point-set-topology

[tacit drift]

is that different at all from being pseudometrizable?

oh i guess it could be

i was thinking in the wrong direction

you can go from pseudometric to metric by quotienting

[ripe creek]

I have started making the PowerPoint presentation now.

[strong lantern]

My favorite example is taking R^2 and taking the quotient which sends the open unit ball to a single point. Then this space is not even Hausdorf

[tacit drift]

i see

how do you show that the space isn't pseudometrizable?

[gaunt linden]

If you let p be the squashed open ball, and q be a point on the unit circle, then p has a neighborhood that doesn't include q -- namely {p} -- but every neighborhood of q includes p.
Thus neither d(p,q)=0 nor d(p,q)>0 are possible.

[opaque scroll]

I think something is pseudometrizable iff you get something metrizable after identifying indistinguishable points.

So for example having distinguishable points that can't be separated would be an obstruction

[sly geyser]

does the poincare conjecture work for (>3)-d spheres?

[strong lantern]

Yes, and the proof is very cool for d>4.

[sly geyser]

what about d=4

[strong lantern]

The proof for d>4 was by Smale using h-cobordisim. It is a really cool proofe.
The case d=4 was by Freedman though I don't know what his methods are.

[sly geyser]

@queen prism why the sweat? this is an extremely obvious followup question

[strong lantern]

The number 4 makes every dif topologist sweat

[sly geyser]

what's so bad about it?

[queen prism]

many theorems that work in every other dimension break down for d = 4 because it's right on the line between low- and high-dimensional geometry/topology

[wide kayak]

allegedly this has something to do with the fact that 4 = 2 + 2 = 2 * 2 = 2^2 = …

[queen prism]

it's because 4 is the number of death

[strong lantern]

For d>4 spaces can be too complex but there's enough room to do stuff
for d<4 spaces can be "simple" even if there's no room to do stuff
For d=4 everything is hard and there's no room for anything

My favorite d=4 fact is that R^d admits only one smoth structure iff d≠4, and for d=4 it admits infinitely many.

[pliant idol]

I read about this in Loring Tu's book but I still have no idea what it means 🥀

[queen prism]

means if you take two overlapping smooth charts from two different smooth structures on R^4 they will never be compatible

(that is to say you cannot smoothly transform one coordinate system into the other)

[pliant idol]

no i mean I get that
its just that I haven't got the point where I've come across that theorem yet so just trying think about how that would be the case breaks my mind a little

[strong lantern]

It is probably better to understand the exotic sphere case in Milnor's famous article "on manifolds homeomorphic to the 7-sphere"

I don't know much about 4-manifold. But in this case I think these is a clear example of how this phenomena presents. Basically it is saying that you can "parametrizes" smoth structures in some way while showing that them being homeomorphic is easy to prove using some condition. In this case Milnor uses Reeb spheres.

[queen prism]

I wonder if this comes down to a difference in terminology, because the way I learned it (from lee) there are infinitely many smooth structures on any R^d, say, the two on R^1 generated by the atlases (atlii) A = {(x -> x)} and {(x -> x^3)}
but in every R^d except for R^4, they are diffeomorphic to one another

I did read that people on this server consider his definition of a smooth structure a bit sus

[strong lantern]

I mean, yeah, being more technical I should have added "up to diffeomorphism".
(I'm doing a lot of complex geometry and seeing x->x^3 being an atlas made my brain hurt)

[strong lantern]

There's probably a way to talk about "maximal atlas" and taking some sort of equivalence classes. Honestly, I wouldn't think about it that much.

[queen prism]

what happens with x^3 in complex geometry?

[strong lantern]

I mean, z^3 is not inyective in the complex plane

[queen prism]

oh true

[strong lantern]

Like, I think the word is "compatible atlas", but i don't know the technical details

[hidden abyss]

I don't think that's equivalent. Otherwise a smooth structure on a connected manifold would be uniquely determined by any single chart which is surely not the case

[dawn frigate]

Y is compact. Pi: X cross Y to X is closed map idk how to do this by lmtube lemma

[tender halo]

just by definition really

[crisp lintel]

Prove the complement is open using the pointwise def

of the image of a closed set I should say

[rotund prism]

So wikipedia says that the neighborhoods of x have to contain at least one open neighborhood of x in order for the space to be locally connected at x. If I'm trying to determine the local connectivity of a subset S of the plane, then the "open neighborhood" only needs to be open with respect to S itself, right?

Ie, consider a unit square in the plane, S. S is intuitively locally connected at its corners.
I can select the first neighborhood N1 as a closed disc with radius 0.1. This neighborhood is closed with respect to the plane, however it is plain not a subset of S.
Then I can select the "open neighborhood" N2 as the intersection of a tiny, open disc and S? Yielding a quarter disc. Which doesn't contain the tiny disc's part of the boundary but does contain the points that were on S's boundary.

With respect to the plane, N2 is neither open nor closed. But its plain open with respect to S, right?

And this is the "open" the wikipedia definition of local connectivity needs? For every neighborhood N1 around x, theres a neighborhood N2 of x which is a subset of N1 and open with respect to S?

or where am i going wrong lol

[gaunt linden]

This definition of "locally connected" is for an entire space, not for a subset of the space. You can view your unit square as a space in itself if you give it the subspace topology -- that means you've forgotten the rest of R², and your N2 open according to the subspace topology.

So it doesn't look like you're going wrong.

[rotund prism]

so a similar argument can prove that the red region inherits a subspace topology that is locally connected? At any point in the red, there will exist a disc open in the plane that, when intersected with the red, will still be open in the red? Which always indicates an eligible N2?

Now this seemingly implies that the union of the open disc and some subset of the unit circle is always locally connected in the inherited subspace topology, which feels weird but i guess is fair enough.

[gaunt linden]

Yes, those examples are locally connected.

There are spaces that are connected but not locally connected, but they need to be weirder than this. The standard examples would be the comb space and the Warsaw circle.

[rotund prism]

if anything, ive been messing around with the inverse, lol. I've basically been trying to figure out a good way to specify subsets of the plane that have nonzero area and are sufficiently "nice", but ive been trying to not impose pretty much any restriction on the boundaries.
Ie any filled jordan curve would be good, any interior of a filled jordan curve would be good. Any disjoint pair of filled jordan curves would be good. The epigraph of y=x^2 would be good. Any points of the boundary can be taken out and still be good.

[gaunt linden]

Perhaps you'd want a condition such as "the interior of the closure of A is the interior of A"?

[tiny obsidian]

"And nonempty" would be a useful addition, since 'non-zero' area was desired requirement

[lusty trench]

For the purposes of this question, a manifold is defined to be Hausdorff (so no line with two origins) and paracompact (so no long line).
Every differentiable manifold can be realized as a subset of R^n for some n, hence is not just Hausdorff, but actually perfectly normal (T6).
Can a mere topological or piecewise linear manifold fail to be perfectly normal?

[rancid umbra]

you can embed any topological m-manifold in R^{2m + 1}

[lusty trench]

It works for topological manifolds too? I knew it worked for differnetiable manifolds because Whitney.

[rancid umbra]

yea, some dimension theory stuff i think

[lusty trench]

I guess it's nice, but also a little disappointing. I want to know in what ways topological manifolds can defy expectations set by differentiable manifolds.

[rancid umbra]

this is from Munkres topology

[lusty trench]

Thanks!

[plush folio]

Is every subspace of R^n T6? I don't think subspaces generally inherit separation axioms, but maybe it's the case for T6 spaces?

[lusty trench]

At least every subspace of R^n is T5, because T5 is a hereditary property.

[plush folio]

I see tbh I have given up trying to keep track of separation axioms, and it doesn't help that all of them have multiple names

And they have stupid names. "Perfectly normal", "completely normal", "normally perfect" etc. etc... And apparently perfect sets don't have anything to do with perfect spaces 😭

[opaque scroll]

I think any metric space is T6

So yes

[crisp lintel]

fairly sure every subspace of a T^6 space is T^6 using a restriction argument

ya

[lusty trench]

At the very least, every closed subspace of a T6 space is again T6.

[crisp lintel]

a space $X$ is T6 iff its T1 and for every closed subset $A\subseteq X$ there is a continuous function $f:X\to[0,1]$ with $f^{-1}(0)=A$. Then if $Y\subseteq X$ is a subspace and $A\subseteq Y$ is closed, we have $A=E\cap Y$ for some closed set $E\subseteq X$. Pick a function $f:X\to[0,1]$ with $f^{-1}(0)=E$. Then $f|_A^{-1}(0)=A$

[gentle osprey]

Blake

[lusty trench]

Ah, I didn't know that characterization of T6 spaces.

[rancid umbra]

wikipedia supports this

the last sentence says that perfect normality is hereditary, so every subspace of a perfectly normal space is perfectly normal

[sly geyser]

so what exactly would a cotopology be like

[gritty widget]

I think there's multiple non-equivalent definitions depending on how you define a topology on a set(even tho the meanings for a topology are equivalent)

[cosmic mirage]

a cotopology is a collection of closed sets, that is closed under arbitrary intersection, finite union, and has the empty set and the entire set

[gritty widget]

You could for example define a topology on X as a subset of Hom(X, {0, 1}) which is closed under the binary operation f+g(x)=min{f(x), g(x)}, and the arbitrary operation taking a set of morphisms from X to {0, 1} to their maximums point-wise, and it contains both constant morphisms

Idk how to phrase that more... categorically?

Then you could probably reverse some arrows and whatnot

[sly geyser]

ok let's state the question more properly
what type of object's category would correspond to the dual category of topologies

with the morphisms between those types of objects being at least somewhat function-like (so a function preserving something or whatever, a homomorphism you know)

[cosmic mirage]

TRUE

open over...

[gritty widget]

Gelfand duality seems to be the closest thing I can find to what you're asking, but that's a subcategory of Top

[lusty trench]

A bit off-topic for a channel called point-set topology, but the category of point-free spaces (aka locales) is usually built by first constructing its opposite (i.e., the category of frames) and then dualizing. A locale is the same thing as a frame, but a morphism of locales is a morphism of frames in the opposite direction.

[sly geyser]

I treat this as if it's the default topology channel smh smh

[prime elbow]

Any hint for part b?

