point-set-topology

[lofty orchid]

Worse bc my prof wasn't very good so I had to teach myself everything from Munkres

[thin scarab]

i learned topology on my own from lee

[lofty orchid]

Lee is really good, clutch before my finals which I was so scared I was gonna fail

[thin scarab]

then I also had a bad topology professor so I would literally sit in the back and work through other stuff

[lofty orchid]

I'm NOT a math major and I had plans on taking real analysis the semester after topology (class filled up quickly) and my math professor decided to make it not beginner friendly at all 😭

And idk he was really rude if I went to him w questions

[thin scarab]

I read some of sheaves in geometry and logic in that class and it was less confusing than my professor was

[thin scarab]

oh, that's the worst

[lofty orchid]

God forbid someone who's not fully versed with analysis take the class

[tender halo]

lol thats an achievement

[lofty orchid]

That's wild 😭

No when you mentioned Hausdorff I actually had PTSD bc on our first midterm my prof gave me a solid 11/50 because my proofs weren't rigorous enough (and that was never a problem during homeworks???)

Thank god he allowed me to switch it with my finals score

[thin scarab]

professors don't read homework

I know this bc I am a high school teacher and I do not read homework

I think your proofs likely weren't rigorous enough. You see proof solution paths more easily and notice more technical details with mathematical maturity

[tender halo]

they dont pay you to read it they pay you to grade it

[thin scarab]

precisely

[lofty orchid]

they were not, what I got was deserved

I just wish he'd help me get better 😭

Another prof was very happy to teach me properly thankfully

[prime elbow]

why number of distinct families B(x) is not larger than exp | B | ?

[tender halo]

because a distinct family B(x) is a subset of B of certain form

there are exp |B| of B subsets in total

[scarlet turtle]

first time i've seen exp |B| denote 2^|B|

[prime elbow]

I see, thank you

[prime elbow]

So how does this all help me?

[scarlet turtle]

help you with what?

[prime elbow]

i mean why author states this?

[scarlet turtle]

because you already know that the mapping x ↦ B(x) is an injection

[steady gorge]

i know that removing a point would break connectedness on the intervals, but i'm assuming it would also break a homeomorphism. i'm a bit unsure of how to proceed next tho

[ruby delta]

in topology, a cut point is a point in the space that, when removed, the resultant space is disconnected.
show that homeomorphisms must preserve cut points.
how many cut points does (0,1), (0,1] and [0,1] have?

or, rather, how many non-cut points do they have?

[steady gorge]

(0,1) would have no non-cut points, (0,1] only has 1 as its non-cut point, and [0,1] would only have 0 and 1, right?

[uneven bronze]

I have a basic question. Suppose f exists continuously on (0,1) and equals a function g which can be continuously extended to [0,1]. Does f = g on [0,1]?

I know that two continuous functions agreeing on a dense subset agree on the whole set, but here I'm confused, since we don't know if f has a unique continuous extension to [0,1], right?

[ruby delta]

what do you mean? if a continuous function can be extended to a point (that's connected to the domain), the continuity guarantees that the value of the function at that point has to be unique

you're essentially asking if a function is continuous at a point, must the value of the function at that point be unique, no?

[uneven bronze]

Yeah, sorry. 😅 I was quite confused by what I was reading. Indeed, I was wondering if a continuous function can be continuously extended from (0,1) to [0,1], is the extension unique?

[ruby delta]

is the codomain in R, or a general continuous function

[hollow geyser]

and equals a function g
Where? On (0,1) or [0,1]?

[uneven bronze]

the codomain is R or C

[uneven bronze]

on (0,1)

[ruby delta]

in this case it's a standard analytical proof then, no?

[uneven bronze]

Hmm, ok. Do you have a hint?

[ruby delta]

for all epsilon there exists a delta such that |x| < delta implies that |g(x) - g(0)| < epsilon

replace g(x) with f(x)

done

[untold lily]

Is that thing given in the last sentence actually equivalent to the original? I know the equivalence when the union of closed sets has non empty interior, but how do you generalize it to when the union is all of X?

[gaunt linden]

Hmm, since "open dense" sets are exactly the complements of "closed with empty interior" sets, by contraposition the last sentence seems to be saying:

... Equivalently, X is a Baire space if the intersection of a countable family of open dense subsets of X is nonempty.
It's not immediately clear to me how to bridge the gap between "... is nonempty" and "... is dense".

[untold lily]

right, dual of the difference between "union has nonempty interior" and "union is X", I'm guessing it was a typo otherwise this would be too big of an equivalence not to be stated somewhere else lol

[gaunt linden]

A counterexample to the equivalence would be the disjoint union of R and Q. This is not Baire by the usual definition (since we can remove one rational with each of countably many open dense subsets), but on the other hand, every countable intersection of open dense subsets is dense in the R part, so it must be nonempty.

[untold lily]

nice

[gritty widget]

Let T be the union of line segments connecting (1,0) to the rationals of $[0,1] \times \left{ 0 \right} in \mathbb{R}^2$. One question is to show that T is locally connected only at (1,0), so to do that I was thinking of taking some neighborhood of a point x in $T \setminus \left{(1,0) \right}$ defined by $U = T \cap B(x,r), ; r >0$ by bounding r in this (completely moronic) way $|q_k - q_n|<=r < \mu(p_n)$ where pn is the set of points constituting the path on which x is located, and mu is the 1D Lebesgue measure. Can I even use Lebesgue measure here, if so how do I justify its use ?

[gentle osprey]

Dfg

[real needle]

Is there a name for a topological space in which no singleton set is open? Or a certain property that guarantees it? For example euclidean spaces have this property and it is nice

[gritty widget]

Having no isolated points

[unreal stratus]

such spaces are "dense in themselves" i think

makes me think of Kant

[gaunt linden]

Uh... all your points have zero y-coordinate. Should the top of your fan have been (0,1) rather than (1,0)?

[tacit basin]

can i have counterexamples for if X isn't compact or if the C_i aren't necessarily closed?

hrm wait $X=(0, 1]$ and $C_n=(0, \frac{1}{n}]$ works

[gentle osprey]

Sara

[gritty widget]

Yeah

My bad

It’s (0,1)

[gaunt linden]

When you're trying to prove that T is not locally connected at x, you don't really get to choose neighborhood yourself -- in particular you cannot insist on it being ball-shaped; you need to argue that all sufficiently small neighborhoods fail to be connected. You do get to choose what "sufficiently small" means, but that can just be "subset of the neighborhood T\{(0,1)}".
Then when you're given an arbitrary open T-neighborhood U of x such that U doesn't contain (0,1), prove that U is necessarily disconnected.

[gritty widget]

Ah yes fair

Well suppose what I said was correct, how do i justify the use of 1D Lebesgue measure on a subset of R^2, provided i can even use it

[gaunt linden]

I don't really understand what it was you were proposing to use Lebesgue measures for.

Your description has some q_k and q_n popping out of nowhere, and the context in which you're mentioning mu seems to define a ball that, as I said, you don't get to choose yourself anyway.

[gritty widget]

Well nvm that

[gritty widget]

I’ll admit that I did omit the fact it was a negation in my original argument (which breaks the entire argument), but isn’t the definition of local connectedness of X "for any neighborhood U of any point x in X, there exists some connected neighborhood V such that x \in V \subset U", in which case finding a single neighborhood of a single point such that all sub-neighborhoods V are disconnected would be sufficient, and therefore in which case choosing a counterexample would work? Or am I misunderstanding everything ?

[gaunt linden]

Correct -- except V has to be open.
I may have misunderstood your sketch; I somehow got the impression that you were attempting to choose V.
For U you can be generous and pick U = T \ {(0,1)}.

[gritty widget]

No no my sketch was false, I confused the definition for its negation because of inattention

So I was in fact attempting to choose V 🥀

[prime elbow]

why do we define intersection of empty collection to be X?

[lucid ocean]

vacuous truth I guess

intersection over emptyset in the universal set X = { x in X | forall Y in emptyset, x in Y }

= { x in X | true }

= X

[opaque scroll]

Like Ludi says it's just a consequence of the usual definition.

But another reason for why it should be that way is say you have two families of sets
{A_i}_[i in I] and {A_j}_[j in J]

If A1 was the intersection of the first family and A2 was the intersection of the second family you would have that the intersection of A1 with A2 was the intersection of the combined family
{A_k}_[k in I or J]

If this rule should also hold for J empty you would need the empty intersection to be everything.

[prime elbow]

i see

thank you jagr

[tender halo]

also consider that intersection is the same as the glb in P(X)

and union is the lub

glb of an empty collection is the maximum in a poset, since every member is a lower bound

similarly lub of an empty collection will be the minumum of the poset

[prime elbow]

So here, order on P(X) is A ≤ B iff B \subset A ?

[tender halo]

no

the usual order on powerset

A \leq B iff A \subseteq B

[prime elbow]

so how every member is lower bound for an empty set?

[tender halo]

every set is a lower bound for the empty set (of sets)

because its a subset of every element of the empty set

[prime elbow]

Here you mean an empty collection of sets?

[tender halo]

yea

well its the same thing as the empty set

but just to emphasize that we live in P(X)

[prime elbow]

Vacously true, right?

[tender halo]

yes

[prime elbow]

I see

Thanks i got it

[stark veldt]

May I ask a bit of an oblivious question?

