technique to assemble a homeomorphism from that information and the fact the space is countable and has no isolated points

[full quartz]

Can anyone help me? I don't understand why z1 is not an element of E in 2.47.

[pulsar lagoon]

because z < z_1 and z is the sup of (A intersect [x, y]) this means that z_1 is not in A

and also z is not in the closure of B so there is a point between z and the closure of B by completeness

[full quartz]

Thanks!

[umbral hamlet]

Okay I think i have a pretty big misunderstanding of something

A basis for the product topology is where I have the product of basis elements from each factor space up until a certain point and then its just the entire space for each factor after that

[alpine nest]

Well, generally products where only finitely many of the components are not the entire respective space

For countable products you can indeed phrase it the way you did

[umbral hamlet]

thats what im confused about though

Via unions how do I get the open set (X, R, X, R, R, R, ...) in R omega where X is some proper nonempty subset of R

[alpine nest]

I mean, R is an open subset of R; they don't have to be proper subsets.

[kind marlin]

That set is already in the product topology basis

[umbral hamlet]

Oh of course

[kind marlin]

maybe it would be easier to think about the sub basis?

That’s kind of a more natural formulation imo

[umbral hamlet]

Man I overthought that like crazy

[alpine nest]

It can be weirdly tricky to have proper intuitions early on

[umbral hamlet]

I do tend to think about that more but I had to prove in class today that the product of countably many second countable spaces is second countable

And it happened to be pretty direct from the basis i gave

Which is what made me be confused about the basis

[umbral hamlet]

On this note, if I had a set where the first collection of factors did have to be proper, it wouldn't be a basis for the product topology right

[alpine nest]

Eh, it probably would, since each factor space still has to be the union of its basis elements.

It just would be more annoying to work with

[kind marlin]

wdym by first collection

[umbral hamlet]

The basis elements of the product topology are products of arbitrary basis elements in each factor up until the k-th factor for some k, and then products of all of the space in each factor from then on

Here's the proof that I did btw

[kind marlin]

ic, if you enforce the proper subset property then this should be a basis unless one of the factors has a point whose only neighborhood is the entire space

[umbral hamlet]

Makes sense

[kind marlin]

your proof looks good, and there’s a more general fact that the number of finite subsets of a countable set is countable, so you can immediately get a countable collection of countable sets by just picking finite sets of indices

[umbral hamlet]

I see

Man im scared for my topology midterm

This is the first math course thats giving me hell and idk what to do besides just internalize definitions and try to gain as much intuition as I can

[warped helm]

i'm in the same spot don't even fret. i think the path forwards is just immersion and spamming exercises until the concepts become second-nature

[rancid umbra]

what concepts specifically? i never properly took a topology course

just like, the basic ones?

or like, something like the separation axioms or something

[quartz horizon]

i'm also interested

[kind marlin]

Honestly for me topology was a subject that examples were incredibly useful for, to precisely illustrate which properties are compatible and aren’t and how counter examples manifest

[quartz horizon]

-# and counterexamples!

[kind marlin]

I really like how munkres presents its theorems with examples

[quartz horizon]

oh i didn't know munkres went by it/its

[kind marlin]

the more you know

I wanted to send munkres a cookie for writing a math textbook that didn’t feel homicidal towards its readers

but I didn’t see a mailing address 💔

[umbral hamlet]

Its the book we use but I mostly get info from class

I should probably read it tho

[kind marlin]

def read it if only for examples

[gritty widget]

I like most the exercises too

[sonic crane]

lol my class last year used Janich

terrible book

[queen prism]

wait janich doesn't have any exercises tho

[sonic crane]

My prof just made his own homework

[prime elbow]

i see my mistake

[prime elbow]

I didn't know about it

[prime elbow]

terrible mistake

okay, i have to show if X is an ordered set in which every closed interval is compact then X has least upper bound property.

To show this, i am trying by contradiction.
Assuming there is a set A such that A is bounded above but it has not least upper bound, i.e., for each upper bound s there exists upper bound s1 such that s1 < s.

so here i have to use closed interval is compact, so i am trying to make closed interval [s1,s], [s2, s],... but i don't see how it will help me. and also i have to use compactness.

any hint?

[kind marlin]

how about this? construct a union of open sets that are upper bounds of the given set, by taking each subsequent smaller upper bound as the left of an open interval to infinity. Then the complement of this is a closed interval containing our set

you can prove that this closed interval does not contain its maximum, and then you can construct an infinite open cover that has no finite sub cover, giving us a closed interval that isn’t compact

[wide kayak]

nice idea

[frigid rose]

I don't really know what the correct channel for this is but this should be the closest thing i think...
Is it true that if two (real or complex) normed vector spaces on the same vector space (V, ||*||) and (V, ||*||') are topologically isomorphic (ie there's a linear isomorphism from V to V that is also a homeomorphism from (V,||*||) to (V,||*||') ) then the norms ||*|| and ||*||' are equivalent?
I think that it's true but i cant find a way to prove it (mind that im also starting out on functional analysis and stuff so i may not have some more advanced tricks required); i dont even know where to start if i were to search for a counterexample of this...

[crisp lintel]

it is true, the key is the fact that continuous linear maps between normed spaces are bounded

i.e. if you have T:V -> W linear continuous map then |Tv| <= C|v| for some constant C that does not depend on v

Fairly early on in a functional analysis course you learn that for linear maps on normed spaces that continuity at 0 <=> continuity <=> uniform continuity <=> lipschitz <=> bounded

[frigid rose]

Yeah yeah i can see how the homeomorphism's boundedness is crucial to making it follow but how do i translate the ||Tv||' <= C||v|| property to the ||v||' <= C||v|| one i need for norm equivalence?

[frigid rose]

yep we did do that theorem btw

[sly geyser]

what does "dense in itself" mean

[tender halo]

two msgs down from that one

[sly geyser]

ah, makes sense

[sly geyser]

also what is the definition of zero-dimensional here?

[tender halo]

T1 and a base of clopen sets

(and therefore tychonoff)

[prime elbow]

So the countable product of second countable spaces is second countable with product topology.

What if it is an uncountable product of second countable spaces? Then I think it is not second countable, because in case of countable say index is N, then if I count the number of singleton subset of N, it is N and that's why it countable, i.e., number of subsets( of finite size) of countable set is countable so their countable union also countable

But it is not true if I take index, uncountable, the number of subsets of size 1 is uncountable

[quick delta]

Here’s a proof that a non-trivial uncountable product of second countable spaces called X_a (ie one with uncountably many non-indiscrete spaces) is not second countable
Take opens U_i. By considering the usual basis of the product topology, each of these is not the whole space in only finitely many indices. Label the union of all such indices as K. K is countable
Let a be an index which is not in K, let U be a proper non-empty open in X_a and let u \in U. Then there is no open among the U_i inside U x {product of every other space} containing u, so the product isn’t second countable

[rancid umbra]

yea, indiscrete is a good counter example that i was going to say

[prime elbow]

What if I take uncountably many indiscrete space?