[rancid umbra]

these are free homotopies, so the base points can move

[pastel pumice]

can you topologically define differentiability? i made an attempt but it's probably wrong.

[unreal stratus]

Not really, at least by what I would understand by "topologically", since you can have non-differentiable homeomorphisms etc

Any definition has to interact with various smooth structures

[pastel pumice]

circular logic...

[unreal stratus]

How is that circular logic lol

[pastel pumice]

MY ATTEMPT WAS

[unreal stratus]

Ah lol

[pastel pumice]

basically THIS FUNCTION GRAPH IS DIFFERENTIABLE IF ITS A SMOOTH MANIFOLD but there was more.

Whyyyyyyyyyyyyyyyyyyyyyyyyyyyy

[ripe creek]

töpölögy

[crisp lintel]

not really, differentiability is fundamentally about linearity and you don't have any notion of linearity in a topological space

[pastel pumice]

yeah what i did was not only circular logic, it didn't even involve TOPOLOGY.

[iron bolt]

the graph of x^1/3 is a smooth manifold yet x^1/3 is not differentiable at 0

(if you don't believe me it's a smooth manifold, rotate the graph by 90 degrees and look at it again)

[unreal stratus]

What's the joke lol

[rancid umbra]

manifold just likes to laugh at his own messages

[quartz horizon]

based tbh

[safe torrent]

Clopencry

[gritty widget]

Maybe I'm missing something super obvious, but I've no clue what the retraction here is supposed to be... never heard the term "projecting from a point", only projecting as in projection onto a factor. The image isn't really clearing much up for me either. Anyone know what retraction is meant here?

$J^{n-1}$ here is ${1}\times{I}^{n-1}\cup{}I\times\partial{}I^{n-1}$

[gentle osprey]

colimit

[rancid umbra]

the radial projection from the origin is a function R^{n + 1} \ {0} —> S^n taking x to x/|x|

does that help?

[gritty widget]

Oh ok I think I get it now

It maps a point on the n+1-cube to the unique point the line from P to that point meets the subspace right?

[rancid umbra]

yea

[gritty widget]

I see

thanks!

[rancid umbra]

for another example, there is a radial projection that gives rise to a retract of the solid cylinder D x I to (D x {0}) U (S^1 x I)

you project from a point above the cylinder, on its centerline

[gritty widget]

That one's pretty easy to visualize too

thanks

[gentle osprey]

LicensedDumbass

[ruby delta]

so in the discrete topology a set is not compact iff it is infinite

so it is enough to show that if a set is infinite then whatever ultrafilter it can be found in has to have a finite set

but then if an ultrafilter consists of only infinite elements you can definitely enlarge it into a bigger filter by adding a finite element (finite subset of any arbitrary thing inside your filter), so it cant be an ultrafilter.
so this is the basic issue, the resultant topology is not compact iff you can find a noncompact open set that has no compact open subsets

[ruby delta]

so there are some non discrete topologies that still work to make the resultant topology compact (any topology on a finite set for instance), discreteness is just a condition that absolutely guarantees this @crimson terrace

also, I don't think X not being discrete implies that beta(X) is not the SC compactification

[crimson terrace]

Sorry I don't think I understand what you mean by "whatever ultrafilter it can be found in... End of msg"

[crimson terrace]

Yea I'm not sure about that
What I was trying to say that it's necessarily true

[crimson terrace]

Like an infinite set can't be in an ultrafilter no?

[ruby delta]

no

[ruby delta]

can you restate to me how the topology on beta(X) is defined again

that might help

[crimson terrace]

By specifying the base opens to be [A] where [A] is the set of all ultrafilters containing A

[ruby delta]

in the case of the discrete topology there is a better description, can you do it

[crimson terrace]

Hmmm

I mean it seems that the definition only relies on sets so

Empty set?

[ruby delta]

ok let me think of how to say this lol

[crimson terrace]

Because we've only talked of filters containing finite sets

Hmm okay yeah I guess a filter can include an infinite set

[ruby delta]

wait, base open?

[crimson terrace]

Like the basis of our topology

[ruby delta]

no no no, that doesn't sound like the right definition to me

[ruby delta]

shouldn't it be base closed?

let me double check

[ruby delta]

because if you define those sets to be the base open sets then beta(X) is homeomorphic to X when X is discrete

i am confused, someone has made a mistake

[strong lantern]

Only if X is finite

[ruby delta]

I am unconvinced, you are defining the principal ultrafilters to be open

[crimson terrace]

I mean

pretty sure its like that everywhere

[ruby delta]

so with this definition, if X is infinite discrete then beta(X) is never compact

[crimson terrace]

I am pretty sure we proved this here

[crisp lintel]

on an intuitive level, the reason is that for discrete spaces an ultrafilter converges to a point iff its a principal ultrafilter

so when you form beta X you only adding what is needed

[crimson terrace]

okay yes I got that
and beta(X) adds a point for those free ultrafilters

but adding those points, wouldn't it give rise to some extra ultrafilters?

[crisp lintel]

whereas for instance [0,1] is compact but it has tons and tons and tons of free ultrafilters

so when you form the set of ultrafilters its just gonna be massive

[crisp lintel]

yes but the topology on the stone cech compactification is designed in such a way that it forces the space to be compact

in particular it isn't discrete

[crimson terrace]

hmm can you maybe elaborate on how?

Because I tried to look at beta(beta(X)) but it gets messy real quick

[crisp lintel]

my favorite way uses some functional analysis, im not sure if there's an easier way

looking at beta(beta(X)) is not the right way

I guess there's a proof up there

[crimson terrace]

I figured

[crisp lintel]

let me remember the definition of the stone topology

[unreal stratus]

Open sets are pebbles and

[strong lantern]

Is this some Hanh-Bannach idea?

[crisp lintel]

not quite, you can identify the stone-cech compactification with the set of multiplicative linear functionals from C_b(X) to the complex numbers

together with the weak* topology

and then by Banach-Alaoglu its more or less immediate that the stone-cech compactification is compact and hausdorff

if X is discrete, say $\mathbb{N}$ to be concrete, then $C_b(X)=\ell^\infty\mathbb{N}$ is just the space of bounded sequences

[gentle osprey]

Blake

[crimson terrace]

Hmm so
When X is discrete, our compactification is adding points for those free ultrafilters, which turn out to be not much, and even after adding them, the excess ultrafilters still converge to some of those points
When X isn't discrete, this (always?) goes wrong because this addition of ultrafilters becomes too much (how? is it because the new ultrafilters made by the extra points end up not converging?)

[crisp lintel]

when X isn't discrete, you are adding too many ultrafilters basically, far too many

the idea is that intuitively a free ultrafilter should correspond to a "place where a sequence can escape to infinity"

however this is only true for discrete spaces

since for a space with a topology, its possible for free ultrafilters to converge

but on a discrete space, free ultrafilters never converge

so every free ultrafilter corresponds to a "point at infinity" in some sense

[crimson terrace]

ahhh okay I see I see
and when we quotient out with the Hausdorff quotient for any topological space, we are relating those escaping to the same "infinity"

[crisp lintel]

well you don't need to quotient if X is discrete

perhaps I'm misunderstanding what you mean though

[crimson terrace]

Sorry let me make it more clear
if X is not discrete, we end up adding far too many
what the haudorff quotient does is relate two filters that diverge to the same infinity point

[crisp lintel]

then every multiplicative linear functional $\varphi:\ell^\infty\mathbb{N}\to\mathbb{C}$ corresponds in a natural way to an ultrafilter on $\mathbb{N}$: Let $A\subseteq\mathbb{N}$, and consider the characteristic function (or characteristic sequence I suppose) $1_A$. Since $\varphi$ is multiplicative, we have $\varphi(1_A)=\varphi(1_A)^2$ which implies $\varphi(1_A)\in{0,1}$. Then you can show the set of all subsets $A$ with $\varphi(1_A)=1$ forms an ultrafilter.

[gentle osprey]

Blake

[crisp lintel]

oh I might not be aware of this hausdorff quotient thing

[crimson terrace]

let me send a ss

[ruby delta]

suppose that X is infinite discrete.
Then for every point in X, its principal ultrafilter is open, and the singleton {x} is contained in a principal ultrafilter iff the principal element is x.
So if i take the union over all the principal ultrafilters, then that is an infinite open cover for beta(X) that has no finite subcover.

what am I misunderstanding?

[crisp lintel]

is this a way of taking a quotient of the set of ultrafilters to obtain the stone-cech compactification

[crimson terrace]

yep!

[crisp lintel]

ah yes then you're exactly right

you're basically fixing the issue of adding too many ultrafilters

any ultrafilter that converges to x in X should be identified with the principal ultrafilter I believe

[crimson terrace]

yep

thats the map we have from X to Beta(X)

[crisp lintel]

yeah that makes sense and is quite a nice intuition I think

[crisp lintel]

the set of principal ultrafilters does not cover beta(X)

unless X is finite

[crimson terrace]

I see I see that makes a lot of sense

When we say going to a point of infinity, we are just saying no convergences. Now, the way classify if two ultrafilters go to the same infinity point is by checking for all compact hausdorff where f^tilde goes to

[crisp lintel]

yeah

[crimson terrace]

is there a specific reason we are checking only compact hausdorff spaces?

