point-set-topology

[opaque zodiac]

But S satisfies WP

I'm wanting a model of HoTT + LEM + ~WP to stare at

that's my ultimate goal

passing to sheaves over a space generally kills LEM expect for really special spaces

[dim perch]

ah

I don't really know how to interpret LEM inside of an ∞-topos tbh

[opaque zodiac]

technically you need some inaccessible cardinals to get a model of HoTT at the end, but that should be completely irrelevant to the topos stuff

[opaque zodiac]

every subobject of a 0-truncated object is complemented

Does that make sense?

a lot of internal logic stuff is confusing but LEM is a really strong assumption for toposes

you know what there might not be a "grotty" ∞-topos model of this

hmm

hm

[dim perch]

hmm

yeah I'm at a loss for examples, if as you say it's too much to ask in sheaf toposes on a space

I think presheaf toposes ought always by hypercomplete

[umbral panther]

Is an infinite product so explicit? There’s an even simpler topos in Dugger-Hollander-Isaksen.

But I prefer 6.5.4.9: the Hilbert cube. Specifically, the dualizing sheaf on the Hilbert cube. This is a sheaf we should care about, but it is not hyper complete. Specifically D(U) = H_*(Q,U)

[opaque zodiac]

What is a dualizing sheaf?

Well the Hilbert cube is also an infinite product though?

[dim perch]

Do you happen to have a section # in dugger-hollander-isaksen for the example?

[umbral panther]

Yes, it’s an infinite product don’t think it’s any worse than the simplified example

I think Lurie created it by starting with the Hilbert cube. I’m not sure what DHI were thinking

[opaque zodiac]

Yeah I'm certainly more comfortable with the Hilbert cube

but I was just questioning your question about the other example being 'explicit'

[umbral panther]

A.10

[opaque zodiac]

Is there a nice characterization of the spaces X for which Sh(X) is hypercomplete?

[umbral panther]

The dualizing sheaf of X is the presheaf of chain complexes that assigns to an open set U the relative singular chain complex (or cone) C_*(X,U). If X is a manifold, this is a Thom construction

[opaque zodiac]

https://tenor.com/view/jfk-clone-high-i-like-your-funny-words-magic-man-jack-black-gif-18659433

[umbral panther]

You ask about homotopy sheaves, which seem to me a lot fancier than this construction. Maybe it’s not obvious that it’s a homotopy sheaf

[umbral panther]

A locally finite dimensional space, such as a CW complex. Or the Zariski spectrum of a finite dimensional ring. But not the etale topos

[opaque zodiac]

Is that a characterization or just some examples?

[dim perch]

does locally finite dimensional mean locally of finite cohomological dimension? or is this an imprecise thing

[umbral panther]

Just some examples, but there isn’t a lot of room between locally finite dimensional and infinite dimensional

[umbral panther]

Yes, cohomology dimension

[dim perch]

this makes sense

[umbral panther]

I don’t know what “locally” means that makes CW complexes locally finite dimensional

[opaque zodiac]

uhh actually if you go looking there is, kind of

there's 'strong infinite dimensionality' which things like the Hilbert cube has

but there are other infinite dimensional topological spaces that don't have this property

[dim perch]

Oh I just assumed you meant covered by a coproduct of finite dimensional boys

[umbral panther]

The obvious interpretation is an open cover, but that doesn’t include all CW complexes

[umbral panther]

Is that a technical term?

[opaque zodiac]

In the dark forgotten art of topological dimension theory, yes

I don't have my copy of Engelking with me

A topological space is strongly infinite dimensional if it can't be written as a countable union of zero dimensional subspaces

The Hilbert cube has this property

[merry geode]

How is theory of dimension this difficult
I am sorry. Don't mind me..

[umbral panther]

A countable union of **finite ** dimensional spaces?