[rancid umbra]

then there are only two bases. the set containing the product, or the set containing the product and the empty set

[prime elbow]

Yes

My bad 🙂

[prime elbow]

Here, are we proving with the help of contradiction? Are we assuming U_i are open basis elements?

[quick delta]

Not really

This is proving “no countable set of opens can form a basis”

Therefore any basis is uncountable

[quick delta]

U_i are any countable set of opens
We’re proving they can’t form a basis

[prime elbow]

Yes I got it, thank you

[gritty widget]

I need a separable space which has subset which is not separable space. i need hint

[quick delta]

Take your favourite non-separable space
Add a point, which we define to be in every non-empty open subset
The resulting space is separable (the new point is a countable set with closure the whole space), but the original space is a subspace of it
I think this (very stupid) example works

(If I can find a non-stupid example, I’ll hint it)

[gritty widget]

Thanks, I am stupid

[plush folio]

lol, most books prefer those pronouns

[unreal stratus]

I guess you can take any non-compact space and consider its one point compactification

[sly geyser]

This logic works with everything but the empty topology right?

Actually

Nahhh

Hmmm

No

[rancid umbra]

the empty topology?

[quick delta]

Indiscrete = only opens are empty and whole space

[sly geyser]

Ok I see what you mean

[rancid umbra]

this is discussed right below the proof lmao

[quick delta]

Indiscrete is essentially the minimal topology on the space

[sly geyser]

Yeah, even if they are not the same, having only the empty set and the whole set causes problems for the proof, so yeah

as the empty set has no element to differentiate it from the base sets

[quick delta]

The problem with the indiscrete topology is that one of the opens will give you the empty set when used in the defn of the basis of the product topology, and the other gives you something indistinguishable from if you hadn’t chosen to take an open from that space

[rancid umbra]

in the sorgenfrey plane, Q x Q is countable dense but the anti-diagonal, {(x,-x)} is not separable

https://math.stackexchange.com/questions/182316/is-the-set-of-irrationals-separable-as-a-subspace-of-the-real-line

[sly geyser]

The truth is, the product of the indiscrete topology with however many copies of itself is just the indiscrete topology

So yeah

[quick delta]

Is that R^2 with the order topology from the “lexical” ordering

If so yeah that’s a nice example

[sly geyser]

Why the anti-diagonal instead of just the diagonal

Huh wait

What's the sorgenfrey plane?

[rancid umbra]

it is the product of two copies of R topologized by the basis of [a,b). i don’t think it’s the same as the lexicographical ordering…

[quick delta]

Yeah kinda similar to what I was imagining

It fails for similar reasoning aa what I was imagining

[sly geyser]

This is a supertopology of the one created by (a,b) right?

[rancid umbra]

yes

yea, you need the anti-diagonal because of the way that [a,b) x [c,d) looks.

it can clip the anti-diagonal at exactly one point

any line with negative slope would have worked the same

[sly geyser]

Ok I see what you're saying

This means that the topology on the anti-diag is just the discrete topology

[rancid umbra]

yea

[sly geyser]

Wait is that even separable?

You mean the one on the lexicorgraphic ordering right?

[rancid umbra]

yea, Q x Q is still dense there

[sly geyser]

Wouldn't that just be equivalent to R copies of R?

Oh Q

[rancid umbra]

the open sets with that topology look like vertical (or horizontal, depending on how you topologies it) slices, semi-slices, and intervals

where a slice is {x} x R

and a semi slice is like {x} x (-oo,a)

[gritty widget]

yes

[sly geyser]

My understanding is, ((a,b),c) is an open set for any combination of a,b and c, and all the other lexicographic intervals can be created from unifying these sets, so, the lexicographic ordering in this case is just Q copies of Q or R copies of R depending on which one you use

No need to delete your reply ((a,b),c) is a confusing as fuck way of writing something

Using the notation as both an interval and a pair plus having the evaluation of a function on a set tossed in

[rancid umbra]

i parsed it eventually lol

[sly geyser]

I could have just written (a,b) x {c} now that I think about it

[rancid umbra]

i think ((a,b),(c,d)) is best

like an interval lol

or (x,y) where x,y in R^2. this one is slightly worse, but still good

[rancid umbra]

so your claim is that R^2 with the dictionary topology is just R x R?

[sly geyser]

Not exactly

More like

[rancid umbra]

sorry

R^R

[sly geyser]

The discrete sum of R copies of R

It is not R^R

[rancid umbra]

okay cool. had a lapse

anyways

so it’s the disjoint union of R copies of R

um

[sly geyser]

Yes

[rancid umbra]

i… somehow don’t think that is true

i mean

as a set

yes

[sly geyser]

Because (a,b) x {c} is an open set for the lexicographic ordering

[rancid umbra]

there is a bijection

[sly geyser]

And any other base set in the ordering can be described as the union of (a, inf) × {c}, (-inf,inf) x {r} for c < r < d, and (-inf, b) x {d}

[gritty widget]

are countable product of second countable topology with box topology, second countable?

i don't think so, take {0,1}^N. Each with discrete topology

[rancid umbra]

wow. that’s weird to me.

i think that showing they are homeomorphic is more illuminating tho, or maybe just showing that a set in the order topology is open iff it’s intersection with each slice is open in R. is that what u were saying?

[sly geyser]

Yeah I think

Also I don't think such a principle works for every lexicographic ordering

Notably, this requires that (a, inf) and (-inf, b) are open in the lower dimensional order topology

Which isn't true in for example 1 + R + 1

Um wait

[rancid umbra]

not sure i understand. [0,1] x R with the order topology is the same

[sly geyser]

I'm wrong

I misunderstood how it was defined

It was defined with {x<a} and {b<x}

Not (a,b)

Yeah in that case it should just be slices for any combination of two orders

[gritty widget]

Honestly, the (standard?) proof that a compact subset of a Hausdorff space is closed is like, idk how to describe it, one of the most elegant proofs I’ve seen I guess?

[rancid umbra]

it has a very intuitive visual

idk if its the most elegant proof i have seen

[gritty widget]

I was moreso thinking the proof itself but I guess that counts too yeah

[gritty widget]

I also haven’t ever taken an advanced math class tho so not speaking of many examples

[quartz horizon]

When I think of elegant proofs in topology my mind jumps to Urysohn’s lemma

[gritty widget]

I don’t recall the proof of that I’ll look rq

[gritty widget]

Let X have a countable basis; let A be an uncountable subset of X. Show that uncountably many points of A is a limit point of A.

say S is the set of all limit points of A in A. If S is countable then A \ S needs to be uncountable. Since X have countable basis, so A has too. Now A has countable dense set.

A\ S is uncountable implies there are uncountably disjoint open sets in A i.e., singletons, but A has countable dense set therefore it can't be happen.
any mistake?

[rancid umbra]

yea this is good

[sly geyser]

can the neighbourhoods around a specific point be themselves be considered open sets in another topology? (I know it can logically I'm asking if there is any reason to actually do this)

[rancid umbra]

like if U is an open nbhd of x in topology \tau and \tau’ is finer than \tau?