[crisp lintel]

and the reason the stone-cech compactification is the "largest" compactification is b/c you are adding "all possible points at infinity"

[crimson terrace]

I guess my intuition tells me that they are the only relevant ones but yeah

[ruby delta]

oh fuck, I know why I made the mistake I reversed the inclusion on the poset of open sets

[crisp lintel]

you can justify it via the universal property, or another way to see it is that compact hausdorff spaces kinda "classify convergence" in some sense

thats a bit vague though let me think about it for a second

[crimson terrace]

I see so other extreme is the one point compactification where we group all limit points that are not in the set at a single point *

[crisp lintel]

yeah exactly

[crimson terrace]

That would be great

[crisp lintel]

well for one it certainly is enough to do the thing I mentioned earlier

that is, if Q is an ultrafilter on X that converges to some point x

then f(Q) will be equal to f(F_x) where F_x is the principal ultrafilter at x

and f is any function into a compact hausdorff space

ah ok ehre's why

pushing forward an ultrafilter is only really well defined for compact hausdorff spaces

[crimson terrace]

f here is f^tilde? the pushforward of the filter defined by f

[crisp lintel]

yeah

[crimson terrace]

OHHHHHHHHH

[crisp lintel]

b/c compact implies the pushforward ultrafilter converges

and hausdorff implies it has a unique limit

and this makes the equivalence relation work even for ultrafilters that don't converge

[crimson terrace]

ah okay so it is a suitable way to classify these infinites

[crisp lintel]

the intuition would be that if Q and F both get pushed forward to the same limit point in every compact hausdorff space, then they should both count as the same "point at infinity"

[crimson terrace]

but now i have one more question
how do we know that there isn't some more general way to relate ultrafilters so that two ultrafilters can be classified as different infinite points

[crisp lintel]

I think there you would probably want to refer to the universal property of the stone-cech compactification if you have seen it

also I think its hard to really come up with a better way to relate ultrafilters

I guess you could say, maybe require that the pushforwards must always converge to the same point if they converge when being mapped to any space

that might follow from this, but im not entirely sure

[crimson terrace]

but I guess for one reason or the other you no longer get compactness due to being too large

[crisp lintel]

or I guess have the same set of limit points if the space isn't hausdorff

[crimson terrace]

I think this follows

[crisp lintel]

p sure it actually might follow using the stone-cech compactification on the target space funnily enough

[crimson terrace]

but what I wanted to say was since this compactification is the largest, it must be the case that any other way we try to get a more general breakdown of those points at infinity, we must get a space that is too large to be compact
does this follow by the universal property?

[crisp lintel]

oh you mean like an even larger compactification

[crimson terrace]

yep

[crisp lintel]

I think it would violate the thing I mentioned previously where you would have an ultrafilter that converges to x but isn't identified with the principal ultrafilter

and I think that kinda breaks things

that's my thinking but I guess maybe in theory you could come up with a relation that doesn't impact that property

ah ok here's an issue

for every compact hausdorff space includes itself

actually wait this might not be an issue let me think

oh yes it is an issue

if X is a compact hausdorff space, then x=y in X if and only if the images of x and y are the same under any continuous map into another compact hausdorff space are the same

which is basically completely trivial because you can just take the identity map to itself

so philosophically, if we want to built a compact hausdorff space then this property ought to be preserved with the points

that is, if two ultrafilters have equal image in all compact hausdorff spaces, then they ought to be the same if we want our compactification to be compact and hausdorff

if that makes sense

i guess technically we're talking about maps out of X instead of beta(X) so this isn't technically true its more like philosophically true

also this is more or less the proof that the universal property characterizes the stone-cech compactification by the way

[crimson terrace]

ahhh I see! So it's to guarantee that beta(X) is compact hausdorff

[crimson terrace]

was planning to read it tmr but this got too interesting to pass 😅

[crisp lintel]

if Y and Z were 2 stone-cech compactifications of X, then you would obtain a maps from Y to Z and from Z to Y using the universal property, and you can show using some uniqueness trickery that both compositions are the identity

another way to think about the universal property is that if you have any compactification of X at all, then the stone-cech compactification necessarily maps surjectively onto that space

p sure that at least if X is sufficiently nice, every compactification of X must be a quotient of the stone-cech compactification

[crimson terrace]

hmm but doesn't that show that a SC must be of the "same size" but doesn't address whet

[crisp lintel]

if X is bad then the map from X into the stone-cech compactification might not even be injective and things are weird

well that shows its the largest compactification

[crimson terrace]

👀 that's pretty cool

[crisp lintel]

since every other compactification is a quotient

[crimson terrace]

ah okay

[crisp lintel]

in the analogy this would be seeing that the stone-cech compactification puts the minimal restrictions on identifying ultrafilters

if you tried to identify less, then you would lose being hausdorff or compact

https://en.wikipedia.org/wiki/Wallman_compactification for more general spaces you may want to look at this

[crimson terrace]

OHH wait last thing
its because we are mapping to hausdorff and compact spaces and since f is continuous and the pushfoward is the only map we have, we must make this identification otherwise we are saying that the sapce we are mapping to is either not hausdorff or compact

[crisp lintel]

the wallman compactification I believe doesn't have to be hausdorff so it can be larger

[crimson terrace]

but its more restrictive on X

[crimson terrace]

but I think this shows why any other identification fails (which is basically the universal property lol i should've given it a read)

That's everything I have
thank you very much for your help! Have a nice day

[worn mortar]

I’m confused by this passage

I don’t quite understand what the restricted Cartesian product is.

I’m looking at this question here, and it just doesn’t make sense to me. If P is the Cartesian product of the
family of projection maps indexed by M, is the domain not the M fold Cartesian power of Φ?

It seems so to me but I don’t know

[ruby delta]

so in general if we have a collection of sets and functions between them the best we can do is defined the final function to be the individual ones on each component

[worn mortar]

It’s just that in this passage it seems like the Cartesian product H of the family of functions is itself a function between the product of the domains and the product of the codomain.

[ruby delta]

but that makes the cartesian product of functions very inefficient - think about it, for every function we are introducing a whole new coordinate, and that ends up multiplying the size of the domain

if you keep stacking this over and over again the domain is going to explode in size

so, you can try and be a bit more efficient, and if there is overlap between the domains just resue the same thing again

and that is the restricted cartesian product

so what does that look like, let's say f(x) = 2x and g(x) = x+1.
Then the cartesian product of the functions is (x,y) -> (2x, y+1)

but we don't really need the domain to have two variables

we can collapse everything nicely into x -> (2x, x+1), which is the restricted cartesian product of the functions

[worn mortar]

This is still very unclear to me lmao

[worn mortar]

When I read this page, it seems that the Cartesian product H of the family of functions is itself a function between the product of the domains and the product of the codomain.

[ruby delta]

yes

it is

[worn mortar]

So if I have a family of functions with the same domain, shouldn’t the domain of the Cartesian product be the Mth Cartesian power?

[ruby delta]

yes

but the domain of the restricted cartesian product is the domain once

think about the most extreme case, we have the same function f from A to B 10 times

then the cartesian product is A^10 -> B^10

which is very redundant

all the information we really want to know is, how many functions there are, and what everything gets mapped to

so the restriced product is just A -> B^10

[worn mortar]

In this passage the restricted Cartesian product is defined by composing with the diagonal injection.

[ruby delta]

you can even restrict it on the codomain even further but that is not how your textbook has defined it

[ruby delta]

yes, so in this case the full thing is A -> A^10 -> B^10

where the last thing is the original cartesian product

[worn mortar]

I’m guessing in the problem statement then I should be taking the diagonal injection to be from Φ to Φ^M

[ruby delta]

and the diagonal injection at the start is telling you that all 10 copies all actually take in the same inputs

dont think about the diagonal injection lol it's a waste of brainpower

it's just x -> (f1(x), f2(x), f3(x), ....)

adding in a x -> (x, x, x, ...) -> (f1(x), f2(x), f3(x), ....) in between just complicates it in your head

[worn mortar]

I think I understand where I went wrong

[worn mortar]

I know you said ignore this lol

But this is what I have for that exercise so far

[ruby delta]

using f to denote something inside Phi is scuffed

but yeah youre basically there

finish him

[warped helm]

what book is this?

[worn mortar]

I see now

[worn mortar]

General Topology by Hu.

[ruby delta]

yeah, by hu?

they are asking you

[warped helm]

thats some in depth set theory if ive ever seen such a thing

[worn mortar]

It’s a good book

I was also taking notes from chapter 0 of “rings and categories of modules”

[alpine nest]

Abbott or Costello alt account spotted

[oblique sentinel]

Hello! Quick question about Urysohn Lemma. It is said for example in Munkres, that the choice of the intervall [0,1] is arbitrary and the general case follow from that. Is it because we could for the general case just define the sets U_p for whatever we need. For example then we don't just do the counting for rationals from 0 to 1 but rather from maybe 5 to 20 or whatever

[quick delta]

Once you’ve got [0, 1], that makes it much easier to construct other maps you might want
Because you can just define your map by X -> [0, 1] -> Y

Where Y is sufficiently nice/explicitly defined

[gaunt linden]

You could redo the entire proof for a different interval, but you could also just say, construct a 0-1 separating function g using the standard version of the lemma, and then set f(x) = 5+15g(x).

[oblique sentinel]

Oh yeah okay! Thanks 🙂

[brave cipher]

hi so I'm new to topology and I was wondering about the product topology, I know that the product topology of two circles is a torus but when we do the product we get 4d points? ((a1, a2), (b1, b2)) -> (a1, a2, b1, b2) which if I understand correctly is topologically a torus but not geometrically?

is there simple and direct way to project this 4d structure into a 3d torus?

(without using trig functions in the projection)

[young stone]

why don't you want to use trig functions?

[brave cipher]

well the most intriguing part to me was the idea that a structure like a torus can emerge from a simple binary operation between two simple structures

I'm looking for the simplest way possible to do this

the most simple projection

I may be wrong here but from an information pov, isn't the "circle-ness" or "round-ness" information of the torus encoded in the two circles that make it up?

if that's the case then doesn't that mean It's possible get the projection of the torus without using trig functions to add any more of this information?

what I'm saying is it feels a bit like cheating (I know It's not)

[opaque scroll]

I guess you can just do vector addition, like:

(a1, a2, 0) + b1*r(a1, a2, 0) + r*(0, 0, b2)

Where r is whatever you want the smaller radius to be. 1/3 would probably be good.

I'm not really sure how you would use trig functions to improve this...

[brave cipher]

interesting

why this particular choice of projection?

[opaque scroll]

I mean, it's just the standard embedding of a torus in R^3.