Why is topology interesting? On its own it feels very tedious, and rather dry. It all feels kinda pointless

[cosmic mirage]

i guess the answer to your question depends on what topology you have seen so far?

like given you asked in the point set topology channel i would completely understand this if youve only seen like limits/compact spaces/hausdorff spaces/T1 T2 etc spaces/bases for topologies/metric spaces/etc

in my case when I took an intro topology class I was like this is so excruciatingly boring and just happened to take an algebraic topology class due to scheduling conflicts with other interests, and now I have not gone back

[cosmic mirage]

if this is true then the answer is that it is very tedious and dry, but topology at that level isn't meant to be interesting. it is meant to give you the tools to learn and prove more interesting things

like, very very few people study point set topology things for a living, it is meant as a stepping stone to give everyone a common language and set of tools for more advanced and interesting topology

if you are in an intro class, you may end up learning about covering spaces, and that's when it gets more interesting imo. you start to analyze topological spaces with group theory which is kind of wild, and if you go deeper into this you are essentially just doing Galois theory which is crazy

and then manifolds are something i could give an undergrad whos halfway through an intro topology course, and they would understand the definition. but the study of manifolds is extremely rich and cool (according to the manifold people,, i know little about this) and badly behaved in a fascinating way in dimension 4

but for example, to access all the results about covering spaces there's many intermediate results along the way that Require being Very Careful about point set issues

[stark veldt]

I am taking my first topology class

Technically I took but I am still in exam season and feeling rather burnt up

[cosmic mirage]

very fair

[stark veldt]

I am planning on taking algebraic topology next semester

[cosmic mirage]

i will warn that algtop also can have a similar flavor but can just feel like a bunch of tools for computation, but it will be much more interesting. and that's again to give you tools to do More Fun Things later

[stark veldt]

During summer I am taking differential geometry

Alright, I'll trust the plan

[cosmic mirage]

another fun thing to think about: i assume you've learned about pi_1 then? there are pi_n's for all non-negative integers and these turn out to be very important

also if you just really hate topological spaces -- most people who would call themselves algebraic topologists spent zero amount of most of their days thinking about topological spaces

[normal whale]

everything is a topological space if you try hard enough

[cosmic mirage]

yes but i elect to ignore this hahaha

[stark veldt]

thats what my friends keep talking about? there is this popular notion that topology is abstracted geometry which counts holes of surfaces or smth like that

they arent from a mathematical background

[cosmic mirage]

yeah thats a very reductionist and incomplete view of topology, lol

its good enough if youre a topologist and someone non-mathematical asks what you do and you want them to go away

[stark veldt]

they dismiss the entire field as nonsense so it didnt work well

[cosmic mirage]

i mean, finding an explanation for what you do as a topologist that makes non-mathematicians not think its nonsense is still one of the biggest open problems in this field

[stark veldt]

incredible

[tender halo]

thats like every field from the perspective of a layman

[gritty widget]

Cuz it provides results on extremely large classes of spaces I guess

[scarlet turtle]

It all feels kinda pointless
https://en.wikipedia.org/wiki/Pointless_topology

[radiant stone]

imagine geometry without topology

[tiny obsidian]

Agree with most of what was said so far that it is rather tedious to begin with, but a very useful tool for later. Imo it also is essentially generalised geometry; we care about 'holes' (this is very vague, but is formalised by algebraic topology), 'pieces' (connectedness) and other properties that distance and angles are irrelevant to.

This allows us to generalise beyond metric spaces, and although most topological spaces we care about are metrisable (i.e. we could find a metric so that the topology induced by the metric is the same) e.g. manifolds, working with the metric is really annoying and gets in the way of these properties

[quick crane]

It's useful for applications to Analysis and PDE

[radiant stone]

literally any type of geometries use it

[radiant stone]

also this, and a lot more like everywhere-used

[rancid umbra]

it feels like the the glue for joining different areas of math, honestly

[tiny obsidian]

-# also I wanted to make the pointless topology joke after writing out my more serious answer 🗿

[rancid umbra]

-# now that's pointless

[gritty widget]

Isn’t that just like any geometry before like 200 years ago cause topology is quite recent

[radiant stone]

literally most of maths were like started to develop after 1700-1800 and later so it is not a problem

[gaunt linden]

Interesting that the article doesn't even mention Grothendieck topologies, even though they seem to have related goals.

[stoic herald]

Is a proper map from R^n to R always continuous can we say the same for R^n to R^m?

A proper map is a map such that the inverse image of any compact set in the target space is compact in the source space.

[rancid umbra]

hmm

my first instinct is to say no

[opaque scroll]

f(x) = x + 1/x
f(0) = 0
is proper in this sense

[rancid umbra]

that’s a really nice example

if the domain is hausdorff and the image is compact, then a proper map is continuous

[unreal stratus]

Ngl imo proper should always be assumed continuous anyway

Ah it seems conventions may vary

[hidden abyss]

Maps are continuous by definition anyway

[kind marlin]

topology was highkey my favorite math class 😭

i get why people find it boring but i thought the constructions were super interesting esp bc i didn't know anything abt topology going in, and it generalized/formalized so many intuitive notions for me

[zinc siren]

I think the main motivating thing through basic point-set topology is the metrization theorems. The stuff to get there is neat but not so independently motivating.

[zinc siren]

I still have not found a good way of explaining simply why the topological space definition is good. Like what is a topological space, intuitively? I still don't have a good answer!

e.g. a metric space has an easy answer

i know uniform spaces are a thing but don't know almost anything about them except for vague memories of the definition

e.g. a group (ring, field, vector space, ...) has an easy explanation.

[near egret]

If it's of any interest, there's an interpretation of topological spaces as axiomatizing semidecidable propositions

[zinc siren]

that is of interest!

[near egret]

And so the reason we care is because they encode a lot of good logical data

A good resource for this is Topology Via Logic by Steven Vickers

[zinc siren]

a quick google brings me to https://builds.openlogicproject.org/content/intuitionistic-logic/semantics/topological-semantics.pdf

[near egret]

But basically the idea is: a semidecidable property is a property we can test, which will always be verifiable in finite time but need not be upper-bounded

I.e. is this number within 0.5 of that other number?

We need to go through the decimals to check

And open sets are just sets of properties we can test for in this sense

The false and true properties, corresponding to the empty set and the whole space respectively, are obvious semi-decidable

So there's that axiom

And then we get unions because that's just testing each of them and waiting for one to come back positive, which will happen in finite time

And similarly finite intersections, but not infinite ones, because then we might not get a positive answer in finite time

[zinc siren]

but how would you get arbitrary unions, when you can only dovetail a countable set?

[near egret]

Well, this is an abstraction

With infinite resources, we could in theory check each of them all at once

It just might have to be very infinite resources

Anywho, this provided a lot of intuition for me as to why topological spaces are so weird all the time

It's because they capture logical dynamics and not neccesarily geometric ones

[zinc siren]

so i guess, idea is, topological spaces but where you only do countable unions, is the semidecidable propositions, and then you abstract to arbitrary unions

[near egret]

I'm not sure why we can't take arbitrary unions flat out?

[zinc siren]

Why is this still semidecidable?

If we have an arbitrary set of semidecidable decision problems, we need the "at least one of these is true" decision problem to be semidecidable. If it was a countable set, then I can take a machine that dovetails through the machines for each problem, halting if any of them halt.

[near egret]

Yeah, this isn't really about computation theory in that sense

Which I don't know very much about

You can imagine having as many machines as you want

Or graduate students, if you're tenured

[zinc siren]

@near egret
Imagine Merlin the unbounded agent (who can even solve the halting problem) is trying to convince Arthur, the mortal, that x is in a certain set. Arthur is capable of verifying (in the semidecidable sense) that x is in U for each U in some generating family (which of course we'll want to be a subbase). If x is in an arbitrary union, Merlin can thus convince him of this fact. Likewise if x is in a finite intersection, Merlin can convince him of this fact.

For an infinite intersection, our hapless Arthur will be unable to be convinced, for he cannot verify x in U for every U.

Thanks! It's enlightening.

Another way is a "more intuitive" way of phrasing the kurawtowski closure axioms, via the "touching" relation between points and sets.

1. No point touches the empty set
2. If x in A then x touches A
3. If x touches A or B then x touches A or x touches B

4. If x touches A and every element of A touches B then x touches B

[cosmic mirage]

oh this reminds me, related to encoding good logical information, the definition of a space makes (pre)sheaves work which is a very good property to have

although of course good luck telling someone "a topological space is the basic example of a thing you can have sheaves on" as motivation

[tiny obsidian]

I remember there's an mo post about this

ok found it: https://mathoverflow.net/questions/19152/why-is-a-topology-made-up-of-open-sets

[pallid comet]

Is there a Cantor set in [0,1]^2 that intersects every line passing through [0, 1]^2?

[crystal bolt]

Can anyone suggest a lecture series for starting metric spaces

[quick crane]

I can suggest this book: https://mtaylor.web.unc.edu/wp-content/uploads/sites/16915/2018/04/anal1v.pdf

[crystal bolt]

Thanks

Can anyone suggest a lecture series too

It will be helpful

[final sinew]

I want to do some math projects but dont have any experience.can you tell how should i start?

[pulsar lagoon]

what is the project about lol

[pale plover]

idk what d is, how do i prove ts then 😔

[cosmic mirage]

d is just the usual distance function on R

[pale plover]

as in?

😔

the |x-y|?

[cosmic mirage]

yup exactly!

[pale plover]

okay :3

[cosmic mirage]

(the notation d* probably means its the induced metric on C([a, b]) from the metric on [a, b] / R)

[pale plover]

eh i'm not sure, the textbook i'm using has only introduced proving metric spaces and continuity in metric spaces

[stark veldt]

suppose A a subset of X in a topological space and A is connected, what is the exact definition for A? does it mean there arent two open sets O1, O2 such that A doesnt equal their union or does it mean A isnt a subset of their union

i couldnt find the definition

found it

sorry for the bother...

can someone help me understand this? what does that even mean, they used the same idea elsewhere and it really bothers me. infinity cant be treated like that

[rancid umbra]

a subset A of a topological space X is connected if it is connected in the subspace topology on A

[stark veldt]

they are doing a construction for a new topological space so it is compact, hausdorf and dim(X)=0. my problem is that i have no clue what {infinity} could possibly mean

i can translate the entire problem and proof if it will help

[rancid umbra]

you can think of oo as an alias for any set not in D

[stark veldt]

what does that mean...

the set of all sets cant exist

[rancid umbra]

oo is just a symbol for an element that is not a member of D

[stark veldt]

oh ok

that makes sense

way better than what i had in mind

thats a funny construction

they manually add open sets so it becomes hausdorf

[rancid umbra]

it’s quite common to adjoin elements to a set and give the resulting set a compatible structure

[stark veldt]

yeah but they didnt say which topology they define over it

this course feels like climbing on a 40 degree hill

[rancid umbra]

is this the one point compactification?