[opaque zodiac]

Honestly topological dimension theory is one of those things where you look at the definitions and you think 'there's no fucking way this is going to go anywhere' but then it actually ends up going really far

[opaque zodiac]

that's equivalent

for one of the definitions of topological dimension

uh they're all the same for metrizable spaces iirc

A space is actually n dimensional iff it can be written as a disjoint union of n+1 zero-dimensional subspaces

so like R is the rationals U the irrationals

but yeah topological dimension theory and continuum theory are the two like 'classic' topology topics I'm sad people don't talk about that much anymore

because they're really not like froofy set-theoretic topology stuff but they also really don't fit cleanly into AT (cuz not CW-complexes)

(okay well I do actually like set-theoretic topology, but somehow it doesn't quite feel the same)

[umbral panther]

If you want a paper combining algebraic topology with continuum theory
https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-11/issue-2/Characterizing-k-dimensional-universal-Menger-compacta/bams/1183552180.full

[opaque zodiac]

oh cool now I just need to actually learn any AT at all

[prime elbow]

Let X be a topological space. If A is subspace of X and B is disconnected subspace of A then it is disconnected subspace of X.

Is it correct?

[opaque zodiac]

What is 'disconnected subspace'?

[gentle girder]

probably a subspace of X that is disconnected under the subspace topology

[opaque zodiac]

As in just not connected?

[gentle girder]

yes

my best guess

[opaque zodiac]

The subspace topology is the same either way

Another interpretation is clopen subspace

[prime elbow]

Metrizable space has the first axiom of countability because we have for any x in X, { B(x,r) where r in Q }, right ?

[opaque zodiac]

yes

[gentle girder]

sorry for ping i will respond in a sec.

[tawny cosmos]

Prove that the area metric is not equivalent to the Hausdorff metric on the set of all bounded polygons, but is equivalent on the set of all convex bounded polygons. The area metric is the are of hte symmetric difference of two sets A and B. Otherwise, the area of the Union of A and B minus the Intersection of A and B.

What I’ve done so far is to use the openness criterion to find balls in each metric but couldn’t formulate a equivalence from a point in the area metric to their Hausdorff distance other than the fact that the only property of two sets in the area metric that have something to do with distance is the intersection of A and B which requires A and B to be relatively close together.

[merry geode]

Hmmm

This is quite unique one!

Let me see

[tawny cosmos]

yep

[sick cape]

if you have a delta ball around a bounded convex polygon in the hausdorff metric, and you look at a point exterior to the polygon, you can find an epsilon small enough such that the point must be delta close to the polygon in order for the area difference to not exceed epsilon

[tawny cosmos]

Could you put that in latex terms

I’m lost

[rancid umbra]

@alpine nest let $f:\mathbb{R}+\to\mathbb{R}+$ take $x$ to $1/x$.

then $d(f(x),f(y)) = |x-y|$ and $|f(x)-f(y)| = d(x,y)$ so $f$ is a homeomorphism between the two spaces since it is a bijective isometry

[gentle osprey]

c squared

[merry geode]

Hmm can you see how they are equivalent on the convex bdd polygons?

Or are you struggling with both difference and equivalence

[alpine nest]

Very cunning

[tawny cosmos]

can I dm you so it’s easier to message or nah?

[merry geode]

Sorry, I'd not like DMs for this.

[tawny cosmos]

oh alr that’s fine

[merry geode]

(This channel is usually not crowded)

[tawny cosmos]

What about my help channel? (25)

[merry geode]

Ah okay

[gentle girder]

i spent a long time trying to come up with a counterexample, only to realize that there was a really simple proof of this fact (jpretty much ust straight from the definition of the subspace topology), so yes it's right.

[prime elbow]

Let B is disconnected subspace of A then there exists U such that U proper non-empty clopen set.

Since B is disconnected subspace of A so U is intersection of B and V, where V is the clopen set in A.

Now A is subspace of X, then V is the intersection of W and A, where W is the clopen set in X.
So we can write U at the intersection of W and B. Hence, U is disconnected subspace of X

Is it correct?