You take a circle and rotate around and axis

Like you draw a circle around (1, 0, 0) with radius r. Then spin that around the z-axis. That's the formula I wrote down

[cosmic mirage]

Is it true that any simply connected subset of R^2 is contractible?

this feels like a thing I should be able to prove or disprove but me and another grad student can't figure it out 💀

[quartz horizon]

Apparently not

[cosmic mirage]

I was afraid of this

what does apparently mean

[quartz horizon]

I found an MSE post explaining it

[cosmic mirage]

ah so true

Lol yeah I didn't Google this before asking haha

[quartz horizon]

https://math.stackexchange.com/questions/4447426/sufficient-condition-for-a-simply-connected-subset-of-mathbbr2-to-be-contr

[cosmic mirage]

ahh very interesting, so the Warsaw circle isn't contractible but it is weakly so

hmmm

this was the second thing I was thinking about. Can I get such a space that isn't weakly contractible

[quartz horizon]

https://math.stackexchange.com/questions/548681/simply-connected-does-not-imply-contractible-is-there-a-nice-counter-example-in

[cosmic mirage]

ah cool so these will always be weakly contactible

[worn mortar]

I'm looking at these notes:

https://math.wvu.edu/~jwojciec/teaching_files/2024_Spring-581/node-4.html

I don't understand the point that Proposition 1.4.3 is trying to make

From Proposition 1.4.1 I can see that the intersection of any nonempty family of topologies on a set X is also a topology on X.

I can also see that the topology generated by a family of subsets of a set X is the intersection of all topologies on X that contain the family as a subset.

But I don't understand what exactly the first part of the proof of 1.4.3 is saying. Is it not already obvious from 1.4.1 that the topology generated by a family is indeed a topology on X? Like it seems apparent since we proved that the intersection of an arbitrary family of topologies on X is a topology on X.

[warped helm]

it's giving a characterization of the topology generated by the subbasis

definition 1.4.2 doesn't quite say the same thing

1.4.2 already implicitly states that tau(l) does form a topology, which follows from 1.4.1 as you mentioned

[worn mortar]

In the proof of Proposition 1.4.3 is the first paragraph just saying that tau(l) is a topology on X?

[warped helm]

not quite

that's the subsequent development

the first serves to show that the family of sets mentioned in 1.4.3 (which apriori has still not been shown to be a topology) will necessarily be contained in tau(l)

i guess to simplify how much stuff is "floating around", denote the family of sets in 1.4.3 by tau'

the goal is to show tau' is a topology and tau' = tau(l)

[worn mortar]

Lmao I'm still trying to clarify the first two paragraphs exactly. I can see from the statement that we are trying to give a characterization of tau(l).

But I'm very confused as two what those two paragraphs are saying

Like, it's clear that tau(l) is a topology on X

[warped helm]

yes, that's not part of 1.4.3

[worn mortar]

Hence it contains the full set X and the empty set since X and the empty set belong to any topology on X

Then it says "Moreover, any topology on X containing l contains all unions of all possible intersections of l"

[warped helm]

yes, which shows tau' subset tau(l)

how is "family of subsets" of X defined in this text?

[worn mortar]

I see what you're saying. They could've wrote "Suppose tau' consists of the empty set, X, and all unions of all possible intersections of nonempty finite subfamilies of l. Then because (.......) we have tau' is a subset of tau(l)"

[warped helm]

yeah, the text is a slight bit terse imo

although very rigorous

[worn mortar]

I see it now. Since tau(l) is the smallest topology on X that contains the family l of subsets of X, if tau' is a subset of tau(l) that is also a topology, then it is equal to tau(l).

This is what you were saying

[worn mortar]

When they say that it's clear that X and the empty set are in tau in the second paragraph, I don't quite understand it.

Why do we have to say "together with X and the empty set" if they're (purportedly) already elements of tau

[warped helm]

yeah that part is weird

i'm not sure

i think it might be slightly messed up (?)

[worn mortar]

That has to be the case then

I'll rewrite this whole thing

[warped helm]

munkres defines subbasis as a family of subsets of X that cover X

i think this part is missing from the statement of 1.4.3

[worn mortar]

Well that definition is not the one I'm familiar with lmao...I thought subbasis is just a subcollection of the topology such that the finite intersections of open sets in this subcollection form a basis.

[warped helm]

that's the basis generated by the sub basis, no?

[worn mortar]

Well I think a given subbasis for a topology can generate more than one basis for the same topology

[warped helm]

how?

[worn mortar]

I'll have to come back to this after I understand the content more

[lament steppe]

Ive been trying to show that if X is a topological space, a metric space and X is Lindelof, then X is second countable.

Ive not really been able to nail these details down. Things that I have thought about:

1.) Any subset of X is also Lindelof (so any open sets of X have countable covers)
2.) Given some cover of X that is formed from balls, there is a subset of these balls that form a countable subcover. Tho Im not able to show that this new set of balls is actually a basis (I havent been able to explicitly show that a random open set in X is a union of this collection)
3.) Maybe trying to take the contrapositive of this might somehow be easiser (assume X is not second countable, show that X is not Lindelof)
4.)I could try and show that if X is Lindelof then X is Separable... since I have previously shown that Separable implies Second Countable.. but I was hoping to actually prove things more directly

Anyone able to offer any thoughts that might help out?

[warped helm]

try building up a basis via a countable union of countable covers

[rancid umbra]

given a metric space, can you tell me a basis for that metric space? just any basis, doesn’t have to be countable yet

[lament steppe]

Sure. How about the collection of all the balls. So for B_r(x), r can be any real number and x is any element in X.

[rancid umbra]

okay, so if we apply the lindelof hypothesis, we get a countable subcover. but the sets may not form a basis, since they may not be downward closed under pairwise intersection, .e.g., B_r(x) \cap B_r’(x’) may not contain a ball in the countable cover from applying the lindelof condition

so we want to be able to control the radius of our basic open sets, in the sense that we need to be able to choose them as small as we like

[lament steppe]

Ah so maybe we restrict the values of r to be rational?

[rancid umbra]

well, that isn’t going to give us any extra control immediately

because we already had all real values of r

we want to be able to say that there is some r’’ such that B_{r’’}(x’’) is contained in B_r(x) \cap B_{r’}(x’)

so if, for example, r = 1, and r’ = 1/2, we know that there is some radius r’’ > 0 such that B_{r’’}(x’’) is contained in the intersection

but

it might not be in the open cover that we were given by the lindelof condition

so we should probably apply the lindelof condition again, with a cover of balls of small enough radii

[lament steppe]

Does this logic sorta introduce some recursion here? because these new "small enough" balls can be intersected with each other to introduce another new collection of balls?

[crisp lintel]

I think you'd want to do it one rational at a time and take some sort of union after

basically what josemom was saying

[rancid umbra]

i think i might be explaining what i want to say poorly

the solution i have in mind is to ||apply the lindelof hypothesis to the collection C_n = {B_{1/n}(x) : x in X} for each n||.
||the union of all C_n forms a basis for X||

[crisp lintel]

yeah ok that's what I had in mind as well

[rancid umbra]

but the idea is like, ||i want my collection to be downwards closed under intersection, and the usual open set in the intersection of basic opens that i get from the basis {B_{r}(x) : x in X and r > 0} almost gets me this||.
||i just need to keep throwing in (countably many) successively smaller sets one stage at a time||

[lament steppe]

Awesome. Ill noodle on these thoughts for a while before looking at your solutions. Thanks all!

[bitter coral]

I'm not really a professional, but if I were you I'd prove it this way :
Given a sequence $$(x_n) _{n=m}^{\infinity}$$ which is contained in E, and converges to $$n_0 \in E$$
By definition of convergence, for every neighborhood P of n_0 \text{there exists N such that for all} $$k \geq N, x_k \in P$$. Because
all of the points in the
sequence are in E, every
neighborhood of n_0
intersects E. This implies
that n_0 is in the closure of
E, implies E = E closure,
implies E is closed

[gentle osprey]