[stark veldt]

yes

[cosmic mirage]

The intuition for the ∞ notation comes from, for example the Riemann sphere. Another way to think about it is for example R. You have open sets (a,∞) and (-∞,b), when you add a point at ∞ you get to just glue these together and you get a circle

It's the point where you need to go infinitely far away to access and when compactifying you just now can access

[untold lily]

I know that every compact subset of D(0, R) is contained in cl(D(0,r)) for some r < R

How do you show this for the product? As in the book essentially claims any compact subset of D(0, R) x D(0, R) is contained in cl(D(0,r)) x cl(D(0,r)) for some r < R

ig I've never had to look deeply at the closed sets of a product space

[tiny obsidian]

projection is continuous => preserves compactness, so you can do it coordinatewise

[untold lily]

oh, of course!

thanks

[pale plover]

need a bit of help on this

for the metric space i got $d: X \times X \longrightarrow \mathbb{R}$ defined by $d(x,y)=\max_i{d_i(x_i,y_i)}$

[gentle osprey]

parabolicinsanity

[pale plover]

same for d'

but eh i'm not sure how to show continuity

i tried something crackpot like this though

hold on i keep seeing issues with my writing 😔

[heavy breach]

theres more than one standard way lowk but it ok. if ur using this one i would consider picking the smallest delta

[pale plover]

eh okay

perhaps i'm just a bit

minted on how the thing works 😔

[rancid umbra]

i mean

[pale plover]

but this is what i wrote

Since all $f_i$ are continuous, so for a specific $k: d'_k(f_k(x_k),f_k(y_k)) =\max_i{d'_i(f_i(x_i),f_i(y_i))}$, the corresponding $f_k$ is also continuous. \
$\forall \epsilon_k >0, \exists \delta_k >0: d_k(x_k,y_k) < \delta_k \implies d'_k(f_k(x_k),f_k(y_k)) = \max_i{d'_i(f_i(x_i),f_i(y_i))} < \epsilon_k$

[gentle osprey]

parabolicinsanity

[pale plover]

then like ehh

$d(x,y) = \max_i{d_i(x_i,y_i)} < \delta \implies d_k(x_k,y_k) < \delta=\delta_k$

[gentle osprey]

parabolicinsanity

[pale plover]

or something 😔

[heavy breach]

i know its very tempting to fill a proof with quantifiers and symbols but i would recommend trying to speak things out at this stage

[rancid umbra]

do you know that the metric topology induces the product topology on the product space?

[heavy breach]

i would say thats more advanced

up to point set yes there is one standard way

[pale plover]

😔 i guess i should

[rancid umbra]

only a little bit, i suppose

but then it’s just the universal property

ig it’s instructive tho to symbol bash through it

[heavy breach]

yeah, its also a useful analytic exercise

its good to get intuition for this before the abstraction

its really more analysis than top tho

at least thats when i did stuff like this

[rancid umbra]

yea, i think that’s subjective. this one can be solved both ways

[pale plover]

smallest delta works 😔

thanks

[rancid umbra]

question for you guys that i thought was cool:

Let X be a top. space and p a non-isolated cut-point of X. Given a neighborhood V of p (a set containing an open set containing p), is V - p connected?

[lucid ocean]

since p is a cut point, exist disjoint complementary open sets A, B subset X \ {p}, then A inter V, B inter V subset V \ {p} is a disjoint complementary open set pair separating V - p?

I played around with a few toy examples and I couldn't find one that failed (khalimski line and subsets thereof, intervals, etc.)

so at least for common examples it holds true

[tender halo]

why isnt V contained in A or B

[lucid ocean]

then it wouldn't be a neighbourhood of p?

if V subset.eq A or B, then V subset.eq A union B = X \ {p} so in particular, p not in V

[tender halo]

V \ {p}, sorry

[lucid ocean]

ah okay that's true

hmm, I'm not sure how to continue

but if a counterexample exists, it must have that property of V \ {p} subset A or B

[rancid umbra]

i stumbled upon the k line too when trying to work out this problem. very cool example

the result is true

i prefer the answer by ulli, but both are good. i couldn’t come up with an argument in time ha

[lucid ocean]

that is pretty cool ❤️

[rancid umbra]

yea, right!

glad you thought so too

[lucid ocean]

it'd be really weird if it were false

being a cut-point should feel a bit like a local property

[rancid umbra]

yea

[lucid ocean]

this is also a problem that looks deceptively simple

at least to me, you had to set up some machinery to deal with it

[rancid umbra]

yea

i completely agree

very easy to understand, but a bit tricky to see the nuance at first glance

[tender halo]

let (X,d) be a metric space and d' be another metric on X defined by d' = d/(1+d). then show that (X,d) and (X,d') have precisely the same open sets.

a hint said they have the same open balls with one exception. but I don't see this... how can the open balls be same?

[ruby delta]

the metric is not the same, but the set of all open balls given by the metric is the same

for instance on R, there's the standard metric d(x,y) = |x-y|. I can make a new metric d' = 2d(x,y), and although it is a different metric the open balls I get are the same as the ones I get from the standard metric, just not necessarily with the same center and radius

[tender halo]

x/(1 + x) is injective and monotone and continuous and tends to 0 as x tends to 0

which i think is all you need to prove equivalency of two metrics

[fair thorn]

Is it not all open sets? Claiming same open balls feels too strong to me idk

[lucid ocean]

the open balls generate the topology so that should be fine

[tender halo]

same open balls because you can translate the radius in one metric to a radius in the other

[quick notch]

Is the set of all maximal points in a Priestley space always a closed subset of the Priestly space?

[fossil hill]

I think not necessarily. Consider the cantor space C=2^ω, and fix a point c∈C. Equip C with the order such that for any distinct x,y∈C-{c}, we have c<x and x,y are incomparable. The set C-{c} of maximal points is not closed because the point c is not open.

Suppose x≰y. This implies x≠c, so there exists m,n such that x(m)≠c(m) and x(n)≠y(n). (Possibly m=n, that's fine.) Then U={z|z(m)=x(m) and z(n)=x(n)} is clopen, x∈U, y∉U and U is an up-set because c∉U. Since C is compact, this makes C a Priestley space.

I'm going off Wikipedia's definition of Priestley space.

[quick notch]

OK thanks

New question: under what circumstances is the set of maximal points in the Priestley dual of a distributive lattice a closed set?

[main pulsar]

#help-24 message could probably go here

[prime elbow]

Can I say if F is closed set in topology X, then F = bd(A), for some subset A of X, where bd(A) = cl A intersection cl A^c

cl A is closure of A

[scarlet turtle]

no

equip any large enough set X with the discrete topology. then {x} is closed for any x ∈ X

[unreal stratus]

Ye lol

[scarlet turtle]

but cl A ∩ cl(A^c) = A ∩ A^c = ∅ for any A ⊆ X

(i initially wanted to say [0, 1] ⊆ ℝ is clearly not the boundary of some set but i don't actually know how to show this)

edit: this is why you don't carelessly use the word "clearly"

[unreal stratus]

Me too, but it actually is a boundary — consider [0,1] cap Q

[scarlet turtle]

ahh

[unreal stratus]

I do wonder if it is true for all subsets of R, but I assume not

E.g. cantor or smth

But maybe same argument works there aha

[scarlet turtle]

it feels like the cantor set is the boundary of itself, no?

[prime elbow]

It is true for all subset of R^2, when R^2 as usual Topology

[prime elbow]

Yes but why did you say large enough set X, it is enough X has to be at least two elements

[scarlet turtle]

i was just being careful but i just need X to be non-empty

[scarlet turtle]

how do you show this?

[prime elbow]

I don't know, actually the original question is to show it is true for R^2, then I wondered if that is true for arbitrary topology?

[unreal stratus]

oh nice

[tiny obsidian]

It suffices to show any closed set has a dense subset with empty interior

[scarlet turtle]

following what @unreal stratus did above, i'm guessing the boundary of bd(F) ∪ (int(F) ∩ ℚ²) should do the trick, for any closed set F ⊆ ℝ²

but i'm outside rn so i can't check the details

[lucid ocean]

what are the closed sets that are a closure of some proper subset

[tender halo]

non discrete sets i imagine

[prime elbow]

R is a closed set and closure of Q

[lucid ocean]

yes, I mean is there anything that can be said in general

[prime elbow]

generally it cannot be true, because we can take discrete topology

[lucid ocean]

what cannot be true, I never stated any claim

[unreal stratus]

lol

[prime elbow]

i mean in discrete topology you can't find a closed set which is closure of some proper subset

[unreal stratus]

I do wonder when F = bd(F \cap Q^n) or smth lol

like ||cl(F \cap Q^n) = cl(F) \cap cl(Q^n) = F, and cl( R^n \ (F \cap Q^n)) = R^n since R^n \ (F \cap Q^n) contains R^n \ Q^n||

intersect and you get F

[prime elbow]

i found something on mse

[prime elbow]

any hint?

[scarlet turtle]

try showing it's sequentially closed, i.e. if x₁, x₂, x₃, ... ∈ ∪C_λ are such that xₙ → x ∈ X, then x ∈ ∪C_λ. because being closed is equivalent to being sequentially closed in any metric space

i don't want to hear any yap about the axiom of choice

[prime elbow]

so say x_n is convergent to x, then x_n also cauchy so for eps ( which is given ), there exists m in N such that for all n,k>= m, d(x_n, x_k) < eps, so given condition implies x_k in some C_lamda, for fixed lamda, otherwise there will be contradiction, and since C_lamda closed so x will be in C_lamda so it will be in union, right?

[tender halo]

C lambda is locally finite

for obvious reasons

the end

[prime elbow]

i have to see what is the definition of locally finite, i forgot

[scarlet turtle]

yep

[tender halo]

every point has a nbhd that only intersects finite amount of elements

union of a family of locally finite closed sets is closed

[smoky sigil]

I doing a proof about a relation on R. I have to prove something about an arbitrary point in R^2. What I did was start the proof by saying let x be in R^2. x=(x1, y1) for some x1, y1 in R. I’ve seen other people start the proof by saying let (x1, y1) be in R^2. I’m a little confused because the first approach we say “for some” x1, x2, and the second approach x1 and x2 seem to be arbitrary. I’m not sure how both of these could be correct? Thank you!

I’m getting confused because the first approach says “for some” which means “there exists”. Since “there exists” wouldn’t x_1 and x_2 not be arbitrary in the first approach

[tender halo]

they wouldnt be, the x is arbitrary, x1 y1 are just the "unpacking" of the x

the "there exists" is referencing the fact that all elements of R^2 have a certain shape - namely, being pairs of elements of R

the second approach just starts with picking out arbitrary x1 y1 because thats how cartesian product works i guess, taking an arbitrary point of the cartesian product is the same as arbitrarily choosing each coordinate

[smoky sigil]

So are both approaches proving the same thing just different ways?

[tender halo]

they are different (equivalent) ways to arbitrarily choose an element of R2

[smoky sigil]

Ohhh ok. I guess I’m having trouble seeing they are equivalent because of how the first approach the coordinates aren’t arbitrary but the second approach they are

[prime elbow]

Yes i proved this

[tender halo]

so you just need to prove that c lambda is locally finite

[prime elbow]

Thank you

[prime elbow]

Yes

[restive vortex]

If it's not a bother, if you're an undergraduate student, can you send me your point set topology worksheets, my dm's are open and I would really appreciate it

[opaque scroll]

I'm not an undergrad student, but here are some topology work sheets for undergrads
https://wiki.math.ntnu.no/tma4190/2024v

[restive vortex]

Thank you that is exactly what I am looking for, I want to get as many different worksheets as possible from different universities!
You have been of great help

[prime elbow]

thanks for sharing, were you the tutor for this course?