[tender halo]

disconnected subspaces are just spaces disconnected in the subspace topology

i wouldnt even say your statement needs a proof really

[rancid umbra]

i thought this meant B is a disconnected subspace of X, not A is a disconnected subspace of X

did you mean A is a disconnected subspace of X?

[prime elbow]

Yes B is disconnected subspace of X

[opaque scroll]

This is correct, but note that being disconnected is a topological property. So it's a property of B completely independent of whether B is a subspace or not or of what

[rancid umbra]

U should be a non-empty, proper clopen (in the subspace topology on A) subset of B

im confused as to what your V is… U is already clopen in A.
i am similarly confused as to what your W is…

try this:

U open in A means that U = OA for some open subset O of X
U closed in A means that U = CA for some closed subset C of X

now you need to show that B contains a non-empty, proper, clopen (in the subspace topology on B) in B

rest of proof: ||U is a subset of B so U = UB = CAB = CB and U = UB = OAB = OB. so U is clopen in B, it is non-empty, and it cannot be all of of B since U is a proper subset of B ||

[prime elbow]

But its open set depends on the underlying set?

[prime elbow]

Actually I made a mistake here

[opaque scroll]

Well, it doesn't, but I guess that's what you're showing in the proof anyway

[prime elbow]

How V is clopen set in A

[opaque scroll]

It should be that V = Acap W = A cap W' with W open and W' closed

[rancid umbra]

what do you expect V to do? what is its intended purpose?

i do not think you need it as it serves no purpose

[prime elbow]

How U is clopen in A ?

[rancid umbra]

if B is a disconnected subset of A, then there is some non-empty, proper, clopen subset of A contained in B

by definition of disconnected

[opaque scroll]

Not necessarily, like think about B = {0, 1}, A=[0, 1]

[rancid umbra]

right. um. why is this definition giving me a hard time lol

lemme fix what i said earlier then

B is disconnected in A if B is disconnected in the subspace topology on B inherited from A

so B contains some non-empty, proper clopen (in the subspace topology on B inherited from A) subset U = BC = BO where C is closed in A and O is open in A

C = AC’ for some C’ closed in X and O = AO’ for some O’ open in X

rest of proof: ||U = BC = BAC’ = BC’ and U = BO = BAO’ = BO’, which says that U is clopen in the subspace topology on B inherited from X||

[tawny cosmos]

I have a metric space problem

Prove that the area metric is not equivalent to the Hausdorff metric on the set of all bounded polygons, but is equivalent on the set of all convex bounded polygons. The area metric is the are of hte symmetric difference of two sets A and B. Otherwise, the area of the Union of A and B minus the Intersection of A and B.

I’ve been working on this problem for more than 2 days.

I hate it so much.

[rancid umbra]

for any ball around the green set in the hausdorff metric, you can have polygons whose area approaches zero by staying snug to the the cave wall and slowly shrinking the lower left bump to the green set. but in the same ball centered at the green set in the area formula, the only polygons that have distance approaching zero are the ones which wramp around the convex part and slowly close the gap between it and the caved wall

[prime elbow]

Got it, thanks a lot ❤️

[rancid umbra]

none of the sets first mentioned can be in the ball centered at the green polygon for sufficiently small radii

[tawny cosmos]

The picture doesn’t explain much because the question is in the set of polygons

[rancid umbra]

just make it polygonal

[tawny cosmos]

What is the cave wall in the picture?

Is it the outer lining of the green set?

[rancid umbra]

the part that where the green set caves in on itself; the only part on the green set where it fails to be convex

[tawny cosmos]

hm

Any chance you could prove this using the openness criterion?

[tawny cosmos]

AKA making a generalization of balls in one metric and proving that there is another ball in the other metric centered around each ball iin the first metric?

[rancid umbra]

polygonal version of the set we are talking about

do you want a proof for the convex case? or a disproof in the bounded case?