[gritty widget]

i don't know, if X is countable then is it necessary that topology on X is always second countable? I am trying to find an counter example for it

[alpine nest]

There is an example of a countable space that isn't even first countable, but it's not something I'd expect a normal person to come up with as an exercise.

https://en.wikipedia.org/wiki/Arens–Fort_space

I lowkey love the fact that the Arens-Fort space is perfectly normal.

Imagine looking at this and going "yep, a perfectly normal space"

I was about to ponder whether your question might be slightly easier because it asks for second countability, but then I realized that if X is countable, then first countability and second countability are equivalent.

[gritty widget]

thanks, what is the origin of this topology?

[gaunt linden]

10 imaginary internet points say its main purpose is to be a counterexample.

[tender halo]

its main purposed is to be fucking cool

but the wikipedia like omits stuff? arens fort is a subspace of another space that is sequential but not frechet

engelking has the construction of that one

[tender halo]

(and therefore not first countable or second countable)

[unreal stratus]

Which proof do you mean

[gritty widget]

For any arbitrary point x in X\C, take disjoint neighborhoods U_c of c and V_c of x for all c in C. U_c cover C, so we can pick finitely many which still cover c, hence the intersection of the corresponding V_c’s is a finite intersection of neighborhoods of x, and is disjoint from C, so x is not a point of closure of C.

[alpine nest]

Yep, it's a neat proof

I'm not sure "elegant" is the word that springs to my mind, but it is a very good illustration of how to use compactness, since this kind of argument is used very often (find a neighborhood of every point in your compact set, use compactness to get finitely many of them, profit from the finiteness)

[gritty widget]

It can also be (very slightly) rephrased, by an equivalent formulation of compactness that someone here showed me.

Namely, a space is compact iff. for an property P of sets, P holding locally(holds for a neighborhood of each point) and P being preserved under finite union imply P holds for the space itself.

So we can rephrase the proof as:

Let C be a compact subset of X, and x in X\C.

Then, let P be the property such that P(A) iff x does not belong the closure of A. This is preserved under finite unions because the closure of the union is the union of the closures. So all that’s left to do to prove this holds locally is Hausdorffness! Since, for every point c in C, there is a closed neighborhood of c not containing x, then we have that P(C) holds.

[quartz horizon]

ah yes this is

https://www.drmaciver.com/2015/03/topological-compactness-is-an-induction-principle/

i've also been investigating whether it makes sense to try to frame continuity in terms of predicates and preserving locality

[gritty widget]

Wdym?

[quartz horizon]

ok so i can try and sketch the ideas

this was actually inspired by understanding filters as providing a notion of truth

first, you want to use the correspondence between subsets and predicates - the ideas i'm talking about make most sense in predicate-world

[sonic crane]

I should re-read what you were saying to me about predicates

I forget the context

[quartz horizon]

say we're working on the real line, and have some predicate $p : \mathbb{R} \to {0, 1}$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

we can use "p holds at x" as an alias for "p(x) = 1"

then i think it makes sense to define "p holds $\textit{near}$ x" iff there's some $\epsilon > 0$ such that $p$ holds on $(x - \epsilon, x + \epsilon)$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

in subset-world, this is essentially the definition of a neighbourhood of x

open subsets can be defined as those which are a neighbourhood of all their points (they "surround" every point within them, there's wiggle-room in an open subset, etc etc)

in predicate world, you could say that a predicate p is "local" iff whenever p holds at some x_0, it necessarily holds near x_0

e.g. the predicate "x < 3" is local - if some x_0 is less than 3, then values near x_0 are also less than 3

[gritty widget]

Trivial observation, probably not relevant to your discussion:a predicate is continuous iff it is constant on connected components

[quartz horizon]

mhm

you can then translate the properties of open sets to local predicates

if p(x) and q(x) are local, then the predicate "p(x) and q(x)" should be local as well, which corresponds to binary/finite intersections of open sets being open

if $(p_i(x))_{i \in I}$ are a family of local predicates, then the predicate $\exists i \in I, p_i(x)$ should be local as well, which corresponds to arbitrary unions of open sets being open

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[quartz horizon]

the constant-1 predicate is trivially local, and the constant-0 predicate is vacuously local

finally, continuous functions are interpreted as "locality-preserving", in that their behaviour is determined locally

so if $f : X \to Y$ is continuous and $p(y)$ is a local predicate on $Y$, then $p(f(x))$ should be a local predicate on $X$

[gentle osprey]

Pseudo (Cat theory #1 Fan)

[gritty widget]

Wow this is some genuine cooking

[quartz horizon]

p(y) being local means that if it's true at y_0, it'll be true near y_0

[gritty widget]

First like cool original idea in topology I’ve seen in a long time

[quartz horizon]

p(f(x)) being local means that if it's true at x_0, it'll be true near x_0

but so long as f is continuous, values near x_0 get mapped (under f) to values near y_0

you then just define f to be continuous iff this holds for every local predicate

so in summary, the main idea is that:

• try to view things in predicate-world as opposed to subset-world
• a topology on a set X tells you which predicates are "local" and which aren't, subject to some consistency conditions
• continuous functions are precisely those that "preserve locality", or are otherwise "determined by local behaviour", meaning they map local predicates to local predicates

does that make sense?

[gritty widget]

Yep!

[quartz horizon]

the main reason this works is that, in predicate-world, preimage is by far the most obvious operation

since it just corresponds to substitution of y = f(x)

[gritty widget]

I wonder if there’s a neat way to view the closure of a set(predicate?) or closed sets in general as predicates, like maybe a way of viewing Kuratowski closure axioms(equivalent to a topology) in a predicate way?

[quartz horizon]

yeah i haven't tried that, but potentially!

i was just trying to make precise the way in which open sets capture the notion of "locality"

and this feels like a neat way to do that

by viewing them as predicates, not as subsets

[unreal stratus]

Yeah to me this is more like a nonelegant thing lol

[gritty widget]

Maybe you could define the closure of a predicate by saying it holds locally like, very strongly? E.g. it holds of some point in any small area around x? You could probably reformulate the kuratowksi closure axioms fast

[gritty widget]

R\times... countable copies of R, with box and product topology is regular in both cases, correct?

[magic siren]

Oh cool, I didn't know anyone actually still cared about these posts of mine. 😅

[quartz horizon]

holy shit you're the original guy

thanks so much for your post

[magic siren]

Very welcome!

[quartz horizon]

i did one recently about connectedness as an induction principle, actually

it's interesting how both of these concepts fall into the same "local to global" idea

[magic siren]

How does the connectedness one work?

[quartz horizon]

oh it might be easier if i just link my post :P

[magic siren]

Please do!

[quartz horizon]

https://pseudonium.github.io/2025/09/17/Connectedness_As_Induction.html

[magic siren]

Oh, cute. Yeah, that's a nice way to look at it.

[quartz horizon]

[gritty widget]

so induction on R is equivalent to infinite transitive induction?