[opaque scroll]

I was not

[prime elbow]

okay

[umbral hamlet]

this is the first question im asked in the first topology course ive taken ever so i wanna make sure im getting this right

the union of all such U is A and by the definition of the topology that would mean that A is in the topology and is thus open on X

[gaunt linden]

Yes.

[ruby delta]

this is a good lemma to keep in mind, by the way

it's very useful for proving that sets are open

[umbral hamlet]

i know we used it pretty often in real analysis 1

its pretty interesting how much topology drives the interesting results in analysis

[ruby delta]

the basic ones... yeah sure

it divereges after a bit

[umbral hamlet]

i see

it was an introductory course to be fair

so thats all the exposure i have thus far

im taking analysis 2 now so we will see how much topology overlaps there

seems like its a lot from what my prof says but i imagine once i get to measure theoretic analysis itll be different

[prime elbow]

i don't get it, linearly ordered 'distances from x' have been replaced by the partially ordered nhoods, and what is the closeness in metric space? yes in topology, it is not necessary that if y is in every nbhd of x then x has to be nbhd of y, so what is the notion of closeness in metric space?

[queen prism]

first time seeing "nhoods"

[ruby delta]

if U <= V are both neighbourhoods of x, then points in U are closer to x than points in V

a lot of the ideas from topology are motivated from metric spaces

1 is closer to 0 than 2

so natrually (-1,1) is inside (-2,2)

[prime elbow]

me too

[prime elbow]

how?

[ruby delta]

considering the standard metric on R

[prime elbow]

yes

i see

got it

[prime elbow]

are these important exercise/

[gaunt linden]

4E looks somewhat important -- it's part of showing that topology can also be axiomatized by making neighborhood filters instead of open sets the core concept.

4C and 4D are more "here are some funny but not individually important examples to practice your understanding of the concepts with".

[prime elbow]

okay

[prime elbow]

so they define C = { (x,y,z) | max{|x|,|y|,|z|} = 1 } which is the cubical surface of side 2 centered at the origin, so it is cube, right?

[gaunt linden]

Sounds right.

[prime elbow]

any hint to show every local homeomorphism is an open map?

say f: X\rightarrow Y is local homeomorphism

now i am taking U is an open set in X

[unreal stratus]

Show the more general claim that being an open map is local (on the source), like can be checked on a base for X

[tender halo]

there is like 1 thing to try

[prime elbow]

i know for every point of X, there exists open nbhd V such that f(V) is open and restriction of f to V is homeomorphism f(V)

[prime elbow]

what?

[prime elbow]

okay

[tender halo]

just do the obvious thing and itll work

[prime elbow]

so you are saying if f is an open mapping and continuous then f is local homeomorphism?

[tender halo]

no, they are saying that if a map is an open map locally (every point has an open nbhd such that the restriction to that nbhd is open) then its open in the usual sense

[prime elbow]

let U is an open set, now for any x in U we have open set V_x such that f(V_x) is open in Y, and f restricted to V_x is homeomorphic mapping to f(V_x).

since U intersection V_x \subset V_x also open in V_x, so f(U intersection V_x) is open in f(V_x), because restriction is homeomorphism.

since f(V_x) is open in Y it implies that f(U intersection V_x) is open in Y.

since f(U) = f(union over x in U, U intersection V_x ) = union over x f(U intersection V_x), and later one is open in Y, so f(U) will be open in Y.

is it correct?

[unreal stratus]

Yes this. And then any local homeo is "locally open" as homeos are open

[prime elbow]

is that one correct?

i used the same idea that restriction is open map

[unreal stratus]

It's good. Though when you say restriction is a homeo I guess you mean is open

[prime elbow]

yes in definition of locally homemorphism restriction is homeomorphism but yes we just need open map

[unreal stratus]

Oh I guess you mean like

"The restriction is a homeomorphism"

It sounded to me like you were saying the restriction map is a homeo or smth

[prime elbow]

yes this is the definition

thank you for the help, bussy beaver and potato

[prime elbow]

so i am proving that manifold's definition is equivalent to a, so if x is a point in X, and it has nbhd U ( here they define nbhd as open set) such that U is homeomorphic to open set E of R^n.

so since E is open set in R^n, we can find the open ball A inside in E, and then take preimage of that ball under that homeomorphism it give us that there is open set V in U which is homeomorphic to A and U is open so V is open in that topology.

so now we have open nbhd V of x which is homeomorphic to open ball in R^n.

other direction is trivial
is it correct?

V is that preimage

[umbral hamlet]

this is my proof that given a family of topologies ${\mathcal{T}\alpha}$ on a set $X$, the unique smallest topology on $X$ containing each $\mathcal{T}\alpha$ is the topology generated by the subbasis $\bigcup \mathcal{T}_\alpha$

is this a good proof?

[gentle osprey]

hiidostuff

[prime elbow]

Proof is correct but I want to address something, First why you wrote element of P(X) as x, it creates confusion, if it is set then denote as capital.

Second, you said if T is any such topology containing all such T_alpha. So it contains the intersection of all finite collection of elements of all T_alpha, here for me it is not clear that if U and V belongs to different topology so whether their intersection belong or not, ofcourse they belong but it is not clear from your proof.

And third, yes you assumed if there exists topology which is smallest then it will be equal to T_U, but don't you think you can directly show T_U is smallest because you already show any topology which contains all T_alpha it will contains T_U, and you can show T_U contains all T_alpha, so T_U will be smallest

[umbral hamlet]

idk what V ur referring to on ur second comment

[rancid umbra]

T is a topology containing all T_a, so of course, if U and V are elements of T, then U \cap V is in T. this also holds in the case that U is in T_a and V is in T_b, since T_a and T_b are subsets of T

[umbral hamlet]

for the third comment, the entire last step is to specifically show that the smallest topology is unique

the middle steps show that it is lesser or equal to every other topology

but the goal is to show there isnt some other smallest topology thats also lesser or equal to

[rancid umbra]

i don’t understand your last step. your second to last step is that T_U is contained in any topology T which contains all T_a.
if there is any other topology T’ such that T’ is a subset of T for any topology T containing all T_U, then T_U and T’ are equal.

btw, there is some basic category theory going on here. T_U is initial among topologies containing T_a.
you are proving the universal property of T_U

[umbral hamlet]

yeah i mean thats essentially what i was writing

maybe it goes without saying but i guess i dont wanna assume too much

[rancid umbra]

you don’t need \tau

like, i think you are overcomplicating the last step

[umbral hamlet]

gotcha

[rancid umbra]

for uniqueness

[umbral hamlet]

ill rewrite it then

[rancid umbra]

yea, was just pointing it out

[prime elbow]

yes i know, but it is not clear for me from his proof

[rancid umbra]

its tautological

what kind of explanation are you looking for?

[prime elbow]

all T_alpha is not clear, if we write \bigcup T_alpha then it is clear

[prime elbow]

same explanation, just little bit clear proof

[jovial badge]

Given two topological spaces $X$ and $Y$, is the statement "$A\subset X$ is homeomorphic to $Y$" talking about the subspace topology of $A$?

[gentle osprey]

struct ∅ {};

[jovial badge]

Hence if A isn't open in X, the statement can still hold?

[ruby delta]

yes!

[ruby delta]

for instance consider the euclidean topology on R.
Then [0,1), which is not open, is homeomorphic to [0,infinity)

[barren glacier]

hey guys topology question here, take the space of continuous functions from a compact interval [a,b] to [a,b], then using the sup metric, the space of functions that are no t surjective is open

I used contradiction to solve it, can yall give a direct one?

I feel like I'm missing something

[scarlet turtle]

i don't really know what you mean by a direct proof. what if you just took any continuous non-surjective function f : [a, b] -> [a, b] and "wiggle it a bit" wrt your metric?

because it's continuous and non-surjective you know that sup f < b or inf f > a (or is this the point which you want to avoid because it uses LEM?), so just wiggle it a tiny amount in the appropriate direction

[tender halo]

just take the radius leq max(d(sup f, b), d(inf f, a))

[full inlet]

Anyone have recs for introductory topology textbooks?

[gritty widget]

Topology & Groupoids, or Munkres

[full inlet]

hm I’ll have a look - thanks!

[queen prism]

I didn't know about the first one, sounds cool

[slender glen]

I can vouch for Kroom

[umbral hamlet]

this check was a lot and i just wanna make sure i didnt miss anything

and also that my reasoning is correct

[full inlet]

Ok, ty!

[scarlet turtle]

what's the topology of $\mathbb R_K$?

[gentle osprey]

GoslingGang

[umbral hamlet]

its the topology generated by the basis of all sets of the form (a,b) along with all sets (a,b) - K with K = {1/n} for all n \in {1,2,3,...}

[scarlet turtle]

you might want to double-check if $\mathcal T_5$ contains $\mathcal T_3$. How would you write $\mathbb R \setminus {0}$ as a union of sets of the form $(-\infty, a)$?

[gentle osprey]

GoslingGang

[umbral hamlet]

ah

then would it be that T3 contains T5?

maybe i got that switched around on accident

itd be for the same reason that T1 contains T5

no that cant be either

[scarlet turtle]

is the complement of the open interval $(-\infty, 0)$ finite?

[umbral hamlet]

of course not

so then the two topologies are not comparable

[scarlet turtle]

mhm

[umbral hamlet]

ok im somewhat convinced now that both T3 and T5 dont contain any of the listed topologies

T5 cant contain T1 because no elements of T5 have finite measure

but T1 obviously has such elements

same with T2

we just showed why T3 isnt contained in T5

T4 is the same case as T1

maybe measure isnt the right verbage as i havent quite learned about it yet

but thats how im intuitively thinking about it

no sets with infima can possibly be in T5

[scarlet turtle]

the intuitive way you're using it (for the lebesgue measure) seems fine by me. but yes, the word "measure" is a technical term

you haven't misused it though

[umbral hamlet]

cool

im excited to learn about that but thatll not be for another year

topology is pretty awesome anyways so im content with where im at

[scarlet turtle]

yeah measure theory is great

[scarlet turtle]

so, if i'm getting this right,
$\mathcal T_2 = {,(a, b) : a < b,} \cup \left{,(a, b) \setminus \left{\frac{1}{n}\right} \ : \ a < b \text{ and } n \in{1,2,3,\dots},\right}$

is this what you mean?