Mατchα

I'm not really a professional, but if I were you I'd prove it this way :
Given a sequence $$(x_n) _{n=m}^{\infinity}$$ which is contained in E, and converges to $$n_0 \in E$$
By definition of convergence, for every neighborhood P of n_0 \text{there exists N such that for all} $$k \geq N, x_k \in P$$. Because
 all of the points in the
 sequence are in E, every
 neighborhood of n_0
 intersects E. This implies
 that n_0 is in the closure of
 E, implies E = E closure,
 implies E is closed
```Compilation error:```! Undefined control sequence.
l.50 Given a sequence $$(x_n) _{n=m}^{\infinity
                                               }$$ which is contained in E, ...
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.```

[bitter coral]

This is so annoying but I hope you get my point

[dusk ferry]

yes this works too, but I am not sure whether the logic of my proof is correct

[dusk ferry]

yes, I got it.

[bitter coral]

Honestly I'm not very sure, I don't really understand your proof. Maybe you could rewrite it?

[dusk ferry]

okay

I tried to use the equivalence of a and c in my proof

In my proof I first assumed that E is closed and then I had to prove a statement for every convergent sequence in E, so I assumed xn to be some arbitrary convergent sequence in E, from this can I conclude that there exists a sequence in E which convergest to limit xn? If I could conclude this then using (c) and (a) I can conclude that lim xn is an adherent point of E, which by definition means lim xn is in the closure of E, but as E is closed, lim xn must be in E, which concludes the proof of the first implication

This is the defintion of closure

[bitter coral]

Ic, ty for the context! I'll reread your proof

Your proof seems all correct

[dusk ferry]

Okay, thanks!

[alpine nest]

What is the definition of "closed set" you're working with here?

[dusk ferry]

this one

[vocal dawn]

Sorry to interrupt if you're all having a conversation about something, but I'm trying to start topology and I want some good material, and please, don't give me introductory textbooks, I can handle higher level.

[tender halo]

munkres is a good starting point

[crisp lintel]

Munkres prob

[vocal dawn]

Is that high level?

[warped helm]

yes

[tender halo]

i mean

what does high level mean

its an introduction to the subject

[vocal dawn]

Something else? Because I know that one book can't cover everything.

[vocal dawn]

I don't want an introduction, I want something that is terse, proof-heavy, abstract, and generalizing.

[tender halo]

thats like

[quartz horizon]

Maybe “topology: a categorical approach”…?

[tender halo]

you are describing what a math textbook is

except hatcher

[warped helm]

why do you want terse?

i understand the rest but specifically wanting terse is strange

[crisp lintel]

Learning math isn't about showing off by reading "hard books"

[warped helm]

munkres is rigorous and general

[vocal dawn]

It trains my brain for mathematical understanding, which I think I will need for later, naturally terse subjects, it isn't for showing off.

[warped helm]

“naturally terse subjects” 🤔

[vocal dawn]

Analytic number theory is an example.

Non-linear PDEs is another.

[warped helm]

are you sure you know what “terse” means?

[tender halo]

ok i suggest you try reading tom dieck then you give up and read munkres

[crisp lintel]

in math introductory doesn't mean basic or without rigor

it just means like... teaching you the subject

an advanced topology book would be a book covering advanced topics for people who already know all of the basics

[warped helm]

munkres is plenty terse i would say

very little exposition

[crisp lintel]

also you don't need to read terse books to train mathematical thinking

in fact the opposite is often goood

[tender halo]

what no half of munkres is waxing philosophical

[crisp lintel]

a better written book will explain what the key tricks and key ideas are

[tender halo]

if you want terse read idk

engelking

kuratowski if you are out there

[vocal dawn]

Ok then, all notes taken, I'd suppose Munkres it is, but what should I follow it up with?

[quick delta]

Tbh in general I feel maths books are too terse
I’m much more likely to end up thinking “would adding one more diagram have killed you?” than “why are you spending a page on a 3 sentence proof?”

[crisp lintel]

depends what you want to learn

munkres is like 500 pages so you got some time

[vocal dawn]

What will I need for functional analysis?

And algebraic-differential topology?

[crisp lintel]

real analysis, measure theory, topology

for functional

linear algebra

as well of course

[vocal dawn]

What will i need in topology for that.

[crisp lintel]

basically everything up to the algebraic topology section in munkres

also nets idk if munkres covers nets

you were posting about real analysis the other day so honestly I'd do that before even doing topology

[warped helm]

he does briefly

more as a supplementary topic though

in which you do most of the development in exercises

[crisp lintel]

ah ok fair enough

[vocal dawn]

Believe me, I am doing everything simultaneously, it may look like a mess, but I think it'll help me see how it all clicks in the end.

[wild lichen]

munkres does cover nets iirc in one of the supplemental sections

i don’t remember which one

[prisma garnet]

the first one

and not as a supplement

it's in the first chapter

[gritty widget]

Topology & Groupoids by Ronald Brown

[marsh lotus]

Would you have an exemple of a Gdelta of a Baire space that wouldn’t be a Baire space (I proved that if it is dense it is a Baire space, and I want to be sure it’s not true in general)

[alpine nest]

There's an example here: https://math.stackexchange.com/a/3003831

The upper half-plane but with only rational points on the x-axis is Baire, the rationals on the x-axis are its closed subset (thus G-delta, because we're in a metric space), but aren't Baire

[marsh lotus]

Thx !

[worn mortar]

I'm trying to show that a countable product $X= \prod_{\alpha =0}^{\infty }X_\alpha$, $X$ is sequentially compact if and only if all of the $X_\alpha$ are also sequentially compact, where each of the $X_\alpha$ are topological spaces.

[gentle osprey]

calebuic魏凯布

[worn mortar]

I’m trying to figure out how to word this to pull a convergent subsequence of (s_i) out

[quick delta]

You want to choose your subsequences more carefully than that
I’m not sure how to explain more without just handing you the solution

Essentially the problem is you want them all to converge simultaneously, but you’ve currently only got them converging individually

[worn mortar]

Let me think about this some more. I see the problem now.

Thanks

[worldly radish]

how does one attain this handwriting 🙏

[worn mortar]

I think simply writing a lot and intentionally

[north ore]

With enough practice you can attain this handwriting

[radiant stone]

day 0 math: beautiful writing
day 365*10 math: unreadable

[north ore]

mine was unreadable from day 0

[radiant stone]

Based

[cosmic mirage]

at what point

[rotund badger]

hi new to topology
what does it actually mean to say two lipschitz equivelant metrics produce the "same" open sets?

it surely doesnt mean an open ball wrt to one metric is equal to the open ball wrt to the other

[warped helm]

you can fit open balls of each respective metric inside the other

[unreal stratus]

I.e. they have the same set of open sets

[rotund badger]

i see i guess that makes sense. with intuitive notions of distance and metrics this seems obvious i guess non lipschitz equivelant metrics dont seem intuitive to me

[rotund badger]

ohhh yeah that makes sense

[quartz horizon]

an open ball in one metric is a union of open balls in the other metric

[unreal stratus]

Could have the metric on R^n where distance between any two distinct points is 1

This is very different from the usual one

[warped helm]

so if you have a ball in the metric (d_1), (B_{d_1}(x, r)), for each (p \in B_{d_1}(x,r)) you can find a ball in the metric (d_2) such that (B_{d_2}(p, \epsilon) \subset B_{d_1}(x, r) ) and vice versa

[gentle osprey]

josemom2

[rotund badger]

yeah its just spaces like this are just hard to conceptualize

[unreal stratus]

Ye

[rotund badger]

ah okay i see. so is there also always a ball of a different raidus in the other metric thats = to the ball in the first? no right

[warped helm]

think about R^2 and open disks versus open rectangles

within each point of an open disk D in R^2, you can fit an open rectangle R around the point that sits inside the disk

for each point inside an open rectangle, you can likewise find an open disk that contains the point and lies within the open rectangle

[rotund badger]

ah okay that makes sense, cause a union of open balls isnt always a ball
but any open set in one space is also open in the other, just not always a ball ✅ ?

[warped helm]

you can write the ball as a union of open rectangles

[queen prism]

they won't necessarily have the same open balls but they will generate the same open sets

[rotund badger]

i see that makes sense

[rotund badger]

i get why you can fit each open set in or around an open set in the other (from the LE defintion), but how does that imply an open set in one can be written as a union of open sets in the other?

[warped helm]

Let (\rho) be the euclidean metric on (\mathbb{R}^2), that is for two points (x ,y \in \mathbb{R}^2) we have [ \rho (x,y) = \sqrt{(x_1 - y_1)^2 + (x_2 - y_2)^2} ] and let (d) be the square metric on (\mathbb{R}^2), that is [
d(x,y) = \max { |x_1 - y_1| , |x_2 - y_2| }. ]
Let (x) be a point in (\mathbb{R}^2) and (B_{rho}(x, r)) be an open ball around (x). For each (p \in B_{\rho} (x ,r)), we can find an open rectangle (B_{d}(p ,\epsilon_p) ) where [
p \in B_{d}(p, \epsilon) \subseteq B_{\rho} (x, r). ] It follows, as you should check, that [
B_{\rho}(x,r) = \bigcup_{p \in B_{\rho}(x,r)} B_{d}(p, \epsilon_p) ]

[gentle osprey]

josemom2

[warped helm]

there's nothing special about the 2 in R^2 here, the same argument works for any natural n

[rotund badger]

oh for each p right that makes sense then

thank you so much !!

[warped helm]

the technical details are to actually show they're equivalent

i.e actually saying that such an epsilon_p exists

but this is a readily seen estimate from the definitions of the metrics

[rotund badger]

which exists because theyre lipschitz equivalent right

we get that from the constants bounding the two metrics?

[stray zodiac]

proof that discrete metric always satisfies triangle inequality

for any x, y, z in M (the space), lets say x = y

then d(x y) + d(y z) = d(y z) = d(x z) >= d(x z) thus it satisfies the inequality

and lets say that x != y

then d(x y) + d(y z) = 1 + d(y z), which is trivially greater than or equal to 1 (since value of d can't be nagative)

and as the value of d can be 0 or 1, d(x z) should be less than or equal to 1

combine both together, we get d(x y) + d(y z) >= 1 >= d(x z) thus it satisfies the inequality

(it was in my exercise... seems correct)

[bitter coral]

Seems all correct to me

[vocal dawn]

Since I work with pdfs of books, and by consequence can't find included exercises, could anyone tell me where to find exercises that accompany munkres?

[quick delta]

Most textbooks have exercises in them
I’d be very surprised if Munkres doesn’t

[vocal dawn]

PDFS

[quick delta]

Yes
PDFs of a book are usually identical in terms of content to the book

[vocal dawn]

Ok then, I suppose I'll look.

[quick delta]

Munkres seems to have them at the end of each chapter

[vocal dawn]

And are all the previous exercises from findable textbooks or websites?

[quick delta]

Wdym by previous exercises?

[vocal dawn]

Those shared here.

[quick delta]

Some exercises people ask about here will be from books
Some will be from problem sheets set by professors
Some will’ve come about more “naturally

[unreal stratus]

Ig potentially by definition

[quick delta]

I’ve seen bad digitisations
But most (especially for well known stuff) is well done

[lament steppe]

Im thinking about what locally Euclidean means. What my book says about this property of a topological space, is that every element of that space has a neighborhood that is homeomorphic to a subset of Euclidean space.