[quartz horizon]

well induction on R is an instance of "topological induction" in this sense

[gritty widget]

so which one is more generalized version, topological induction or transfinite induction

[quartz horizon]

i feel like they're different generalisations?

transfinite induction is really "well-founded induction"

but topological induction as i laid out in that post isn't really about that

you don't need any kind of order on your topological space for it to work

hence why i called it "diffusive" induction

[gritty widget]

i see

i have a question related to second countability of R^{\omega} with box topology

[rancid umbra]

hey! chain connectedness made an appearance

[lucid ocean]

oh damn hi

I also loved your article on compactness

[rain mason]

continuity method goes brrr

[quartz horizon]

continuity method?

[rain mason]

Let $X, Y$ be Banach spaces, $L_t : X \to Y$ be (smooth) maps for $t \in [0, 1]$. Say we want to find $x_1 \in X$ such that $L_1(x_1) = 0$. Say we have $x_0$ such that $L_0(x_0) = 0$. We can show such an $x_1$ exists by considering the set
$$I = {t \in [0, 1] \mid \exists x_t \text{ s.t. } L_t(x_t) = 0}.$$
We have that $0 \in I$. Thus, it suffices to prove that $I$ is open and closed.

[gentle osprey]

shingtaklam1324

[rain mason]

this is a common technique in PDE theory

Openness usually follows from the set-up

Closedness is what you need to work hard for

[quartz horizon]

ooooh

that's quite interesting

i guess that's an application of topological induction?

[rain mason]

in some sense

I think there's a bit more input in this case than just induction

[alpine nest]

Transfinite induction typically applies when you're inducting over a set which is in some natural way well-ordered, which the reals are not.

[rain mason]

in this step, you need some form of compactness

[quartz horizon]

i see

[rain mason]

but I guess you could encapsulate all of this into "the inductive step"?

https://en.wikipedia.org/wiki/Method_of_continuity

in the linear case, what you need is a uniform in t bound

[quartz horizon]

makes sense

[magic siren]

Thanks!

[gritty widget]

can we construct compact Hausdroff topology on any non empty set?

[lucid ocean]

well order it by the successor of a limit ordinal and then order topology maybe

[unreal stratus]

I guess given a set X you can take the discrete topology on X \ {x} and one point compactify it

[alpine nest]

Does anyone have an example of a situation where the most "natural" characterization of a topology is via the Kuratowski closure operator?

I.e. one where it's the easiest to work with it via that perspective rather than some characterization of open sets or continuous functions.

[tender halo]

work hmm idk

I like the definition of uniform topology on the space of continuous functions where you just say the closure is adding all uniform limits

and then you prove that its metrizable

it looks a little more motivated because open balls in that topology are a little tricky

from toy examples hmm particular point topology and variations thereof are much nicer if you think about the closure operator

like the Fort space

[alpine nest]

Yeah, spaces where the natural starting point is defining convergence were what I was thinking of, although that will only work in situations where closure is the same as sequential closure.

With the weak topology on operators you do start with convergence, but "add all limits in the sense of weak convergence" won't give you weak closure

[alpine nest]

Good point, but I was hoping to motivate the closure axioms somehow, by pointing out a scenario where it's actually a good starting point for your topology

As in an actual scenario in some area of mathematics where that's done

[tender halo]

maybe topologies related to convex structures? they love closure systems

idk i didnt get far into the book about convex structures lol

[alpine nest]

Yeah, my mathematical work involved spaces which were either compact metric (so easymode, topologically), or some kind of weak topology induced by a family of functions, so those are my natural points of reference.

Heck, even the weak topology was often metrizable, although that was one of the situations where working directly with the metric wasn't really that useful a lot of the time.

[prime elbow]

I know I am wrong here, but I don't why it seems true to me, a well ordered set with ordered topology is discrete topology because in well ordered set for any x( if x is not smallest and largest ) se can find y < x < z such that (y, z) = {x}.

And for the smallest element and largest element we can do the same just replace y of z accordingly with x

[alpine nest]

Why can we find such y and z?

What if your uncountable set is omega*2 (i.e. {0,1,2,3,..., omega, omega+1, ....}) and x = omega? What are those y and z then?

[prime elbow]

ah, in that case we can only find z = omega

I see

So that's true in well ordered we can find a successor not a predecessor

[alpine nest]

omega isn't greater than omega, but I suppose you could take z = omega+1

[prime elbow]

Yeah, typo

[alpine nest]

Yes, every ordinal has a successor, but not every ordinal has a predecessor

[prime elbow]

Thanks Outsider

[alpine nest]

(also for easier intuitions it might be worth keeping in mind that every well-ordered set is order-isomorphic to an initial segment of the ordinal numbers, so it suffices to think in terms of ordinals as such)

[prime elbow]

What is the initial segment of the ordinal numbers?

I know construction of omega

[alpine nest]

https://en.wikipedia.org/wiki/Ordinal_number#Von_Neumann_definition_of_ordinals for example these.

[prime elbow]

Thanks

I read it out but still I don't think I fully understand how ordinal works.

[alpine nest]

Similarly to how cardinal numbers describe cardinality; ordinal numbers define the order structure of well-ordered set; the key observation is that given any two well-ordered sets, one is always order-isomorphic to an initial segment of the other. As a result ordinals are a sort of "reference" family of well-ordered sets, on which you even have well-defined operations like addition, and also any transfinite construction can be indexed by the ordinals in the standard way.

It sort of is what you refered to as the construction of omega, except extended indefinitely

(and you lose being able to visualize what's going on fairly quickly)

(also you can define cardinals in terms of ordinals, as the least ordinal number of a given cardinality)

[prime elbow]

And how can I show every well ordered set is order isomorphic to one of ordinal numbers?

[alpine nest]

This is the proof (from Jech's book) that given any two well-ordered sets, one is order-isomorphic to an initial segment of another, surprisingly short:

Then you do this:

[prime elbow]

Okay i am going to read this book

[alpine nest]

Just that chapter, really

(which is a very short chapter, and you don't really need much ordinal theory for learning topology)

[prime elbow]

No that's not the case, I am getting interest in set theory since the last few days, but I don't know how to start it from scratch

[tender halo]

dont read jech first

well you can read small jech first

which i dont remember the name of

but big jech is a subpar introduction to the subject

[prime elbow]

Some days ago, I went to the college library and found a book on lattice theory and i don't know but I wanted to read it

[prime elbow]

Okay

[valid patio]

i cant find it can u reply later if u find out name of book ?

i wanna look at it

[tender halo]

its called introduction to set theory

(as opposed to "set theory")

[valid patio]

oh got it i assume 300 paged one then thanks !

[hasty thicket]

are these notes any good for self teaching some topology https://www.math.toronto.edu/ivan/mat327/?resources

[hexed steppe]

all this nonsense before compactness and connectedness

[hasty thicket]

😂

[hexed steppe]

im going to say no

[hasty thicket]

should i just stick w munkres then

[hexed steppe]

just use munkres imo

yes

munkres is amazing

[hasty thicket]

are there any videos or smth

alr thank u

[hexed steppe]

sure im sure you can find just looking on youtube

do you already know what metric spaces are

if not it might be worth first studying that from eg rudin

[plush folio]

connectedness in chapter 18 out of 19 is crazy can you really cover all first 17 chapters without mentioning connectedness once?