[umbral hamlet]

yes

munkres calls it the K-topology

but it seemed strange

[scarlet turtle]

sorry, i mean T_2 is the topology generated by the set on the right-hand-side

[umbral hamlet]

yeah i assumed so

but yes thats the basis for the K-topology

[scarlet turtle]

then yes, by definition, T₁ ⊆ T₂.

but also

are the sets of the form (a, b) \ {1/n} also open in T₁?

[umbral hamlet]

yes

[scarlet turtle]

so you can actually do better than this

[umbral hamlet]

ooh i see

T1 and T2 are the same topology

[scarlet turtle]

yep!

[umbral hamlet]

its hard to make concise the union of sets im looking for

im trying to say its the union of open sets (x_i, x_{i+1}) with x_i being elements of the infinite partition of a and b with the elements that arent a and b being all K-elements between a and b

[scarlet turtle]

hang on, infinite partitions?

[scarlet turtle]

do you mean
so, if i'm getting this right,
$\mathcal T_2$ is generated by the basis $${,(a, b) : a < b,} \cup \left{,(a, b) \setminus \left{\frac{1}{n} : n \in {1,2,3,\dots}\right} \ : \ a < b,\right}$$

[gentle osprey]

GoslingGang

[scarlet turtle]

?

because this is different to what i wrote before

because now you can look at the set
(-1, 2) \ {1, 1/2, 1/3, 1/4, ...} and this is no longer open in the standard topology (look at the point 0)

[umbral hamlet]

yeah i know a partition is definitionally finite

but what im thinking is

take the set (a,b) - K for some a,b

then thats equal to (a, x_1) union (x_1, x_2) union ... union (x_n, b)

where x_1 through x_n are all distinct elements of K between a and b

now that list could be infinite but were taking unions so thats ok

[scarlet turtle]

im not following tbh. do you mean if we have (a, b) \ {x₁, x₂, ...} but in such a way that we can still write it as an infinite union of open intervals?

[umbral hamlet]

Yes

But that set were subtracting is a subset of K

[gritty widget]

What in the world is this from anyway

Who in their right mind uses T_1-T_5 in topology for something that isn’t a separation axiom lmao

That’s like using pi as some other functor than the fundamental group

[tiny obsidian]

Well it is mathcal so distinct

[scarlet turtle]

i feel like this isn't too bad, since i often see \mathcal{T} for topologies anyway

[scarlet turtle]

and there isn't really any chance of confusion here, just like how you wouldn't confuse π when used for projections vs. π when used for the fundamental group

[gritty widget]

True

[umbral hamlet]

They arent using T on its own

They're doing \mathcal{T} I just didnt wanna type it out

[stable vapor]

Just began reading topology and groupoids myself

[gritty widget]

One of my favorite books!

[gritty widget]

Icl I wouldn’t notice the difference without being told anyway

I need new glasses

[hexed steppe]

imo using this notation for separation axioms is bad style

better to just write the adjectives

[gritty widget]

True

[full inlet]

lmk how it goes!

[stable vapor]

Yeah sure. Liking the presentation so far.

[tender halo]

they mean different things though

usually unless stated otherwise normal/regular/completely regular does not imply T1

Tychonoff does imply T1 though usually

[lucid ocean]

yup, and unless if you enjoy writing hausdorff a lot more often, probably should just stick with the Ti's

[hexed steppe]

?

[hexed steppe]

writing T_2 is really bad

the others are excusable i guess

but everyone knows what hausdorff means

[lucid ocean]

personally I always prefer using the Ti's

and by hausdorff, I mean "regular hausdorff", "normal hausdorff", "completely normal hausdorff" and "perfectly normal hausdorff" for T3-6

[hexed steppe]

just makes it harder to read imo

[lucid ocean]

it's way easier to read for me

[hexed steppe]

generally when writing the goal is to help the audience understand

[lucid ocean]

it also works out in the favour of the Ti's because they're all ordered by logical implication, so T6 => Ti for i in {0, 1, 2, 2 1/2, 3, 3 1/2, 4, 5} for example

[lucid ocean]

I guess we'll agree to disagree then, because I find it easier to understand with the Ti's

[hexed steppe]

im pretty sure most mathematicians would only know the definition of T_2 off the top of their heads

maybe T_1

[kind marlin]

is there a topology where (0,1) is compact

i think im overthinking this

oh i guess the indiscrete topology would work, but are there any infinite topologies (as in, an infinite number of open sets) where the space could be compact?

for some reason i cant find this online

oh wait does the subspace topology of "closed intervals" from R, like [a, b], work

hmm but then (a, b) intervals are still open bc they're unions of [ ] intervals, and u can clearly make an infinite union without finite subcover

ugh this feels like an obvious thing to prove

[quartz horizon]

Yes

[kind marlin]

can you give the example or hint towards it

specifically one with an infinite number of open sets

[quartz horizon]

Transport of structure

[novel acorn]

it has like 3 elements

[kind marlin]

like 3 open sets?

[novel acorn]

yesh

[kind marlin]

.

[novel acorn]

ohhh

[novel acorn]

okay I have one with a countable number of elements

[quartz horizon]

(Did you understand my hint)

[novel acorn]

and pseudo's hint is good enough

[kind marlin]

No but I’m googling LOL

[novel acorn]

think about shifting things left or right

oh wait

my example fails

hm okay let me think

err no it doesn't okay I think we're good

yeah it has a countable number of elements and you shift stuff left and right to get it

actually wait

it might not have a countable number of elements lol

god cardinality is weird to think about

[quartz horizon]

Did you figure it out?

[gaunt linden]

There's the trivial toplogy too.

Though if you want compact Hausdorff you'll need something like Pseudo's suggestion.

[plush folio]

How about taking the euclidean topology only on [1/4, 3/4], then adding (0, 1/4] and [3/4, 1) as open sets?

[gaunt linden]

(The key point is that there exists a bijection between (0,1) and [0,1]).

[kind marlin]

this is finite still though

[kind marlin]

hm this feels mostly related to what im asking so ill stick with that

[kind marlin]

this works! but i think maybe compact hausdorff is more interesting

[tender halo]

i mean do you want it to have some kind of interplay with the structure of the real numbers

or is the question essentially "are there compact spaces of cardinality c"

(with choice) enumerate all real numbers by ordinals from 0 to 2^\aleph_0 included, take the order topology, its hausdorff and compact

[gaunt linden]

Ah, I didn't notice that condition, sorry.

[tender halo]

alternatively cantors set straightforwadly has power 2^omega

[gritty widget]

(0, 1) is in bijection to [0, 1]

You can define a topology on it via this bijection

Which would be a homeomorphism under the topology

[opaque scroll]

I guess if you want a topology that is easy to describe just in terms of (0, 1) (like a bijection with [0, 1] is slightly annoying) could be

The open sets are the usual ones, except if an open set contains 1/2 it must also contain an open ball around 0 and 1.

This would just comes from a bijection with (0, 1) and a figure 8

[gritty widget]

That’s pretty creative dang

[kind marlin]

I’m getting a lot of tags sorry if I didn’t respond to anyone but I appreciate all the input ! I will try to construct one myself before looking at the solutions

[kind marlin]

whoa i love this

i ended up just doing a bijection construction but that’s so elegant omg

it almost feels like a compactification the way I’m visualizing it

Is there a name for this kind of thing, it feels very generalizable :0 like if you had the open unit disk i think you could just fold the edges up and inward to connect to a concentric circle?

[gaunt linden]

What Jagr did could be dignified by describing it as "take the one-point compactification, and glue the new point at infinity to 1/2 so we don't need any new points".

if you had the open unit disk i think you could just fold the edges up and inward to connect to a concentric circle
That will work too -- you could say that you take the closure of your open disk in the ambient R^2 and then glue the new points to a circle that you already had.

[kind marlin]

that makes sense, is gluing a formal/existing notion? and is this something you could do for any topological space (the way compactification can be)? I’m trying to think of more abstract examples but I’m not sure what would be interesting here tbh

I think I’m understanding gluing point A to point B to be “any open set containing A must also contain an open set containing B and vice versa” but that seems like a bad definition since 0 and 1 aren’t actual points in our space, is there a better / existing notion for this

[gaunt linden]

The formal textbook term for gluing is "quotient space", but speaking of "gluing" points together is very common.

[quartz horizon]

yeah there's a notion of quotient topology

[kind marlin]

ooh okay

I think we skipped that chapter when I took topo so this makes sense LOL

I will read it now

[unreal stratus]

I feel like gluing usually refers to gluing different things together I.e. forming a coproduct then quotienting

Eh okay I guess this was a different use of gluing here lol

[gaunt linden]

Hmm. Perhaps. But if you just have "a torus is a square with the sides glued together like this", then there's no coproduct involved.

[kind marlin]

Oh quotient topologies came before compactness and I’m probably not smart enough to make the connection myself 😭 I will maybe do some googling for a bit or see if they talk about this at all in the book later on

[gaunt linden]

In these examples, compactifying came first, by adding additional points to the space, and the quotienting/gluing was just used as a way to brute-force satisfy the (somewhat unnatural) condition that we want a space with the same underlying set as what we started with.

[kind marlin]

ohh wait that makes perfect sense

that answers all my questions tyy

[kind marlin]

oh wait just to check in with this, does transport of structure refer to defining the open sets of a new space by the images of the open sets in the old space under a bijective mapping (basically creating a homeomorphism by construction)?

i think that was the method i was able to get to by myself

[gaunt linden]

Yeah.

[kind marlin]

okay awesome

[pallid hill]

how useful is this

im taking topology

[queen prism]

it's pretty important

[rancid umbra]

useful for what?

[pallid hill]

Like for anything

[gaunt linden]

Pretty much everything in postgraduate math either depends on topology to even state its central questions, or at least has important arguments that use topological methods.

(With the possible exception of algebraic number theory, but that might just be because I don't know it well enough to have come across the essential applications of topology there).

[unreal stratus]

Things from topology are very important in alg number theory

[gaunt linden]

Well, there you go. :-)

Oh wait, p-adics count as algebraic NT.

Yes, plenty topology there.