Lets say we have a topological space X and a homeomorphism f : U -> B_r(x) such that U is an open subset of X (and B_r(x) is an open ball in R^n)

Since f is continuous and there are an infinite amount of other balls that are contained in B_r(x), then the inverse images of all these balls are also open in X.

So if we have space that is locally Euclidean, it's never really that there is just 1 neighborhood that has a homeomorphism (for every point in X), but rather an infinite amount of said neighborhoods. Is that correct?

[gritty widget]

If you have a neighborhood that is homeomorphic to R^n, you can just take some open subset of that equivalent to (0, a)^n for arbitrarily small a to get arbitrarily small neighborhoods homeomorphic to R^n yeah

This is true even if you have only some points contained in that neighborhood, not necessarily locally Euclidean everywhere

[queen prism]

-# homeomorphic to an open subset of*

[versed flower]

Are the elements of the subbase of the weak topology open?

[gaunt linden]

Elements of a subbase are open by definition.

[versed flower]

doh

thanks

[prime elbow]

So disjoint union topology is co-product in category of topological space?

[quick delta]

Yes

[prime elbow]

Okay thank you

[sly geyser]

can someone explain the "by elementary topology part"

here's the whole thing
https://people.math.osu.edu/costin.9/6212/tychonoff.pdf

[quick delta]

If m is not in F, then X \ F is an open neighbourhood of m which doesn’t intersect F

(Tbh I think literally just explaining it would’ve been shorter and more enlightening than what theyve said)

[alpine nest]

Equivalently: those intersections of preimages of the O's are basic neighboorhoods of m in the product topology. If every basic neigbhoorhood of m has nonempty intersection with F, then m is in the closure of F. But as F is closed, being in the closure of F is the same as being in F.

[final token]

im trying to prove that in a first countable space $X$, $x \in \overline A$ implies there is a sequence $x_n\to x$ in $A$. the general proof i wanted to do was create a countable basis at x, and order it to be decreasing wrt inclusion, so $B_1 \supset B_2 \supset \cdots$. but im not sure if that is valid? my intuition is that a countable basis is a bunch of sets that shrinks down to be smaller than every nbhd of x. so could we wlog pick it to have a largest (and second largest and third largest etc.) set?

[gentle osprey]

isabelle

[tender halo]

it is unclear that you can order an existing basis to be that way

but you can transform any countable basis at a point into another one that is decreasing wrt inclusion

[final token]

mmm

yeah i can see that transforming an existing one wouldnt work, cause like ${(x-1/n, x+1/n)} \cup {(-n, n)}$ is a perfectly fine countable basis with no largest element

[gentle osprey]

isabelle

[final token]

but my intuituion is that the increasing sets are like, extra and can be removed no problem

[final token]

(countable basis of x=0)

[tender halo]

its not a basis at 0

or at anything really

[final token]

wait what? shouldnt any nbhd of 0 contain (-1/n, 1/n) for some n

[tender halo]

ohh i mean

sorry i read it wrong

[final token]

oh sorry yeah i probably didnt do the notation well

[tender halo]

yea its a little confusing

sure its a basis at x

[final token]

ok yeah i think i understand, a general basis isnt well-ordered wrt inclusion but you can just pick one that is

ty !!

[tender halo]

well its less pick and more transform, you have to make one explicitly i think

by taking intersections

[final token]

right yeah

mm that makes sense

like make a new basis where the nth element is the intersection of the first n from the original basis? or something like that

[tender halo]

yea exactly

[reef mural]

Hi

I can I ask a question about metric spaces here?

Okay

[gentle osprey]

Poetic justice

I working on a problem the say let x and y be distinct points on a metric space M. And I am to prove that there are disjoint open sets U and V in M such that x$\in$U and y$\in$V

[reef mural]

I'm wondering if I could use the fact that their intersection, which is the empty set since it's open and a subset of M we can show that there exists such sets U and V

Or I would have to explicitly construct a V and U instead?

[quick delta]

Can you actually write down what you mean here?
It doesn’t make sense

[crisp lintel]

You should try to construct U and V using the metric

[reef mural]

Okay. I mean using the fact that the empty set is open, and is a subset of M. And the intersection of finite open sets must be open as well. Hence there must exist 2 disjointed open sets such that their intersection is empty but they

[reef mural]

I will try and do that. Thank you

[quick delta]

There are not necessarily two disjoint non-empty open sets that their intersection is open
If you have only a topological space, it fails (and even if your metric space has only one point, it fails)

[reef mural]

Thank you.

[reef mural]

I think I should mention that there are two distinct points we know of in the metric space in the exercise, x and y

[fringe thorn]

drawing a picture might be really helpful here

[reef mural]

Thank you

[gritty widget]

Althought this is quite simple as is, it is interesting that Hausdorffness can be rephrased as “given points x, y, there is a neighborhood U of x such that y is not a point of closure of U”

[stone kayak]

Indeed this view ties Hausdorffness to unique limits closedness of compact sets and hints at generalizations in function spaces

[gritty widget]

never thought about it like that

[iron bolt]

mildly interesting thing I recently realised: usually, when people talk about "locally closed set" they mean a set A ⊆ X such that each point x in A has a neighbourhood U in X such that A ∩ U is closed in U, or a set satisfying one of many conditions that are equivalent to that

but analogous to how the right notion of things like locally connected and locally compact is "each neighbourhood of a point contains a smaller neighbourhood of that point that is compact/connected/whatever" and not "each point has a compact/connected neighbourhood", also can also happen that the property you actually need is "each neighbourhood of a point contains a closed neighbourhood"

and if you think that through, that condition is actually just equivalent to... the space being T3?

[unreal stratus]

Lol

[iron bolt]

I'm surprised I've never heard T3 spaces getting called locally closed spaces anywhere

[unreal stratus]

Isn't that a different notion though to the point where it may be confusing

Idk maybe I've misunderstood what you mean

[iron bolt]

it is different, the same way that "a set where every point has a convex neighbourhood" is different from the notion of a locally convex vector space

[unreal stratus]

Yee sure

I realise now that I have only ever thought about locally closed stuff in the context of AG where T3 never happens unless you are discrete or smth lol

[iron bolt]

the only reason it came up to me was some technical stuff I'm formalising in lean, where I had a function f and two sets A, B such that B ⊆ ∂A\A and f is "continuous at x within A" for every x ∈ B in the sense that the filter of all intersections of neighbourhoods of x with A tends to the neighbourhood filter at f(x) along f, and was wondering whether that meant that f had to be continuous on all of A ∪ B

the only proof I came up with so far requires that the space codomain of f is locally closed in this sense

[red folio]

try drawing boobs

[umbral hamlet]

Me when someone asks me what diagram demonstrates that a space is Hausdorff

[alpine nest]

In those cases I tend to draw balls

[umbral hamlet]

What about the points though

[alpine nest]

pointless topology

[radiant stone]

tfw Grothendieck topology has no points so it's really a pointless topology /j

[stone kayak]

Careful if grothendieck was alive he would put you on his manifesto

[sonic crane]

Balls

[queen prism]

open or closed?

[lucid ocean]

clopen

[sonic crane]

clopen balls

Sounds painful.

[queen prism]

well that just means it's all of space

[hidden abyss]

Are balls necessarily connected

[finite token]

Not in an arbitrary metric space

[alpine nest]

Consider the discrete space

Totally disconnected balls

[kind marlin]

is there a disconnected metric space where every metric ball is connected

[quick delta]

No
Take disjoint opens U, V
Take x \in U, y \in V
Then B(x, 2d(x, y)) intersects both U and V so is disconnected

[kind marlin]

oh right

[finite token]

is there a connected metric space where every metric ball is disconnected

[slender glen]

yes, I have an idea in my head lemme think of the correct words to type it out with

our space is going to be a subset of R2, and it is the union of two smaller spaces, the first is (R\Q) x R and the second is {(f(n),n) : n \in Z} for some bijection f: Z -> Q

[finite token]

R/Q is chosen arbitrarily with AOC?

[quick delta]

We take Y = {(x, y) \in ([0, 1]/0 ~ 1) x [0, 1] | x = 0 or y = p/q is rational in lowest form and x < 1 - 1/q}
this works

[slender glen]

wait no this is R \ Q

setminus

I'm not 100% sure this works but it feels like it does

[finite token]

surely that's not connected, its a subset of R x Q which clearly involves points from multiple connected components

[finite token]

with the metric i expect you would use this is totally bounded

so if it's connected, a large enough metric ball will be connected too

[slender glen]

you're of course correct, I meant R\Q x R

I'm sorry for being sloppy

it's like, a line at each irrational, and then an integer point at each rational

[finite token]

oh nice

yeah that sounds like it would work

is there a path connected metric space where every metric ball is disconnected

[kind marlin]

i was just typing that out LOL

[finite token]

maybe we can just do (R\Q) x R union with x=y

ah no

oh yeah that should work

[rancid umbra]

here is another one, in similar spirit.

the standard map t |-> (exp(2 \pi i t), exp(2 \pi i a t)) for a irrational has dense image in the torus.
the image is path connected since it is parameterized by the real line, but each open ball (with the subspace metric inherited from R^3) is disconnected, save the ones which completely contain the torus.
the fix is to glue infinitely many of these spaces together in a row so that the resulting space is unbounded.

[finite token]

ooh thats cursed

[rancid umbra]

lol

[finite token]

im not sure i understand how to interpret the coordinates but i can picture it

they look complex

[rancid umbra]

i guess im thinking of this space as a subspace of C^2 and R^3 at the same time

but i should have just chosen one

so we can inheret the metric from C^2 and nothing changes

like, the map is R -> S^1 x S^1 subseteq C x C

alternatively, we could just pass through an embedding of the torus into R^3, so
R -> S^1 x S^1 -> R^3

[pastel pumice]

is every discreet space hausdorff? if so why?

[crisp lintel]

well think about what a discrete space means

[pastel pumice]

every subset open

[crisp lintel]

So if you have two distinct points and want to separate them, it's enough to just separate them by any subset

in other words if x and y are distinct points in some set, can you find two disjoint subsets one containing x and the other containing y?

[pastel pumice]

sorry my keyboard broke wait

did hausdorff mean for every distinct points, you can find neighborhood for the points, such that their intersection is empty, does every neightborhood have to be this way?

shit feels a lot more different when you are

not just googling stuff and looking at wikipeida

[crisp lintel]

Not every neighborhood, just need existence

[pastel pumice]

ohhh then that makes sense

i am so afraid of saying stuff that are wrong or dont make sense, so much so that i recorded my heart rate after asking this and i got 127 bpm.

examples to non-t2 spaces?

[crisp lintel]

One is the cofinite topology on an infinite set

That's a topology where the open sets are sets where the complement is finite

[pastel pumice]

if the set is infinite, how can the complement of any subset be finite?

[crisp lintel]

well for example the subset {2,3,4,...} of the natural numbers has complement {1}

[pastel pumice]

so the complement of some, although not all, infinite subsets, can be finite?