[hexed steppe]

yeah guessing the prof is a set theorist

with no interest in math beyond set theory

[hasty thicket]

yes i covered abbotts book recently

[hexed steppe]

yep

[hasty thicket]

it’s enough for topology right

ts so challenging lowk i havent encountered this much abstraction in school

[hexed steppe]

yes

[tiny obsidian]

😢 "nonsense"

(though I agree that for self teaching topology they're not good)

[hexed steppe]

wait does this do metric spaces

[hasty thicket]

Yes the end chapter

Ch 8

[hexed steppe]

looks a bit sparse on examples

anyway it's probably fine

metric spaces arent really a prerequisite they just form a central example providing a lot of intuition

but just doing analysis on R might convey the same intuition

[hasty thicket]

i think i suck at analysis probably need to go over ideas or do more exercises

[hexed steppe]

exercises are essential

same with all of math

[hexed steppe]

a lot of it isnt nonsense but it's like wildly out of order

[hasty thicket]

its too demotivating im in highschool rn and doing quite well w highschool maths but if its abstract stuff its so over

[hexed steppe]

theres no reason to learn about 100000 separation axioms and urysohn lemma before compactness and connectedness

they dont even do quotient topology

[tiny obsidian]

this is precisely why I say this is bad for self teaching topology

it actually is a reasonable order for a course that focuses only on analytic topology

in fact it's pretty similar to my fourth year course on it

[hexed steppe]

what is analytic topology

[alpine nest]

Yeah, I don't find this order particularly alarming either

[tiny obsidian]

(whereas just topology was 2nd)

[tiny obsidian]

focus more on just the set theory parts ig? will think of a better definition

the metrization theorems in particular

[alpine nest]

But also I work with Cantor spaces so connectedness is entirely irrelevant to me

[hexed steppe]

i think it's also a matter of building mathematical maturity in the students

the metrization stuff is significantly harder

than elementary connectedness/compactness arguments

[desert vortex]

watch some videos on youtube to supplement and focus on the basics, no need to burn out over this abstract nonsense 🙂

I learned topology in highschool aswell, it took a few months to get the hang of it, make sure you do as many exercises as you can, if you need more exercises look in other books if you must

[smoky sigil]

Can someone explain why in an indexed family of sets we can have two set A_1 and A_2 be in the set but A_1 and A_2 be the same set even though elements of a set are distinct?

[quartz horizon]

This essentially comes down to what the definition of “indexed family” is

Are you considering indexed families of open sets, for example?

[smoky sigil]

My definition is coming from Munkres Topology

[opaque scroll]

An indexed family is just a function from J to curly-A. Functions aren't always injective.

Are you asking why indexed family is defined this way?

[smoky sigil]

I think I was just thinking of the definition incorrectly. Reading it again I think I understand better. I was thinking the an indexed family would look like this:
{A_1, A_2, A_3, …} but it’s just a collection curly A with the function J. J isn’t injective so J can map say 2 and 3 to the same set

[valid patio]

function f:J->C, where J is the index set and C is a collection of sets
The function f is not required to be injective (one-to-one). The terms A1 and A2
are simply outputs of this function, f(1) and f(2). Since the function is not injective, it is perfectly allowed to have two distinct indices (1 and 2) map to the identical set

[smoky sigil]

Thank you! Your explanation makes sense to me!

[valid patio]

np , notjagr said same thing esentially

[smoky sigil]

Yeah that makes sense

[tender halo]

i mean what do you need it for

it has one important application which is IVT

which is not very relevant to pointset

[tiny obsidian]

Connectedness is a very nice property outside of IVT as well (e.g. domains are important in complex analysis) especially because of the equivalent formulation with clopen sets, and there's probably quite a bit of pointset flavoured stuff you can do regarding a) components b) the various (locally) ____-connected properties and c) a and b together

[hexed steppe]

?

guy who just passed calculus 1: getting a lot of intermediate value theorem vibes from this

[tender halo]

what do you need it for in the intro to pointset

[hexed steppe]

the point of the intro is to learn it…

intro to point set is not research level math

it’s prerequisite collecting

[tender halo]

what i mean is

its not a foundational thing to the course, more of a side topic

[hexed steppe]

lmao

[tender halo]

if anything disconnectedness is much more important to pointset

[hexed steppe]

nice bait

[tender halo]

what topic in intro pointset relies on connectedness?

[hexed steppe]

it’s called “connectedness”

that is the topic

and it should occupy a significant fraction of any reasonable intro point set topology course

[tender halo]

what interesting theorems do you know about connectedness that are appropriate for an intro course?

that are not specifically about like zero dimensionality or whatever

[hexed steppe]

idfk everything in intro point set topology is utterly trivial

nothing is a “theorem”

but it’s important to learn fundamental definitions and get a feel for how to reason about them

[tender halo]

sure, what i mean is that it is a separate topic that is not particularly connected to the rest of the course

and you can slot it at the end if you so desire

unless you are doing dimension theory which you probably aren't

[quick crane]

the only connected subsets of $\mathbb{R}$ are intervals, defined as the convex subsets of $\mathbb{R}$.

[gentle osprey]

L

[tender halo]

cmon thats a best a note, not even a proposition

[rancid umbra]

is there a generalization of path connectedness where the domain isn’t I?

maybe this isn’t interesting since I is initial somehow

[gritty widget]

I think there was something done similar in an MSE post with the long line(with endpoints) instead

There’s probably plenty of generalizations with other domains they just might not be super interesting

[tender halo]

@quartz horizon mentioned something about I being a universal space

[rancid umbra]

what, like (\omega_1 + 1) x I?

[rancid umbra]

yea. this is what i was referencing

[tender halo]

i think to the long ray with endpoint you do one-point compactification instead?

[rancid umbra]

i see

is there a continuous function R —> R whose zero set is the cantor set?

[tender halo]

distance to the cantor set?

all closed subsets of a metric space are zero-sets

which precludes metric spaces from being funny

[quick crane]

I don't think it's that obvious

[hexed steppe]

wdym

suppose A is connected and not an interval. find x,y in A and z not in A such that x<z<y

and use (-infty,z) and (z,infty) to disconnect A

[rancid umbra]

it’s the other direction that is non-trivial

you have to use some properties about R like sup and inf

unless u use the fact that interval implies path connected

[quick crane]

the proof that path connected implies connected relies on the fact that [0, 1] is connected

[rancid umbra]

true, that is a bit circular

[hexed steppe]

what

well yes

but i wouldnt really call this nontrivial

[smoky sigil]

Why is this theorem proved using induction? I am getting confused because in the theorem it states for some n in the positive integers. I’m not sure where we are proving a proposition for all n in the positive integers.

[queen prism]

this looks better suited for #proofs-and-logic but you can read the theorem as, let n be a positive integer and suppose A has cardinality n; then ...

[prime elbow]

So in order topology with least upper bound property we can say every order topology is locally compact, right? Because for any x, we can find x in (a,b) \subset [a, b], and using the least upper bound we can conclude [a,b] is compact

[young stone]

What is a universal space?