[tawny cosmos]

I would like a proof for the convex case (using openness criterion if possible) and you could explain which part of the proof applies to the convex polygons exclusively and not the non-convex polygons.

[rancid umbra]

ill have to think on it a bit. pretty early over here and i didn't sleep. i can get back to u later tho

[tawny cosmos]

yea sure, would you like to add me to make it easier for later or just @ me?

[rancid umbra]

this is what i was trying to say

the rgb polygons share the obvious part of the black polygon, and begin to break off towards the left of the screen

r here is some fixed radius. in the hausdorff metric, the rgb polygons all have distance r from the black polyhon, but their area metric is approaching zero. for the ball of radius r/2 centered at the black polygon, none of the colored polygons can be in in the r/2 ball in the hausdorff metric, but you will always find polygons like the colored ones in any ball centered at the black polygon in the area metric

you really don’t even need the black shape the be convex like that. it can be a square

it’s the approximating polygons being non-convex which allows us to obtain this picture

this is the basic idea for the first part

[tawny cosmos]

huh

How are the rgb things polygons

[alpine nest]

all the polygons have the same right part, the rgb things just indicate where they differ

[real granite]

Teaching is done in an order so that we haven't done homeomorphisms at this point. I know what an isometry is from geometry but I try to use methods that are taught in that module rather than use "extra" ones from others.

Can this be rephrased as follows?

$x\in X$ is a limit point of $A$ if $\forall r>0, (B_r (x) \cap A)\setminus {x} \neq \emptyset$.

[gentle osprey]

Douglas

[lethal oxide]

what is compact neighbourhood?

[gaunt linden]

A set that happens to be both a neighborhood (of the point in question) and compact.

[umbral panther]

But there are two definitions of neighborhood

[tender halo]

with the other definition you will have a neighborhood with a compact closure in the definition of local compactness

[real granite]

.

[kindred cairn]

yes it's correct

[opaque scroll]

Is there? Like you're saying some people always require neighborhoods to be open?

[real granite]

Would appreciate some help with this.

1. How do we go from $B_r(x) \cap (X\setminus A) =\emptyset$ to no point of $A$ is a limit point of $X\setminus A$? I agree with the former, but a priori we could have a situation where all $x_n \in X\setminus A$ but the limit point is outside, i.e. $x_n\to x\in A$.

2. How do we go from no point of $X\setminus A$ being a limit point of $A$ to there existing such open balls?

There seems to be a connection I'm not appreciating between the mutually exclusive nature of $A$ and $X\setminus A$ on one hand and the existence of open balls on the other.

[gentle osprey]

Douglas

[bright acorn]

Hi so, there's a question I have in mind that I am sort of having trouble formulating precisely, but that at least for me sounds interesting if properly formalized.

So, consider an elliptic complex on a (compact) manifold.

when is it the case that the cohomology of this complex only depends on the homotopy type of the space ?

e.g

the cohomology associated to the de rham complex only depends on the homotopy type of the manifold

while the Dolbeault complex associated to a choice of integrable almost complex structure on your manifold doesn't (in the sense that the Dolbeault cohomology of a manifold endowed with two distinct integrable almost complex structures might be different)

How to make this distinction precise? And are there works studying this problem of when elliptic complexes define homotopy invariants ?

The trouble I'm having trying to formalize this is that the De Rham complex is naturally associated to a manifold

while, for instance, the Dolbeaut complex needs additional data in order to be defined

same goes for a general elliptic complex

Hmmmm

[bright acorn]

This is actually a bit more subtle

Since the De Rham complex actually needs additional data external to the homotopy type of the space in order to be defined (a smooth structure)

and you can have topological manifolds with the same homotopy type carrying non diffeomorphic smooth structures.

So the De Rham complex also seems to depend on additional non-homotopical data in order to be defined.

[real granite]

Yeah I think I get it.