[unreal stratus]

Does combinatorics use topology much? I suspected that to be the "least topological" pure math I can think of

[unreal stratus]

Yes, completion are super important, plus the Galois group has a topology and NT has lots of geometry in general

[gaunt linden]

Hmm right, that might be a better exception.
Graph theory shares at least some terminology with topology (such as connectedness, and I think even in technically matching ways if we view a graph as a CW-complex with no 2-cells). But that's not particularly deep.

[unreal stratus]

Yeah and it does depend on what one counts as combinatorics lol

Like there are definitely more geometric parts of it for example

[gaunt linden]

We did come across an incidental connection in math-discussion earlier: counting the possible preorders of a given finite set turns out to the the same as counting possible topologies on it!

[hexed steppe]

is there an explicit bijection?

also can you link to the relevant messages

[hexed steppe]

how do you know the converse?

like every preorder encodes a topology

it sounds true tbh

because it encodes the convergent sequences

[gaunt linden]

The open sets are the upwards closed ones. (Or downwards, depending on which way you encode the preorder).

Since the set is finite, to be a topology we only need binary unions and intersections, and upwards/downwards closed sets easily satisfy either.

(In fact, now that I think about it, it looks like upwards closed sets even in an infinite preorder qualify as a topology).

[hexed steppe]

ig im wondering how you know it’s a bijection

[gaunt linden]

At this point, just vibes and intuition.

[hexed steppe]

like this should be equivalent

fair enough

[gaunt linden]

But it doesn't feel like it should be hard to prove that the two conversions are inverses.

[hexed steppe]

yeah

i asked partly bc i saw a lot of oeis discussion so was curious if that was used

[gaunt linden]

The OEIS results were what made us aware that the same sequence represented "number of preorders" and "number of topologies".

Hmm. We can also reinterpret "x is in the smallest neighborhood of y" as "y is in the closure of {x}".
Unless I'm mistaken, that means the same preorder->topology correspondence generates the Zariski topology on an affine scheme from the inclusion order of prime ideals. (But in the infinite case not every topology is generated by a preorder this way).

(I might be mistaken, though. It's late and I'm half asleep).

[prime elbow]

So here the underlying set is the same?

What is meant by disjoint union?

[prime elbow]

Okay so the underlying set is not the same

Here they define X as the disjoint union of X_alpha, why? Why can't we write X as union of X_alpha?

https://math.stackexchange.com/questions/2928350/definition-of-disjoint-union-spaces

So the disjoint union of X_alpha means the set { (x, alpha) | x in X_alpha }

[gaunt linden]

You can think of it intuitively as "this definition only works if the X_alpha are already disjoint; if they are not, you'll have to rename some points in ond or more of them to make them disjoint first".

The (x,alpha) thing is just a systematic way of doing that rensming proactively everywhere just in case there’s overlap, but it's not how one would think about the resulting space.

[prime elbow]

Oh

[prime elbow]

I think I got what you mean, so if they are not disjoint we can just rename some elements

[gaunt linden]

Precisely.

[prime elbow]

And rename doesn't change topology

So is it final topology?

[hexed steppe]

nice

[prime elbow]

Oh co product topology

[prime elbow]

I want to show that a topological space is a 0-manifold if and only if it is a countable discrete space.
I don't want a solution.

I want to understand what it means by 0- manifold?

I know manifold is a second countable, Hausdorff topological space with locally Euclidean dimension n, so when n = 0, what does it mean?

Every point in X has a neighborhood which is homeomorphic to what?

[gaunt linden]

Each point must have a neighborhood homeomorphic to R^0.

R^0 is the set of all lists of real numbers of length 0.

Therefore it contains exactly one element, namely the empty list.

[hexed steppe]

alternatively it’s the zero dimensional real vector space

[prime elbow]

So every point has neighborhood homeomorphic to discrete topology with one element

[gaunt linden]

Yes.

[gaunt linden]

Right. I assumed more explanation than that would be needed.

[prime elbow]

So it is clear that it is discrete space but I want to show countable

[gaunt linden]

Your definition of manifold sounded like it requires second countable.

[prime elbow]

Yes it required, i got it

[prime elbow]

Btw is there any other definition of topological manifold?

[pulsar lagoon]

some people require paracompactness but it turns out to be the same in the case of second countable haussdorff spaces

[gaunt linden]

It's not universal to require second countabilty.

[prime elbow]

I see

[gritty widget]

Is there any application or even just somewhat interesting use of looking at certain filters on spaces containing neighborhoods, perhaps where the complements are "small" in some topological sense too, say, nowhere dense? I don't really have anything more clear than this in mind, just thought it could hypothetically be interesting in some cases, maybe, for example, something to do with some notion of topological dimension, because for example, when I think of, say, sorting R^2 into "small" and "large" subsets, the most obvious way is lower dimensional = small, higher dimensional = large, although this obviously isn't a filter, it has a similar intuition behind it.

[prime elbow]

I want to show that a locally Euclidean space is first countable.

Let x in X, now we know that there exists an open set U in X such that x in U and U is homeomorphic to R^n.

Since R^n is first countable so U is first countable, that means if I pick any element y in U it has a countable neighborhood system.

So it has a countable neighborhood system around x, say \cal B is a countable collection of open sets in U such that if any other open set V in U contains x then there exists an element B in \cal B such that B \subset V.

Since U is open so every element B in \cal B is open in X.

Now if V is any open set in X containing x then V intersection U is open in U containing x, so there exists an element B in \cal B such that B \subset U intersection V \subset V.

Hence, \cal B works in X, so X is the first countable.

Is it correct?

[uneven bronze]

Does this theorem have a name? I'm currently looking for a text where I may have encountered this theorem before.

E.g. in Folland's text?

[prime elbow]

I don't get it, that purple arc can be written as { exp(iy) | a < y < b }

[gaunt linden]

Yes, but what of that?

[prime elbow]

I don't know how I can write in that form

This condition always holds for any two topologies we can take U as an empty set so U will be contained in every open set .

If we add "non empty" condition then also it is not necessary that T = S

[rotund halo]

exp(it)=cos(t)+isin(t) is often the definition of exp(it) (or equivalent to it).

[prime elbow]

I know this definition but I want to understand how I do write an arc in that form, in interval form?

[rancid umbra]

i don’t understand what you are confused about

[gaunt linden]

Do you agree that each point on the arc can be written as e^it for some t?

[prime elbow]

Yes

[gaunt linden]

And for neighboring points on the arc you can choose t's that are next to each other?
(Handwaving alert!)

[prime elbow]

Sorry, i don't get it

[gaunt linden]

I'm pretty confused about what confuses you. Do you inagine the t values that do produce points on the arc are randomly intermixed with those that don’t?

[plush folio]

interesting exercise 🤔 I guess the point is that two metric spaces M, N induce the same topology if you can nest an open ball of M inside an open ball of N and vice versa, but the same thing is not true for open sets in general

[prime elbow]

I know every exp(it) lies on the circle and every point of the circle can be written as exp(it)

[gaunt linden]

But you don't see that the t values you need make up an interval? Hmmm....

[prime elbow]

Yes it is not true in general, but I am not sure about basis elements

[prime elbow]

Yes I have to show that t makes an interval

[plush folio]

I don't think it's true for basis elements either, because you can just take the topology itself as basis

[prime elbow]

I mean I am not sure about open balls

[gaunt linden]

I'm sorry I don’t understand why that is even difficult, so I find myself unable to help...

[plush folio]

the open balls part is true, see https://en.wikipedia.org/wiki/Equivalence_of_metrics

[prime elbow]

No problems i will try again, thank you for help

[gaunt linden]

I mean it seems obvious to me that the endpoints of the interval must be the two t such that e^it are the endpoints of the arc, and you need exactly the t values between those endpoints.

[prime elbow]

Here, it needs that for any x in X and for any B'(x,r) there exists B(x,s) \subset in B'(x,r).

But say there is B(y,s) \subset B'(x,r) but how can I say that ball B(y,s) contains x?

[prime elbow]

Yes

[plush folio]

uh, you can't, but I'm not sure why you need to say that. What are you trying to prove?

btw, are you sure you want to have two conversations at the same time? 😅 You can focus on Tropo's explanation if you want

[prime elbow]

Say we have those conditions, for any open ball B(x,r) in T there exists open ball in S contained in B(x,r) and vice versa

[prime elbow]

I mean yes the endpoint of intervals must have those t such that exp(it) are on end point on arc

[gritty widget]

For a counterexample, even to “a non-empty subset such that …”, consider the two following topologies on R:

Define the topology T on R by taking as a basis the set of all lower-closed intervals in R, e.g. [a, b) for a≠b in R.

For S, take the regular topology on R.

Clearly, every set of the form [a, b) contains a set of the form (x, b), for some x, simply take the midpoint of a and b.

Similarly, (x, b) contains some point c between them(unless it’s empty), so that [c, b) is open in the T and is contained in the open set

[prime elbow]

we can do discrete set also

[rotund halo]

Why not intersect S^1 with the closed ball? The intersection will be compact and try and muck about with the argument function? It is continuous except on the real axis. If S^1 lies fully inside the other circle then the intersection is just exp[[0,1]]. If not wlog let the point outside the other circle lie on the real axis (by rotating). Now take the arg of the intersection and apply evt and ivt to recover the interval you're looking for.

[gritty widget]

If T is discrete, every singleton is open. If S has a non-empty open set contained in every singleton, S should be discrete too, no?

[rotund halo]

Now for the open ball intersection just exclude the endpoints of whatever interval you get.

Realistically, I'm thinking even all of this is too much.

The proof is written in a way where it seems like they expect you to be comfy enough with analysis/geometry to just see this.

[prime elbow]

let X = {1,2,3,4}, take T = { {1,2,3},{1,2}, empty set, X}, S = {empty set, {1,2}, {1,2,3,4} }

i said discrete set

[gritty widget]

I was trying to think of an example with rational-endpoint closed intervals earlier; that might be more illustrative too

Let T be the topology generated by letting [a, b] be a basis element for all a≠b in Q. Then, clearly, no open set in T has an irrational endpoint that it contains, although (x, y) will be open for irrational x and y, [x, y] will not.

Let S be the topology defined by letting [x, y] be open for all x≠y in R that is not in Q. Then, clearly, no open interval has rational endpoints which it contains. Yet, obviously, every closed rational interval(open in T) is contained in some closed irrational interval(open in S) and vice versa

[gritty widget]

What do you mean by “discrete set”?