another question, if every point of a set is isolated, does that make it discreet?

[crisp lintel]

so a point x is isolated if and only if {x} is open right

[pastel pumice]

this was another thought i had

[crisp lintel]

If all of the singletons are open, you should be able to prove every subset is open using the axioms of a topology

[pastel pumice]

an epsilon ball around x contains only x?

[crisp lintel]

In a metric space yeah that's what it would mean

[pastel pumice]

but there is no metric in topoolgy

[crisp lintel]

In topology an isolated point is just a point where {x} is open

[pastel pumice]

oh makes sense

[crisp lintel]

Which you can convince yourself is the same as the metric space definition

For metric spaces

[pastel pumice]

boundry points

yeah

anyways so if every point of a set is isolated

does that make it discreet?

[warped helm]

yes, since every single point {x} will be open

isolated gives you a neighborhood U around x such that U doesn't intersect A at any point other than x

O is open in A iff O = A \cap U for some U open in X

therefore {x} = A \cap U

[crisp lintel]

yeah we already covered that, I think they just need the last step

and for that you just need to use properties of open sets, namely that the union of open sets is open

[pastel pumice]

i am new

best video and book sources for PST?

[warped helm]

book: munkres, wilson sutherland intro to metric spaces and general topology

i think covering metric spaces first will be best

[pastel pumice]

i started by reading wikipedia articles and i know i didn't learn much.

[warped helm]

(in fact i think a good course in analysis in R and R^n should ideally come first)

[pastel pumice]

they told me there were no prereqs other than basic set theory

[warped helm]

who is "they"

and while that's technically true, i would not reccomend it

much of the techniques you use to actually prove things are probably best learnt in concrete settings

[pastel pumice]

any good video sources that are free?

[warped helm]

not really that i know of, most videos i've seen on youtube are just eh

[pastel pumice]

what is the best?

[warped helm]

a class

second to that, a textbook

if you can find a lecture series that might be good

[pastel pumice]

MIT OCW

[warped helm]

sure, that's probably good

[pastel pumice]

if my adhd let's me

[warped helm]

@pastel pumice there's also this free online textbook "Topology without tears" https://www.topologywithouttears.net/

it's not great in my opinion since it doesn't cover some "big" and intricate examples like munkres does, but it has a very easy going coverage of basic pointset material

[pastel pumice]

with a magic trick i am going to get munkres for free as well

[radiant stone]

i mean this is technically true if and only if someone has enough mathematical maturity

[pastel pumice]

do i

[radiant stone]

idk

[pastel pumice]

clearly not

[radiant stone]

how do i know lol

[warped helm]

no offense, but i dont think so

[pastel pumice]

i know

[warped helm]

but this is why i say you should check out metric spaces and perhaps analysis in R first

i just don't think "only set theory" is good advice

[radiant stone]

yeah metric spaces motivates why we should care about general ones

good examples as well

[pastel pumice]

metric spaces first ok

munkres has it?

[warped helm]

uhhhhhhhh

i would not use munkres to learn about metric spaces for the first time

[radiant stone]

get an analysis book

[warped helm]

the sections are perfectly fine, but they're a bit brisk since they assume you've seen analysis

[radiant stone]

rudin has metric topology on chapter 2 (btw i dont recommend this), and i heard abbott has these in the last chapter or something

idk about other analysis books

[warped helm]

metric spaces first, yes, but analysis on R firster

[radiant stone]

tfw proceeding with metric spaces by force

[warped helm]

really you just want to be able to draw pictures

the real line and the plane are good places to do this

n = 3 if you're a good artist or something

[pastel pumice]

R^4

[radiant stone]

n=4 if you're built different

[radiant stone]

it's okay to say wrong stuffs
you can just fix what you have in mind as you try to learn

it's very normal to think/say about stuffs in the wrong way and that's not the wrong thing to do

[austere flare]

That's what math is all about

[sonic crane]

bruh

[worn mortar]

There technically are no prerequisites other than set theory, but many examples needed in topology would be understood better with a firm background in real analysis.

One early example I can immediately think of is showing that the collection of half-open lines determined by the rational numbers, indexed by the rational numbers, forms a countable sub-basis for the real line R, showing that R satisfies the second axiom of countability.

Topology Without Tears is a good book I’d say, and if you work through the appendices (for example, working through the appendix on equinumerosity prior to chapter 2, as advised) you can get away with it.

I’ve been working through “Introduction to General Topology” by S.T. Hu, and so far not much real analysis has been needed, although some of the basic examples, such as understanding the usual topology on the real line, is needed.

Keep in mind though, that although I haven’t worked through a full real analysis book (I’ve been reading Stancl’s “Real Analysis and Point-Set Topology”) I had already worked through a good amount of group theory and basic mathematics, such as sets, maps/functions, binary relations and Cartesian products, partially ordered sets, Peano’s axioms, cardinality, quotient sets, and the Axiom of Choice and equivalent formulations of the statement, such as Zorn’s Lemma, the Hausdorff Maximal Principle, Zermelo’s Theorem, the Well-Ordering Theorem, and other statements.

Another option is “Introduction to Topology” by Mendelson, which kind of builds up ideas from analysis in tandem during the metric spaces chapter. I’m still working through this too though.

IMO, the bare minimum you should have is some familiarity with the field and order properties of the real numbers, properties of the absolute value, and understanding the Archimedean Property and Completeness Property of the real numbers. It’s also nice to have seen some construction of the real numbers, such as Dedekind cuts.

[worn mortar]

Yea Abbott goes into those details in the final chapter

[worn mortar]

The interesting thing is that this book has a whole sub-book on topological groups 😍

[swift sluice]

how can this exercise be requested at a point where the author explicitly says they have not explained how to topologise a subset of R^n using the standard topology?

I know what the subspace topology is, but when the author makes a point that they have not explained it, I don't know how this exercise is supposed to make sense

[warped helm]

have they mentioned anything about sets that are open relative to a subset A of a topological space ?

[kind marlin]

It says Euclidean metric topology, I think that’s the construction for both

[warped helm]

adding to this ^, if you take any point p on S^1 then the set of all points q on S^1 with d(p,q) < r is just a way to say B_r(p) \cap S^1

which is the sort of open set you would "expect" to see if the subspace topology were mentioned explicitly by the author

[swift sluice]

They have remarked that given a topology on X, the collection of open sets contained in a particular open set U \subset X is a topology on U, i.e. the special case of the subspace topology on A \subset X where A is open in X

Oh I somehow didn't register that the author says "with the Euclidean topology"

But this doesn't really fix the issue. At this point in the book it's not clear what "the Euclidean metric topology on S^1" means

[swift sluice]

Sorry, I don't understand what you mean

[warped helm]

A subset Y of a metric space X is a metric space in it's own right. A ball of radius R around a point p in Y is the set of all points q in Y with d(q,y) < R

[swift sluice]

Yes

Oh

So you mean that we don't know how to equip S1 with a topology, but we're expected to know how to equip it with an induced metric and what the metric topology coming from that is?

[alpine nest]

I mean, it's not even an induced metric, it's the same metric

[swift sluice]

Does induced normally mean something else?

[alpine nest]

Well, on a pedantic level it's not because the domain is different

[swift sluice]

I live on the pedantic level

[alpine nest]

Probably not, but induced still makes it sound like it's a new object somehow

Which it is but only on a very pedantic level

[warped helm]

idk how else to phrase it beyond "it's the natural thing to do" since all you're requiring is that the points that are less than R away just also live in Y

though it's weird if the author hasn't mentioned it

[alpine nest]

That is fair, ideally there would be at least a mention that if you have a metric space, then any subset of it, equipped with the same metric, can be considered a metric space in its own right

[swift sluice]

I think this is the intended interpretation

[swift sluice]

yeah

He does!

I just forgot or missed this

Anyway thanks for entertaining my pedantry everyone lol

[kind marlin]

Tbh I think it’s simpler than that

if you just think of the Euclidean metric as a metric, and apply it to points in S1 by interpreting them as points in R2, you get the metric topology on S1

You don’t really need to think of S1 as any sort of subset or subspace topology, you just apply the same basis construction to points in S1 and consider distances between points in S1

Since it tells us that the S1 points are really just R2 points, nothing here is added machinery

[normal swift]

they're the same tho

if the subset has induced topology from a metric space, its topology is induced by the restricted metric. notice how they both have "induce"

thats literally kinda the definition of a subsp[ace topology

[kind marlin]

yes

[normal swift]

like, not only homeomorphic

theyre like fundamentally the same concept

[kind marlin]

yes 😭

[normal swift]

liberal snowflake

[kind marlin]

right

[normal swift]

[kind marlin]

what you’re saying is basically what I’m getting at, subspace topologies are super natural in their construction, especially for metric topologies

[quartz horizon]

naturality mentioned?

[swift sluice]

Have I shown that the closed unit ball in R2 is a topological 2-manifold with boundary?

Put $B\coloneqq{(x,y)\in\mathbb{R}^2;:;\Vert(x,y)\Vert\leq 1}$ with the usual topology. Since $R^2$ is second-countable Hausdorff, we only need to check that every point q in B has a neighbourhood which is homeomorphic to an open subset of $H\coloneqq{(x,y);:;y\geq 0}$. Define the map $\varphi:B\setminus{(0,0),(0,1)}\to H$ by $$\phi(x,y)=\left(\frac x{1-y},\frac1{x^2+y^2}-1\right).$$ This is a homeomorphism. Hence, it is a local homeomorphism, which verifies the claim for all q except the origin and (0,1). For the origin, consider that the open ball of radius 1/2 is a neighbourhood of (0,0); translating by one unit upward verifies the claim for q=0. For q=(0,1), first apply reflection across the x-axis (a homeomorphism), then use the homeomorphism $\varphi$.

[gentle osprey]

person2709505

[swift sluice]

I should find phi^-1 and show it's continuous, but that's not hard

And the first coordinate of phi is wrong. I should replace x and y by x/r and y/r, where r is the norm of (x,y). I'm just doing stereographic projection to all the closed spheres of radius 0<r<=1

[warped helm]

for the part in 25 where i'm supposed to say that K is separable, can't I just use the result of 24?

Since K is compact, every infinite subset will have a limit point

[quick delta]

That works
You get the same proof by following the hint as you’d get by “unfolding” the proof of 24

[warped helm]

yeah pretty much\

[swift sluice]

What's that book?

[warped helm]

baby rudin

[swift sluice]

joy

[void patrol]

need a study partner on this course

[gritty widget]

Is there any practical method to determine the interior of a set?(In normed spaces Especially) In some cases is it obvious, but in other cases it's not clear at all .

[normal swift]