[prime elbow]

if w_1 is the first uncountable ordinal number, then i can write the topology on w_1 + 1 as [a, \omega], where a is the smallest elemenet of w_1, right?

for ref, w_1 + 1 defined as w_1 cup {\omega}, x < \omega for all x in w_1

then w_1 + 1 is compact because it is order topology and it has lub property and it can be written as closed interval

[opaque scroll]

You can write it as [0, w_1] yeah

[hasty thicket]

yes that’s very true thank u

i should just cope lowk and get on w it

[prime elbow]

Okay

I read about real induction and topology induction, and it seems interesting that we can prove some introductory proof of real analysis by real induction so why is it not mentioned in most books?

[quartz horizon]

oh yeah real induction is quite neat

I think often there are alternative ways to do proofs that can be done with real induction, using suprema/infima or sometimes with a bisection argument

[alpine nest]

Yes, and most textbooks employ those alternatiwe ways which is why real induction isn't often covered as such

Even if the argument (which is basically a compactness argument, really) is used in proofs, it's not labelled as a specific thing

That's probably why it's not covered, the framing of "real induction" requires a linearly ordered set, so an argument phrased in those terms doesn't generalize to R^n or to general metric/topological spaces.

[quartz horizon]

As for “topological induction”, I think it’s more common but usually just not named as such

Instead you’re just appealing to connectedness

[alpine nest]

Is that also basically compactness; the kind of thing we talked a while ago?

[quartz horizon]

Compactness is a slightly different induction principle

[alpine nest]

We talked about that recently, that's what brought it to mind

[quartz horizon]

https://www.drmaciver.com/2015/03/topological-compactness-is-an-induction-principle/

Here’s the post for compactness

[alpine nest]

Ye

[quartz horizon]

https://pseudonium.github.io/2025/09/17/Connectedness_As_Induction.html

Here’s the one for connectedness (by me)

[alpine nest]

Yeah, that's what I thought of when you mentioned connectedness

Linking points via a chain of intersecting open sets

[quartz horizon]

Yeah that’s chain-connectedness

[alpine nest]

I don't get much use out of that, working as I do mostly in Cantor spaces

Very compact, very not connected

totally not connected, in fact

[quartz horizon]

The way I think of connectedness as an induction principle is to show a set is the whole thing, you show it’s nonempty, open and closed

[quartz horizon]

Hm, given how much you work in compact spaces, how would you explain compactness to a student

Or maybe, what does compactness “feel like” to you

[alpine nest]

Points can (edit: can't) really "run away" in a compact set, is my intuition; in practical terms the three consequences of compactness I use the most is the attainment of maxima/minima by continuous functions; uniform continuity of continuous functions; and existence of convergent subsequences for all sequences.

For purposes of coming up with proofs I basically conceptualize my space as [0,1], the closed unit square or the Cantor set, and think of how I'd do things there

And then try to adapt my proof so that it wouldn't use anything specific to those spaces, of course.

I've often heard the intuition that a compact set is "almost" finite, but I never really grokked that.

-# Side note, I hate that Musk ruined the word for the rest of us

[quartz horizon]

Interesting

[prime elbow]

What do you mean by run away?

[quartz horizon]

Do you ever work with compact but not sequentially compact spaces?

[alpine nest]

Literally never outside of teaching, since all the spaces that occur in my work are compact metrizable (either Cantor spaces, or unit balls in the dual of a separable space, with weak* topology).

[alpine nest]

I'm mostly thinking of the property that every sequence in a compact metric space has a convergent subsequence, but also the fact that if every point of your space has a neigborhood where something happens, you can pick finitely many of those neighborhoods which will cover the whole space (so every other point will be "covered" by those finitely many points)

In metric terms compactness implies total boundedness

[crisp lintel]

I think of it as "topologically finite"

[gaunt linden]

Hmm, I would have phrased that as "points can't really run away"...

[crisp lintel]

I assume that was a typo

[alpine nest]

Oh duh, yeah, that was a typo

[quartz horizon]

I see - it seems like the properties you mean apply more to sequential compactness

[alpine nest]

Yep

[quartz horizon]

Hm

The extreme value theorem works for both compact and sequentially compact spaces, right?

By which I mean - suppose X is compact or sequentially compact, and f : X -> R is continuous. Then f has and attains its maxima/minima

[alpine nest]

Yes, because continuous maps preserve both compactness and sequential compactness.

So the image of X will be a compact subset of R either way

[quartz horizon]

Interesting

Uniform continuity only really makes sense when X is metrizable, I assume, and then compactness and sequential compactness are the same

[alpine nest]

Well, there's a slightly broader class called uniform spaces: https://en.wikipedia.org/wiki/Uniform_space

[quartz horizon]

Ah I see

[alpine nest]

I don't think I've ever seen them mentioned "in the wild" (i.e. outside topology textbooks), but of course there's many areas of mathematics I'm not familiar with.

[quartz horizon]

I guess I’m trying to understand the extent to which compactness and sequential compactness are different concepts

[alpine nest]

Yeah, since for metric spaces they're equivalent, I admit I also find the distinction tricky to intuit abot.

[quartz horizon]

I know that compactness is equivalent to “every net has a convergent subnet”, maybe that helps

[alpine nest]

Similarly for closure and sequential closure.

[quartz horizon]

Hm, is it always true that the sequential closure is contained in the closure?

[alpine nest]

Yes (if x is the limit of a sequence of elements of A, then every neighborhood of x contains points from A, and that's one of the equivalent criteria for being in the closure)

But I meant that since in metric spaces closure is equal to the sequential closure, I find it easy to forget that an element of the closure might not be the limit of any sequence from your original set.

For example in l^1 the weak closure of the unit sphere is the unit ball (as it is in every infinitely-dimensional normed vector space), but since in l^1 weak convergence is equivalent to norm convergence, you won't find a sequence of unit sphere vectors which would converge to 0 in the weak topology.

You will find a net, of course, although I have no idea what it would look like.

Ah, of course someone has already asked and answered this: https://math.stackexchange.com/questions/3710172/construct-a-net-on-the-unit-ball-in-ell1-mathbb-n-weakly-converging-to-ze?noredirect=1&lq=1

[quartz horizon]

Interesting

[prime elbow]

Yes

[alpine nest]

I mean, that wasn't really a question

[prime elbow]

Yes, if space is first countable then it holds

[alpine nest]

You don't really need to reply to my messages just to state that you know that.

I'm here right now so it doesn't really matter, but if I wasn't I'd get annoyed at the unnecessary pings

[prime elbow]

Okay

I am thinking, which is non-metrizable space which holds sequentially compactness

[alpine nest]

You mean a space which is sequentially compact and not compact (and obviously nonmetrizable, since if it was metrizable it would be both or neither)?

[prime elbow]

Yes, a non-metrizable space which is sequentially compact but not compact

[alpine nest]

Serendipituously, ordinals provide the example.