[prime elbow]

finite set

[gritty widget]

Ah

[prime elbow]

see this one

[kind marlin]

i think you are just defining discrete weirdly

[prime elbow]

my mistake

[hexed steppe]

i think you mean “comfy enough to supply a proof without much effort”

[gritty widget]

No I think “discrete” usually refers to finite sets(or countable ones) in like idk discrete mathematics, it’s just that I’m used to the topology-usage of it. This isn’t the first time I’ve heard “discrete” used like this, fwiw

[hexed steppe]

you have to be able to make these things rigorous

[rotund halo]

Yeah fair

[gritty widget]

Hmm, what about if the containing is made strict(so that one cannot have the same open set in both as an example of the hypothesis)

[hexed steppe]

no it basically always refers to the discrete topology when used in a technical sense

[kind marlin]

discrete meaning countable i agree with

[hexed steppe]

no

you shouldnt

[gritty widget]

Wait shoot, then X would have to be strictly contained, which isn’t possible

Nvm

[kind marlin]

i mean im okay with that 😭

[prime elbow]

oh

[kind marlin]

moreso than finite

[hexed steppe]

theyre both equally incorrect

[gritty widget]

Nvm it wouldn’t even make sense that way 💔

Fun exercise tho

You got any more?

[hexed steppe]

why dont you just consider the discrete topology on a two point space

and a topology consisting of three elements

[gritty widget]

Too simple

[prime elbow]

i have one exercise but i am not sure is that fun or not

[gritty widget]

Post it and we’ll see

[prime elbow]

so question is can Q be complete metrizable? means is there any metric distnace on Q such that it induce topology as subspace topology on Q of R as standard topology but it is complete

[kind marlin]

i think context matters but yes this is what i meant originally, i do think when we enumerate things we typically are viewing them in a discrete way

[hexed steppe]

isnt a countable complete space discrete

[kind marlin]

but obviously yes discrete and countable are independent concepts

[gritty widget]

What’s complete mean again

[tiny obsidian]

Cauchy (which is a metric property) sequences converge

[hexed steppe]

no sorry

[gritty widget]

Can’t you use the Baire category theorem here

Cause Q ain’t baire

[rotund halo]

I have a fun one if you want?

[gritty widget]

Sure!

[prime elbow]

counterexample ?

[hexed steppe]

a limit

of a sequence

i mean a convergent sequence

[prime elbow]

i see

[hexed steppe]

is a complete non-discrete space

[prime elbow]

i can't

[prime elbow]

yes

[rotund halo]

Fix a subset A in an arbitrary topological space. Prove there are only at most 14 possible sets you can produce by repeatedly taking closures and/or complements of A. Find a space where the maximum number of 14 is achieved.

[hexed steppe]

how do you do this without baire

[gritty widget]

Darn

[rotund halo]

This is called the kuratowski closure/complement problem. Idk if it's useful anywhere but it's vaguely nifty.

[gritty widget]

I mean you could just slightly alter the proof of BCT for complete metric spaces and call it something else..

[prime elbow]

Yes I know it is in Munkres

[hexed steppe]

yeah

[prime elbow]

[rotund halo]

Oh yeah, this one is one of my favs.

You can find some nice relations on closure/complement to apply.

[gritty widget]

anything less laborious?

I might do this later

[rotund halo]

Then you can make a kind of simple combinatorial arg based on something like a cayley diagram or whatever from algebra.

[prime elbow]

Can you prove there are infinitely prime numbers using topology?

[gritty widget]

I’ve seen the proof but I don’t exactly recall it

Which topology do you give N again?

I forgot 💔

[prime elbow]

Arithmetic

[rotund halo]

I'm kinda also curious where the kuratowski problem actually started popping up in books

P sure it's in engelking and also kelley

(Besides munkres)

[gritty widget]

I might go pick up Counterexamples In Topology, skip to the last pages, look at the chart listing all the spaces and their properties, and just try to solve some interesting unlisted properties as an exercise.

I should start a group chat doing this to divide the labor so we could get it all done in a reasonable time frame.

[gritty widget]

Or, alternatively, maybe try to prove the listed properties without looking at the given proof? Probably easier

[uneven bronze]

I'm confused. A cylinder set at level $n$ in $\mathbb{R}^{\mathbb{N}}$ is a set of the form $A\times\mathbb{R}^{\mathbb{N}}$ where $A\subset \mathbb{R}^n$. Now it is claimed the product topology is the smallest topology that contains the cylinder sets. I have learnt that the product topology is the topology generated by sets of the form $\pi_{\alpha}^{-1}(U_i)$ where $\pi_\alpha$ is a projection onto the $\alpha$th coordinate, and a basic open set is therefore $\cap_1^n\pi_{\alpha_j}^{-1}(U_j)$. \

What confuses me is that cylinder sets seem to start from the very beginning, i.e. $$U_1\times U_2\times \mathbb{R}\times\mathbb{R}\times\cdots,$$whereas for basic open sets in the product topology we may have $$\mathbb{R}\times U_1\times U_2\times \mathbb{R}\times\cdots.$$ Are these two topologies that I seem to distinguish the same? Why?

[gentle osprey]

psie

[hexed steppe]

R is an open set

[uneven bronze]

ah ok. So you're saying $$\mathbb{R}\times U_1\times U_2\times \mathbb{R}\times\cdots$$ is a cylinder sets. Ok, makes sense. I get it now I think. The basic open sets in the product topology (defined via projections) contain the cylinder sets, and conversely, the cylinder sets contain the basic open sets. So the cylinder sets are a basis for the product topology. Thanks!

[gentle osprey]

psie

[hexed steppe]

the definition of cylinder set you gave is also not correct afaict

i guess A is supposed to be open

in which case you can just write A as a union of sets of the form U_1 x … x U_n where the U_i are all open subsets of R

and hence it suffices to check that A xR^N where A is of the form U1 x … Un is in the product topology

[prime elbow]

Actually I don't get it

[opaque scroll]

Let's do N instead of Z.

X = N u {n* : n in N}.

The topology is discrete on N and indiscrete on the star marked N.

Then f(n) = n-1, f(0) = 0*, f(n*) = (n+1)* is a bijection

So open sets in X are any subset of X that either contain all the *-numbers or none of them

[weary temple]

Subspace of a 2nd countable space is 2nd countable, right?

[gritty widget]

Well, what’s the definition of a second countable space?

(This isn’t rhetorical I’m just making sure it is second countable base right?)

[weary temple]

Yes

[gritty widget]

Right

So

The subspace topology on a subset A of, say, a second countable space X, is just the topology who’s elements are intersections of open sets of X with A

Say U is an open set of X, so that UnA is open in A. U is then a union of base elements from the countable base, so equivalently you can simply take intersection of each of those base elements with A individually, then apply the union operation

Can you think of any way to make a base on A from that information, which is countable?

[prime elbow]

So in this setting f is continuous but f^-1 is not, because the inverse image of N under f^-1 is N u {0*}.

Thanks jagr

[gritty widget]

Let R be the set of real numbers with the topology generated by the base of all sets [a, b] where a<b are irrational. Then, clearly, this is a finer topology than the regular real numbers, and is obviously not connected. The main difference I can find so far between this topology and a similar one for a, b rational is that this space is not second countable.

Proof:Let W be a countable basis for this space. Then, by the assumption of countability, there is some irrational number p such that no element of W begins nor ends at p, e.g. neither [p, a] nor [a, p] is an element of W for any a. Then, clearly, [p, a] is an open set, yet cannot be a union of elements of W, as all of them that contain p do not begin in p, so it must contain some numbers before p, otherwise it would begin at p.

Is this argument correct? Anything too hand-wavey or not clear enough?

[gritty widget]

It is, however, still separable, because any interval still contains a rational

[tough scroll]

It's correct

In other words, for each $a \in \mathbb{R} \setminus \mathbb{Q}$ there is $B_a \in W$ such that $a \in B_a \subseteq [a, a+1]$, then it's straightforward to show that $B_a \neq B_{a'}$ for $a \neq a'$ so $|W| \geqslant 2^{\aleph_0}$

[gentle osprey]

Adayah

[gritty widget]

This part I got wrong tho, since I just now realized the base for the topology I mentioned earlier is actually a subbase, so each irrational is isolated

[umbral hamlet]

I have to prove that for functions f: R to R, the eps-delta definition of continuity implies the open set definition

im struggling a bit tho so i think i need some guidance

so far i have that given an open set O of R such that $O \cap f(\mathbb{R}) \neq \emptyset$, let $x_0 \in R$ such that $f(x_0) \in O \cap f(\mathbb{R})$. Then there is an open interval $(a,b)$ such that $f(x_0) \in (a,b) \subset O$.

[gentle osprey]

hiidostuff

[umbral hamlet]

Then we have by assumption that for any $\varepsilon > 0$, there is a $\delta > 0$ such that $x \in (x_0 - \delta, x_0 + \delta) \implies $f(x) \in (f(x_0) - \varepsilon, f(x_0) + \varepsilon)$

[gentle osprey]

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

[umbral hamlet]

From here im pretty sure this implies that every element $x$ of $f^{-1}(O)$ lies in some open interval and thus the preimage of O is the union of elements from the basis generating the standard topology on R and so the preimage of O is open

[gentle osprey]

hiidostuff

[gritty widget]

I’ll list some (somewhat easily provable) equivalent definitions of continuity if that helps

1. F:X to Y is continuous iff for every subset A of X, and point x in X, x in cl(A) implies F(x) in cl(F[A])

2. The preimage of a closed set is closed

3. Given F:X to Y, if for every open subset U of Y, and a point p in U, the preimage of U contains an open set which contains the preimage of p

[rancid umbra]

take an open set U around f(x0) for x0 in dom(f).

since U is open, there is an 𝛆 > 0 such that the interval (f(x0) - 𝛆, f(x0) + 𝛆) is contained in U. see if you can go from there.

[umbral hamlet]

i see

Suppose that f(x) is in that open interval for some x in R. then by the eps-delta definition of continuity, x in (x0 - delta, x0 + delta) for some delta

each x in the preimage of U is thus in some open interval so the preimage of U can be given as the union of open intervals and is therefore open

wait that last part isnt true at all

[rancid umbra]

well, the epsilon-delta definition gives you a 𝛅 > 0 such that if x is in the interval (x0 - 𝛅, x0 + 𝛅), then f(x) is in the 𝛆-ball around f(x0).