Suppose $A$ is open. Then $\forall a\in A,\ \exists r>0\ s.t.\ B_r (a) \subseteq A$. Because $A$ and $A^c$ are disjoint, so too are the open balls $B_r (a)$ disjoint with $A^c$, i.e. they share no elements.

Consider a limit point $x$ of $A^c$. Either $x\in A$ or $x\in A^c$. If the latter, then, because $A$ is open, we have $B_r (x) \subseteq A$ for some $r>0$. But, by virtue of being a limit point, there is a sequence ${x_n}\subseteq A^c$ for which $x_n\to x$. Convergence entails $x_n\in B_r(x) \subseteq A$, which contradicts each $x_n$ being an element of $A^c$. Therefore every limit point of $A^c$ is in $A^c$, meaning it is closed.

[gentle osprey]

Douglas

[cedar pebble]

right but the miracle is that the de Rham cohomology itself is only sensitive to the homotopy type of the smooth manifold

[umbral panther]

Lots of people require neighborhoods to be open. But it’s rare for them to say “compact neighborhood.” There are yet other people who require neighborhoods to be open, except in compact neighborhoods

[lethal oxide]

that is defined in same way on my book

[real granite]

@crimson lily Is this right for the second bit?

$A$ closed means $A' \subseteq A$, so no points in $A^c$ are limit points, i.e. every point of $A^c$ is not a limit point of $A$. Not being a limit point is just $$\neg [\forall r>0,\ (B_r(x) \cap A)\setminus{x}\neq\emptyset]=\exists r>0,\ s.t.\ (B_r(x) \cap A)\setminus{x}=\emptyset$$. By the assumption that $x\in A^c$, we must have $B_r(x)\cap A =\emptyset$ too (otherwise $x\in A$, contradiction), thus the open balls $B_r(x)$ are entirely within $A^c$, i.e. it is open

[gentle osprey]

Douglas

[bright acorn]

Is it possible to have two topological manifolds with same homotopy type, where one is smoothable and the other is not?

[lethal oxide]

i think no

[bright acorn]

Yup, and my general question concerns why this miracle happens in the first place

When is the cohomology of an elliptic complex sensitive only to the homotopy type of the underlying space where it is defined?

[lethal oxide]

but ı not sure ı mean homotopic; two differant functions on same space

[lethal oxide]

how can two of them be differant and equivalent at the same time?

we will should ask a question on here these two manifolds( smoothable and the other) can be equivalent

[umbral panther]

Yes, it is possible to have 2 topological manifolds that are homotopy equivalent, with only one smoothable. It is also possible to have PL manifolds that are homeomorphic but not PL isomorphic, with one smoothable and the other not

[rancid umbra]

i think it’s pretty rough to try and do it like that. you can figure out other radii in terms of a quadratic equation of the other, but it’s kind of annoying

[real granite]

FIgure out the radii?

[umbral panther]

Actually, it is easy to give examples. If a manifold M^n is not smoothable, then MxR^n is not smoothable (n not 4). But it embeds in R^2n and a neighborhood is smooth. But MxR^n and the neighborhood are homotopy equivalent to M

[real granite]

Idk if you're imagining I'm doing this for a geometry module or something, but to be clear, the module is called "metric spaces and topology". I'm not aware of needing to compute the value of a radius, just that there exists an r>0 sufficient for this or that method of proving something

[lethal oxide]

oh yes you right @bright acorn sorry bro

[real granite]

E.g. there exists r>0 s.t. B(x,r) is contained in A

[bright acorn]

At some point i need to learn these things related to the distinctions of various categories of manifolds.

Sounds really tricky

[bright acorn]

also, @umbral panther (sorry for pinging) do you happen to know anything about this?

[rancid umbra]

you’re trying to show that the metrics are equivalent.

for each positive radius r and x in R_+,
you need to show that there are r’, r” > 0 such that
B_r’(x; d) subset B_r(x; euc)
and
B_r”(x; euc) subset B_r(x; d)

u need to get r’ as a function of r in a sense

what ends up happening (at least when i did it) is that r’ is a quadratic in r

[real granite]

I mean, my method seems to work alright I think?