Specifically, the set of ||all countable ordinals|| with the order topology

[prime elbow]

I see

Yeah now i remember that Munkres mentioned it in the limit point compact section

[gritty widget]

Got it, thank you

[brittle cedar]

I think I've proven (i) but I don't understand the statement of (ii)

(X,d) is a metric space. it's saying that X' is the same set as X, but with a different topology?

[crisp lintel]

Any topology

but such that the metric is continuous

and then you want to show it has more (or equal) open sets

[brittle cedar]

ok right I see. I'll think for a minute

[crisp lintel]

I think the question is phrased pretty awkwardly

I probably would've said something like "let tau be a topology on X such that the metric is continuous wrt tau, show that tau is finer than the metric topology"

[brittle cedar]

yeah I was confused by the wording but I think I fully understand the statement now so I'll give it a good try before asking for a hint if I'm stuck. thanks for reiterating it though 🙂

[smoky sigil]

This theorem is from Munkres topology. I understand the proof but I’m not understanding the technique they used. I’m still not understanding why induction was used here. The theorem says for some n in the positive integers which I’m interpreting as “there exists n in the positive integers”. I would expect induction to be used on a statement that is saying something about “all positive integers”. Can someone explain why induction is being used here when the theorem says “for some n”.

[gritty widget]

It’s for some n, so he’s just proving that if the n is 1, then it holds, or 2, etc, I think?

[smoky sigil]

Yeah that makes sense. It’s just not clicking why we can used induction here. It seems like the author is proving a different statement since he is proving something for all the positive integers

[orchid perch]

if X = {0} U {1/n : n in N1} how do you create a filter for p = 0
in the subspace topology on (R, Teucl)

[dawn frigate]

Is this proof right so far

[prime elbow]

X is not Hausdroff means there exists two point such that any open set containing them must intersect to each other, not there exists open set U and V which intersects

[dawn frigate]

Oh…my bad

Guess I need to fix it

Anymore mistakes

How to finish it?

[valid patio]

The line Every neighborhood of (x,y) has form (U x V) has problems too sets of the form U×V (where U,V are open in X) are a basis for the product topology, not the only open sets. a general open set is a union of these basis elements. while you can correctly use basis elements in a proof, you prolly should must state it properly "Any neighborhood of (x,y) must contain a basic open set of the form U×V" or something

[dawn frigate]

Ok

[valid patio]

the main idea is to prove the contrapositive If ( X) is not Hausdorff, there exist two distinct points ( x \neq y ) in ( X ) such that for any open neighborhood U of ( x ) and any open neighborhood ( V ) of ( y ), we have ( U \cap V \neq \emptyset )
welll to prove that the diagonal ( \Delta = {(z,z) \in X \times X} \subseteq X \times X ) is not closed we can show that its complement ( (X \times X) \setminus \Delta ) contains a point that is not an interior point
We pick the point ( (x, y) ), which lies in the complement of ( \Delta ) because x \neq y and any open neighborhood of ( (x, y)) in the product topology must contain a basic open set of the form ( U \times V ), where ( U) is an open neighborhood of ( x ) and ( V ) is an open neighborhood of ( y ).
Since X is not Hausdorff we know that ( U \cap V \neq \emptyset ). let ( z \in U \cap V). Then ( (z, z) \in U \times V ), and ( (z, z) \in \Delta ). this means that every neighborhood of ( (x, y)) intersects ( \Delta )
Therefore, (x, y) is a limit point of ( \Delta ), but it is not in Delta. Hence, \Delta is not closed in ( X \times X )

[dawn frigate]

Thanks

[dawn frigate]

Do you know a counterexample that disproves continuous image of hausdorff space is hausdorff

[rancid umbra]

consider the codiscrete (or indiscrete) topology on a set Y with more than two elements

when is a function f : X -> Y between top spaces continuous?

[dawn frigate]

Thanks

[dawn frigate]

Show that the subspace (a,b) of R is homeomorphic with (0,1) and [a,b] of R is homeomorphic with [0,1]

so use map from (a,b) to (0,1)?

[supple island]

Yes use a map between the two that is a homeomorphism. If you can’t think of any, think of linear interpolation

[gritty widget]

Prove that (a, b) is homeomorphic to (ca, cb) and (a+c, b+c) for c>0

[sterile jungle]

hi, i want to construct a topological space X such that there is some subset B with the property that B is not closed and if x_n is a convergent sequence with terms in B, then x_n converges to some element x in B. ive seen the arens-fort space as the standard example but i want to intuitively construct one on my own. my prof said that we can make one by making a topology based on what sequences we want to converge to but im not really sure what that means or what to do with that

[lament steppe]

If M is a topological n-manifold, is every element of the basis of the topology homeomorphic to an open subset of R^n?

Based on the definition of locally euclidean I have in front of me, it just states that "some" open neighborhood of every point is homeomorphic to a subset of R^n, and Im not assuming the basis elements have to be those particular subset that do this...

[iron bolt]

do you have a specific basis in mind that you are asking this about? because it definitely won't hold generally

any topology is a basis of itself, and it definitely doesn't hold for every open set

[valid patio]

You can construct such a space by taking an uncountable set like ( X = \mathbb{R} ),
choosing a special point ( p = 0 ), and letting your subset be B = \mathbb{R} \setminus {0}).
Define a topology where a set ( U \subseteq X ) is open if either it does not contain ( 0 )
or it does contain ( 0 ) and its complement ( X \setminus U ) is a countable set.
With this topology, ( B ) is not closed because any open neighborhood of ( 0 )
must be very large (co-countable) and is forced to intersect the uncountable set ( B ),
making ( 0 ) a limit point.
however, any sequence (x_n) ) within ( B ) is just a countable set of points, so you can always construct an open neighborhood of ( 0 ) that specifically excludes every point in that sequence, proving that no sequence from ( B ) can converge to ( 0 ), any convergent sequence from ( B ) must converge to a point within ( B )

[iron bolt]

on the other hand, for any manifold the set of all open subsets that are homeomorphic to R^n is a basis, so that gives you an example of a basis for which the property holds

[lament steppe]

Ah interesting. I was trying to show that every open set of the n-manifold contains a "coordinate ball" (an open set that is homeomorphic to an open subset of R^n) where n is greater than 0... and I think your last statement (which isnt immediately obvious to me) would give me said conclusion.

[iron bolt]

I would say it's the other way around, that what I said follows from what you are trying to prove

I don't want to spoil too much, but it can be proven relatively directly. you basically need to pick coordinates once, transport your open set across the homeomorphism, and then pick a second smaller ball inside that set

[lament steppe]

Ohhh I see what you are getting at 👀
That makes sense to me! I guess when a set is specifically homeomorphic to R^n, since R^n is a metric space, there is always a smaller ball... so its open subsets all the way down.

[alpine nest]

Apart from Yuxuan Wang's example of the cocountable topology, another example would be space of all countable ordinals together with the least uncountable ordinal, with the order topology. The set of countable ordinals is sequentially closed in this space, but not closed (the remaining point is in its closure but can't be reached using any sequence).

[rancid umbra]

cool characterization of $\RR$ that i found in an MSE post:$\newline$
The real line is characterized up to homeomorphism as a connected, locally connected, separable, regular space $X$ such that for any $x \in X$ the subspace $X-{x}$ has exactly two connected components.