[umbral hamlet]

right

so i worded it wrong

[rancid umbra]

yea

you need to show that f^-1(U) is open

[umbral hamlet]

i shouldve said that there exists a delta such that for each x in (x0 - delta, x0 + delta), f(x) in (f(x0) - eps, f(x0) + eps)

[rancid umbra]

so, the 𝛅-ball around x0 should be contained in f^-1(U)

yea, exactly

and the since x0 in dom(f) was arbitrary, then ur done

[umbral hamlet]

i see

the last thing i said was moreso incomplete

[rancid umbra]

mhm

[umbral hamlet]

its that since each x in the preimage of U can be found in an open set entirely within the preimage, the preimage is the union of open sets

nice

[rancid umbra]

ye

[umbral hamlet]

thank you

[prime elbow]

Part a, it is not true, right? Because if we take A = Q and X = R, so bd Q = R but bd R is empty

[prime elbow]

Any hint for 1?

Part a

[prime elbow]

For a, if I take the fixed point Topology on R with fixed point 0, then if A = {0}, and X = {0} and Y = R, so A is precompact in X because cl A in X is {0}, and cl A in Y is R, which is not compact in Y

[opaque scroll]

You can try to think about which spaces you know are homeomorphic to (0, 1)

[prime elbow]

R

But completeness is not preserve by homeomorphism

[opaque scroll]

Yeah, that's what the exercise is showing you

So (0, 1) is completely metrizable, even though the usual metric is not complete

[prime elbow]

How (0,1) is completely metrizable?

[opaque scroll]

Because R is complete and (0, 1) =~ R

[prime elbow]

We need metric d on (0,1) which is complete and topology induced by d is exactly the same standard topology on (0,1)

[opaque scroll]

Yes, you just use the metric on R, moved onto (0, 1)

[prime elbow]

So if f: (0,1) -> R is homomorphism then we have to define a new metric d on (0,1) such that d(x,y) = | f(x) - f(y) |, right?

[prime elbow]

Part a, correct? Any hint for part b?

[opaque scroll]

Part a looks good.

For part b it might be useful to first prove that the in a Hausdorff space, any compact subspace is closed.

And a hint for that could be ||prove that every point in the complement of your compact set is in the exterior||

[prime elbow]

Yes I proved that compact subspace is closed in Hausdorff space

[opaque scroll]

So what does that say about the closure of A in X as a subspace of Y?

[prime elbow]

But closure of A in X already closed in X

[opaque scroll]

I'm not sure what you're trying to say

[prime elbow]

What's your question?

[opaque scroll]

I don't understand what you're saying
"But closure of A in X already closed in A"
Is this a question or a statement? Did you mean to say closed in Y?

[prime elbow]

Statement

Given: cl A \cap X is compact in X, right?
And we have to show that cl A is compact in Y

[opaque scroll]

Yes, so
cl A \cap X is a compact subspace of Y

[prime elbow]

Yes

So cl A \cap X is closed in Y

Because Y is Hausdorff

And cl A \cap X contains A and it is closed implies cl A = cl A \cap X

So cl A is compact in Y

Thanks @opaque scroll

[random mirage]

hi guys, does the "inclusion function" mean that j(a)=a

like is that implied from this statement?

[gritty widget]

Yep

[random mirage]

Gotcha thanks

[near egret]

Super quick sanity check: a map f : X -> C vanishes on x iff it vanishes on y, where x and y are topologically indistinguishable, right?

[kind marlin]

is this saying "if x and y have the same neighborhoods in X, then f(x) = 0 iff f(y) = 0"?

[near egret]

No need to answer now tho

It is trivial, I was just being silly

[kind marlin]

tbh i dont think i understand 😭

there's no condition on f, right? or are you saying f is continuous or something extra o_O

[gritty widget]

Usually “a map” in this context does mean continuous

[kind marlin]

that makes sense

[prime elbow]

For part a,

Since the parabola can be represented by {(x, ax^2 + bx+c ) | x in R, a ≠0 } and line can be represented by {(x,mx+c) | x in R }.

Now i showed that parabola and line are homeomorphic by mapping (x,x^2) -> (x,mx).

And i also showed that every parabola is homeomorphic to parabola y = x^2.

Now I am thinking that if f and g are continuous functions on R, then isn't that the graph of f and graph of g are homeomorphic by (x, f(x) ) -> (x, g(x) ) ?

[cosmic mirage]

yes that is correct

[prime elbow]

Thanks

[prime elbow]

Any hint? Isn't the definition of normal space?

[latent lily]

Let $(X,\tau)$ be a topological space and let $D\subseteq X$. Prove that $\overline D=X$ if and only if $D$ has nontrivial intersection with every nonempty element of $\tau$.

[gentle osprey]

Sour Drop

[latent lily]

any hints to start with this? \overline D is the closure of D

[mellow basalt]

Take G in tau nonempty. Then take an element of G, what can be said about it?

[latent lily]

well, this element g in G is also in X, so g \in G\cap X. by hypothesis, X is an arbitrary intersection of closed sets containing D. so g lives in every one of these closed sets containing D too?

[mellow basalt]

Not necessarily. I'd try proving that if g in cl D then for every open set containing g, it's intersection with cl D is nonempty

[prime elbow]

Do you know the equivalence definition of closure , if x in cl A then every non empty set containing x must intersect with A.

[unreal stratus]

Nonempty open set

[prime elbow]

Yes, my mistake

[gritty widget]

Hint:This is weaker than normality

[latent lily]

no, but i imagine this wouldn't be too difficult to prove from scratch?

[prime elbow]

yes it is not so difficult, and it should be non empty open set instead of non empty set

[prime elbow]

I showed cl C intersect with bd A, it is clear from the image that C should intersect with bd A, any hint?

No

I showed that cl( A intersection C) intersect with cl ( A^c intersection C)

[opaque scroll]

Notice that
C is the union of
A intersect C
and
(X-A) intersect C.

Since C is connected these cannot both be open in the subspace topology

[prime elbow]

Say A intersect C is not open in C, so there exists x in A intersect C such that x in cl (A^c intersect C).

[opaque scroll]

Indeed, and what kind of points live both in A and the closure of A^c

[prime elbow]

I can say if x in A and closure of A^c, then x is in bd A

I got it

Thanks jagr ❤️

To show, if p : X ->Y be a quotient map. Show that if each p^-1{y} is connected, and if Y is connected, then X is connected.

Say X = A u B, where both A and B are non empty disjoint open sets.

Now pick a in A, so a \in p^-1{p(a)}, since p^-1{p(a)} is connected in X so it is contained in A or in B, so it must be contained in A. And a \in p^-1{p(a)} implies A = union p^-1(p(a)), a runs over A.

Similarly for B.

Since A is union it implies C = union p(a) is open in Y, and D = union p(b) is open in Y.

Since A and B are disjoint so C and D are disjoint and Y = C u D, it contradicts that Y is connected. Therefore, X is connected.

Is it correct?

[prime elbow]

Connected components of lower limit Topology on R are singleton sets, right?

[graceful jewel]

Yes ! I think in the Sorgenfrey line (ℝ with the lower limit topology), every basic interval [a, b) is both open and closed, thus the space is zero-dimensional and totally disconnected.

[prime elbow]

So the path component is also a singleton set

[graceful jewel]

Exactly ! Since path components ⊆ connected components

[prime elbow]

Yes

Thanks

[silent knot]

{Path⊆Connected}→True

[silent knot]

p:X→Y, quotient map
∀y∈Y, p⁻¹(y) connected; Y connected ⇒ X connected?
A,B open, disjoint, A∪B=X, A≠∅, B≠∅
a∈A ⇒ a∈p⁻¹(p(a))
p⁻¹(p(a)) connected ⇒ p⁻¹(p(a))⊆A
A=⋃p⁻¹(p(a)), a∈A
B=⋃p⁻¹(p(b)), b∈B
C=⋃p(a), a∈A
D=⋃p(b), b∈B
A∩B=∅ ⇒ C∩D=∅
Y=C∪D
Y connected? ∴ Contradiction
X connected

[silent knot]

$(X,\tau)$ Topological Space, $D \subseteq X$. Prove: $\overline{D}=X \iff D \cap U \neq \emptyset$, $\forall U \in \tau$, $U \neq \emptyset$.

Proof:
$\overline{D}=X \iff (D^c)^o = \emptyset$.
$(D^c)^o = \bigcup {U \in \tau : U \subseteq D^c }$.
If $\exists U \in \tau$, $U \neq \emptyset$, $U \subseteq D^c$, then $D \cap U = \emptyset$.
Contrapositive: $\forall U \in \tau$, $U \neq \emptyset$, $D \cap U \neq \emptyset \implies (D^c)^o = \emptyset$, thus $\overline{D} = X$.
Converse: If $\overline{D} = X$, then $(D^c)^o = \emptyset \implies$ there exists no $U \in \tau$, $U \neq \emptyset$ such that $U \subseteq D^c$. Hence $\forall U \in \tau$, $U \neq \emptyset$, $D \cap U \neq \emptyset$.

Therefore, $\overline{D}=X \iff D \cap U \neq \emptyset$, $\forall U \in \tau$, $U \neq \emptyset$.

[gentle osprey]

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

[prime elbow]

any hint? i want connected space which is not locally connected

[ruby delta]

try and make the connections non-local

there's a construction you can make in R^2 that is sufficient

[prime elbow]

How do I think of that?

[prime elbow]

So R^2\ (Q × Q) is connected, can I do something with it?

[ruby delta]

there are simpler examples that you can work with

[prime elbow]

R^2{0,0}?

[ruby delta]

that doesn't work

[prime elbow]

Yes

So what is the general idea to make non locally connected?

[ruby delta]

the idea is that you want two parts of the space that are arbitrarily close to each other, but their connection is outside the neighbourhood

[prime elbow]

Their connection is outside the neighborhood, means?

[ruby delta]

far away

[prime elbow]

Can you explain it by drawing?

[ruby delta]

no because that spoils the answer

lol

[prime elbow]

Okay I will try it

And one more question

Any hint to show totally disconnected space are Hausdorff

[ruby delta]

connected implies that you can't seperate them with open sets

so totally disconnected means that everything can be separated with open sets

[prime elbow]

https://math.stackexchange.com/questions/1240439/totally-disconnected-space-that-is-not-t-2

[ruby delta]

ah, more double point shenanigans

well that's good to know

[gritty widget]

Here’s a hint for a different example of a space:Think of a space which has a totally disconnected subset, which is just the space minus a point, such that the space is connected, e.g. the point “connects” the space. This is called having a dispersion point iirc

[prime elbow]

R has totally disconnected subset

[gritty widget]

Sorry if I wasn't clear, I meant that removing a single point makes it totally disconnected

That's what I meant by "which is just the space minus a point"

[prime elbow]

different question: locally path connected means give me any point x, give me any open set U of x, we will get open set V containing x, V is path connected and V is contained in U