now that you brought it up, i suppose an algorithm could use the metric structure and keep adding open balls

to approximate

wait no thats a terrible idea

just like all my other ideas

[strong lantern]

Not really. You just kind of go by vibes.

[warped helm]

in a metric space you have that ((E^\circ)^c = \overline{E^c} )

[gentle osprey]

josemom2

[normal swift]

yea but

[warped helm]

so maybe computing the closure of the complement is easier

[normal swift]

finding a closure is just as hard as finding a interior

theyre like

dual concepts

YES I USED THE WORD "DUAL" IRL IM SO COOL

[rancid umbra]

both can be approximated by sequences

[rancid umbra]

i don't think in general there is an algorithm. but usually, there is some property or uniformity about the set you are taking the closure or interior of that you can take advantage of

maybe there is some notion of a computable or decidable set, where the notion of "having a certain property" or "having a certain uniformity" can be made precise

for those sets, there may be an algorithm

but in general, subsets of a topological space, as you suggested, can be wild and non-uniform

so asking for general algorithm seems kind of hopeless

[normal swift]

lets ask the folks in #theoretical-cs whether finding closures is undecidable

[normal swift]

by lets i meant you. im too lazy for this

[rancid umbra]

same

[normal swift]

👍

[alpine nest]

Doesn't this hold in any topological space, in fact?

[alpine nest]

Not really, but a lot of proofs about the interior use the characterization that if E is a set whose interior you're trying to find/prove something about, then x is in the interior of E if and only if there's an open set U such that x is in U and U is a subset of E (i.e. we use the characterization of interior as the largest open subset of E)

So thinking in those terms might be useful

[rancid umbra]

i suppose this is the universal property of the interior of a subset

it is terminal in the poset category of open subsets contained in E, and it is unique if it exists (which it always does). so any open subset U contained in E is contained in int E, and thus int E must be the union of all open subsets contained in E.

[alpine nest]

"An open set is contained in E if and only if it's contained in the interior of E"?

" int E must be the union of all open subsets contained in E" <- I'm used to thinking of that as the primary definition of interior

[rancid umbra]

yea, i guess i was just thinking of how i would define it in terms of a universal property

[cosmic mirage]

i was confused since terminal objects are only defined up to iso but then i remembered posets dont work like that 💀

[rancid umbra]

yea, i think topologies are a nice type of lattice, so it has some good properties here

i don't know much about lattices though

it's just that like, here, we are closed under arbitrary unions/joins

[cosmic mirage]

I mean do you even need that? like all that is being used here is that if a and b are isomorphic in a poset category then they are literally equal

well I guess you need it to actually exist, lol

[rancid umbra]

yea, we need it for the construction, but uniqueness is immediate from what you said

my fault lol

my response wasn't responding to the right thing

[cosmic mirage]

lol nw

but yeah since you have all suprema you win

slash coproducts or whatever

[rancid umbra]

lol yea

posets are funny

[gritty widget]

Like in this particular example second question , I was wondering how would I tell that it is empty if I wasn't ask to prove it

It's hard to use intuition

[quick delta]

It’s kinda like
Hard to give without a specific example
If you wanted to prove it, first you toy around with functions to see that your property is preserved under “small perturbations”, then you set up the epsilon delta mess and continue

[rugged escarp]

Interior is the largest open set

So if it is non empty then there is a point in A around which you can fit a small ball

In this small ball you can imagine that independent of whatever direction you petrub the function ( petrub so you still in the vector space) with small enough norm , you should still be in it

If it is the case you can't fit a small ball then there must be a direction to push the function so you can leave the set

[bitter turtle]

Hello, I was wondering if abstract algebra is good enough for starting point set topology. I’m reading Rafael Lopez’s book

[gaunt linden]

You shouldn't need any abstract algebra for point-set, really.

If you progress to algebraic topology, it's a different matter, of course.

[radiant stone]

just having enough mathmatical maturity, or having some backgrounds on metric topology (at least topology on R^n) is enough to start

no algebra required for pointset

[bitter turtle]

Alright, sounds good. I’m pretty sure a general analysis course teaches that right?

[radiant stone]

yes

[noble axle]

Does anyone have any information for lindelöf space theorem?

[tender halo]

what is that

[noble axle]

That's what I'm talking about, but I don't understand anything

[tender halo]

oh, this one

[noble axle]

Can you explain this in the simplest way possible so I can understand it?

[tender halo]

we take A and B, use regularity to take open covers of A and B respectively, with the property that the closures of the members of the cover of A does not intersect B and vice versa

so, we have a cover of A and cover of B, but they may intersect each other, we want to shrink them so they dont

because then we win, we will have separated A and B

S and T are new covers that have pieces removed so that they dont intersect

because any S_i cannot intersect any T_j

[noble axle]

So, using regularity we place open sets around A and B that do not touch the other set; using the "Lindelöf property" we make these collections countable; then, by inductively shrinking these sets, we obtain two open sets that separate A and B, right?

[tender halo]

yea

[crimson terrace]

Can someone please explain what this proof is trying to do?

Why isn't (b) just a direct consequence of theorem 2.2.6? If it is, what are we proving?

[somber gorge]

youre proving that taking finite intersections of the sets in the subbase form a base, as far as i can tell. the rest of the theorem should follow from 2.2.6 (i think)

[crimson terrace]

But doesn't 2.2.6 guarantee all of that?

T is the smallest topology for which f_i are continuous
which is the same as
T is the smallest topology containing the inverse image of all open (def of continuous)
by thm 2.2.6 the inverse image of all opens is a subbase for the smallest topology containing the inverse image of all opens

[rugged escarp]

I mean the point is, the theorem 2.2.6 gaurentees a sub base but doesn't tell you what it looks like

now basically the proof is trying to show how this is subbase that comes out of the gaurentee looks like

makes cents?

[crimson terrace]

I mean, doesnt 2.2.6/ tell is that any collection of subsets, U, has a smallest topology including it? and also says that given a topology T, a subset, U, is a subbase for it iff T is the smallest topology including it

what you say doesn't line up from what I understand

[rugged escarp]

this is correct

[crimson terrace]

sooo doesn't that hold

[warped helm]

its just giving a description of what stuff looks like, as clemens said

the excerpt does say “Part (b), through the subbase statement, is clear from Theorem 2.2.6”

[crimson terrace]

hmm okay but I am confused about one thing
part (b) follows directly from the theorem, which assert that the inverse images of opens forms a subbase of said smallest topology. All of this is included in the package given to us for free by 2.2.6.

But then we go on to prove that the look like the inverse image of opens

are we not getting how the subbase looks like directly? We already know, after applying 2.2.6, that they are the inverse image of opens

[rugged escarp]

bro you are overreading 2.2.6

where is evne the word inverse image or function mentioned once here

[crimson terrace]

its mentioned here

[rugged escarp]

alright so theorem 2.2.6 is somehow an abstract result. To apply it, what do you need?

A collection of subsets U of X.

What is it here? The preimage of the topology under the function

What does the theorem 2.2.6 gaurentee? We have a smallest topoogy containing the above, and know it we are in the smallest topology if U is a subbase for T

now how do we check if a topology containing this preimage set is the smallest topology containing it? We check if this preimage set is a subbase

now they use this characterization of subbase in terms of a basis

from wiki pedia
https://en.wikipedia.org/wiki/Subbase

[crimson terrace]

Okay... but you still didn't answer this

The smallest topology, T, for which the functions are continous is the same as the smallest topology that includes the inverse image of all open sets by definition of continouity.

By the iff of 2.2.6., since T is the smallest topology including the inverse image of all open sets, then the inverse image of all open sets form a subbase.

[crimson terrace]

In here, you are not using this way of writing out T

[warped helm]

i guess my question is, where exactly did you show that T is the smallest topology such that the f_i are continuous?

[crimson terrace]

where T is the smallest topology containing the inverse image of opens?

[warped helm]

the theorem is about showing it exists

[prime elbow]

So here are we identify i_1(x) with i2(x) or we identify the set { i1(x), i2(x) | x in X0} ?

[warped helm]

the idea that you get it from building up from preimages of opens is correct, but at no point did you actually specify T and verify that its what you want

idk, to me the proof is written a bit confusingly

[crimson terrace]

So me assuming that the smallest topology for which the functions are continous is the same as the smallest topology that includes inverse image of opens is not justified?

[crimson terrace]

tell me about it 😐

[somber gorge]

i dont understand your question. what do you mean by identify?

[prime elbow]

Identification

What is equivalent relation here

[warped helm]

actually, now that i look at it more clearly, theyre showing something is a subbase and then getting the topology T from that

[prime elbow]

Where i1(x) = x1 and i2(x) = x2

[somber gorge]

oops

[prime elbow]

No

[warped helm]

the topology obtained from that subbase will be the smallest topology containing it

i have no idea why they were so like

[crimson terrace]

but all of that is packed into 2.2.6.

[warped helm]

poetic about it

[somber gorge]

@prime elbow

[warped helm]

it still requires you to show something is actually a subbase though no?

the theorem is basically characterizing what the topology generated by a subbase is

[prime elbow]

So that means we are identifying i1(x) with i2(x)

[somber gorge]

yes, if im understanding you correctly

[crimson terrace]

nop?
the smallest topology such that the functions are continuous is the smallest topology which includes the inverse image of opens
by the backward direction of the iff, the inverse image of opens is a subbase for the smallest topology which includes the inverse image of opens

[somber gorge]

well actually i dont think this is quite right. because this need not be an equivalence relation. instead i think you have to take the finest equivalence relation that makes $\forall x\in X_0. i_1(x)\sim i_2(x)$ true @prime elbow

[gentle osprey]

rays

[warped helm]

then i really dont understand what the author was writing about

[crimson terrace]

I am printing the proof and pissing on it then
Thanks a lot!

[warped helm]

idk, it seems like the author wanted to do like a sanity check

since both you and the author acknowledged that it follows from 2.2.6

@crimson terrace what book is this?

[crimson terrace]

Real Analysis and Probability by Dudley

[prime elbow]

So what is exactly it is?

[somber gorge]

its the finest equivalence relation that makes i1(x) ~ i2(x) for all x in X0

oops

[prime elbow]

so what is exactly equivalent relation will be there?

[quartz horizon]

Equivalence relation generated by a relation

Are you familiar with this

[prime elbow]

ah no

[quartz horizon]

Here’s one way to think about it

Think of a relation R on a set X as a graph whose vertices are the elements of X