[gentle osprey]

c squared

[fallen canyon]

@rancid umbra sure its hausdorf

sorry egg o my face, R^2\0 is connected haus but onky 1 component

ya you need separability, hausdorff, regular

[rancid umbra]

oh is there a counter example without the hausdorff assumption?

yes

the space needs to be hausdorff

[fallen canyon]

ya you need hausdorf

[rancid umbra]

https://math.stackexchange.com/questions/5099920/is-ℝ-uniquely-determined-by-the-point-removal-splits-it-into-two-copies-of-itse
the MO link that Mosher references has a paper which says that this characterization assumes X is Hausdorff. it also gives a counter example:
https://mathoverflow.net/a/76139/167473

[fallen canyon]

i meam i guess we talk about cofinite topologies, or something

lemme think about the hausdorf assumption

[rancid umbra]

yea, i haven't seen the counter example yet

[fallen canyon]

i guess regular mean haus on T1

got it: line with two orgins

[kind marlin]

wait so thats connected, locally connected, separable, regular, and removing any point creates 2 connected components?

the first 4 make sense but im confused why removing one of the 0s result in 2 connected components

[fallen canyon]

@kind marlin it's being general on frechet spaces

^^fr eg

[kind marlin]

wait i dont understand what that means 😭 as in the statement only applies to frechet spaces?

[fallen canyon]

no the q was to find a counter for non-hausdorf, now regular on t1 is frechet

so the assumption of regular on frechet and haus are redundant

basically hausdorff implies frechet but not vice-versa

this statement is only true on hausdorff

[kind marlin]

ohh

[balmy nexus]

That is really neat

[rancid umbra]

right!

[balmy nexus]

so you get locally compact and normal for free somewhere in this description?

normal is probably regular + separable or something

[gritty widget]

The Sorgenfrey plane is T_3, Separable, and not T_4

[rancid umbra]

second countable + regular => paracompact
Hausdorff + paracompact => normal

oh shoot

[balmy nexus]

It feels like I will never know this stuff. My eyeballs looked at these properties before

[rancid umbra]

how do you get from separable to second countable here

also, i want to use a metrization theorem some how

will think on this later

[balmy nexus]

generally you cannot

[rancid umbra]

am tired

[balmy nexus]

I think...

[balmy nexus]

huh

[gritty widget]

(https://topology.pi-base.org/theorems/T000275)

But I only know it’s LOTS because it’s R so

[rancid umbra]

right

hmm

[gritty widget]

Not sure how you go from some of those properties to LOTS without already knowing it is R

[rancid umbra]

ikr

very cool, minimal characterization

[tender halo]

i imagine you would be able to get the order relation of the points by comparing the components they split the space into

how do you get orientation i wonder

hmm

[rancid umbra]

yea

like

you could try x < y if one of x's components contains one of y's

but even if you show that its well-defined

[tender halo]

select two points, set one to be lower than the other

[rancid umbra]

you have an issue with orientation

[tender halo]

then the others can be oriented around those two

[rancid umbra]

oh hmm

[tender halo]

well

yea

they can right

[rancid umbra]

i feel like that's legit

[gritty widget]

I’m too tired to put thought into it but that seems true

[rancid umbra]

yea same

[tender halo]

now you only need to prove that components are Dedekind-cut-like

i.e. any two cuts are actually comparable

[gritty widget]

It’s always crazy cool to see how a bunch of properties that don’t look like they should imply an additional property(in this case, being homeomorphic to R) in fact do

[rancid umbra]

baking a cake from scratch is cool lol

[tender halo]

yea the characterizations of Q and the cantor set are pretty cool too

[gritty widget]

Yeah

Irrationals being omega^omega is not exactly what I’m talking about but I’m bringing it up cause it’s cool too

[rancid umbra]

oh i didn't know this one

topology is cool

[balmy nexus]

obviously R is simply a 1-manifold that isn't a circle

[gritty widget]

Yeah and the cantor set is 2^omega

It was always weirdly cool to me that an infinite product of discrete sets is not discrete, even though it’s trivial from the definition of the product topology

[rancid umbra]

there is a proof of the classification of 1-manifolds that i really like

it needs nothing else more than the definition of a 1-manifold

so its super elementary

and you don’t need any differentiability or CW complexes or simplicial complexes

[balmy nexus]

well the simplicial complexes are there for the 2 maniflds and higher it feels

the proof for 1 looks like pure meaningless formalism imo

[rancid umbra]

yea

my bad

[balmy nexus]

you take the line, and then you cut it into balls, and then each balls you cut into a graph...

[rancid umbra]

just going off memory from the toc of ITM

[balmy nexus]

I stared at it yesterday trying to understand what the big ide awas

[rancid umbra]

classification of 1-manifolds: a take home exam

[balmy nexus]

huh that's cool

[rancid umbra]

yea. i remember this was basically the first homework question in my diff top class.
our prof didn't have a solution written up and eventually put it as a bonus question instead because of how difficult it was

[balmy nexus]

it was technical and opaque for sure

[rancid umbra]

yea

but like, ig the main idea is that, in the non-boundary case, any two intervals overlap in either a single connected component or two connected components

[rancid umbra]

this is the classification in ITM?

[balmy nexus]

not sure what ITM is, it's how Lee does it if that's what you mean

ah introduction to topological manifolds?

[fallen canyon]

@rancid umbra i am aalso curious about the next paper in the attachment ; error analysis in complex variables, whhaaa

[rancid umbra]

ye lol

oh ha!

-# this is at the end of the take home exam

-# the next paper

that does seem interesting actually

[warped helm]

@umbral hamlet wanted to ask, how's topology going?

the reason i ask is because i remember about a week or so ago we expressed similar sentiments about the class being a bit of a learning curve

since then i've done a lot more practice and i certainly feel less adversarial towards the topic

[queen prism]

damn it took me more than a semester to attain the same attitude

[umbral hamlet]

Still about the same

Haven't had much time to sit on it, im more busy this semester than id like to be

Im about to do my topology homework though so we will see

[umbral hamlet]

having a bit of trouble showing that $\bar{A} -\bar{B} \subset \overline{A - B}$

[gentle osprey]

hiidostuff

[rancid umbra]

what do you have so far?

[kind marlin]

this was asked recently

i gave a starting point there that might be helpful

[umbral hamlet]

well i know that if x is an interior point then it follows pretty immediately

but the limit point case is whats giving me some problems

wait actually there are four cases arent there

[kind marlin]

you dont really need cases

[umbral hamlet]

x is an interior in both, x is an interior in one but a limit of the other, and x is a limit of both

[umbral hamlet]

ah im seeing what you did there

[kind marlin]

that took me a while to figure out tbh lol

[umbral hamlet]

i just dont work with setminus enough i guess

[kind marlin]

same

[umbral hamlet]

its weird juggling around the quantifiers in this one

thats what was causing me to be so uncertain about what i was doing

[kind marlin]

i drew a few simple examples when i was doing it and that made it a lot more obvious