I know as Outsider said, some care needs to be taken in saying if and only if, but it's perfectly doable to do it that way I think

[rancid umbra]

oh, i haven’t seen yours

sorry about the confusion

[umbral panther]

How many elliptic complexes can you define on multiple manifolds at once? I think this is a finite problem
de Rham
Sign
Dolbeault
Dirac
Probably something about calibrated geometry

[bright acorn]

By "on multiple manifolds at once" you probably mean that the assignment is functorial, right?

[umbral panther]

Functorial for what? It varies between the examples
You can’t really ask the question unless you already have two elliptic complexes, one on each of the homotopy equivalent manifolds

[surreal island]

Covering space of circle

[bright acorn]

So the question really should be "given two elliptic complexes defined on two homotopy equivalent smooth manifolds, respectively, under what conditions do these two complexes define the same cohomology?"

Btw, you are correct that the domain category is changing from example to example

and this is one of the tricky things here

But if we fix the category of smooth manifolds Diff, we could ask when is it the case that for a functor F : Diff -> Elliptic-Complexes, there exists a functor hF : hDiff -> hElliptic-Complexes from the homotopy category of smooth manifolds to the homotopy category of elliptic complexes where:

τ_2 • F = hF • τ_1, and τ_1 : Diff -> hDiff and τ_2 : Elliptic-Complexes-> hElliptic-Complexes are the usual functors sending a smooth manifold to its homotopy class and an elliptic complex to its homotopy class.

[bright acorn]

But i have in mind trying to formulate this for other categories (for the category of complex manifolds, which would include the Dolbeault complex, for the category of spin manifolds, which would include complexes of dirac operators and so on).

[umbral panther]

An elliptic complex invariant under all diff is probably a finite sum of shifts of the de Rham complex

[prime elbow]

I don't understand why they consider a finite set

I think it is true that if X is discrete metric space then it is complete metric space

[rancid umbra]

i think just to give a basic example of a complete metric space?

[unreal stratus]

Yeah idk finiteness seems just distracting here lol

[rancid umbra]

would you mind explaining some of the notation in the “covering space of a circle” thread in this channel? i kinda wanna take a crack at it but i need some more context

[prime elbow]

Oh

[unreal stratus]

Ah this is from Spanier I think which is a textbook I've used

It seems all the notation has been explained there though right?

Like [A,x; B, y] is homotopy classes of maps A -> B which send x -> y

[rancid umbra]

in what sense is the map supposed to be a monomorphism

[unreal stratus]

Well it is an injective homomorphism

The domain being a group via pointwise multiplication of functions into S^1

[rancid umbra]

ah okay, so what is the group structure on the hom set?

idk if that’s the right term; i just don’t feel like typing it out

[unreal stratus]

It should be pointwise multiplication again in the codomain like

If G and H are groups then you can turn Hom(G,H) into a group in the pointwise way

like (f.g)(x) = f(x).g(x)

[real granite]

If $A\subseteq \bar{A} \subset (A\cup A')$, then surely you would have $A' \subseteq (\bar{A})' \subseteq (A\cup A')'\subseteq (A\cup A')$?

Right at the end, they seem to be saying $(A\cup A')'\subseteq \bar{A}$ to get the final set relation on that line, but I don't see how that's true

[gentle osprey]

Douglas

[unreal stratus]

Limit points aren't a subset of the original set in general

[real granite]

We've shown that A union A' is closed in first paragraph

[rancid umbra]

thanks potato

[unreal stratus]

Oh okay

[real granite]

I mean... it obviously is true because closures are closed and we know $\bar{A}=A\cup A'$, but this is what we're trying to prove and we're doing contradiction

[gentle osprey]

Douglas

[unreal stratus]

I don't see how they used what you said they used

Like

Cl(A) contains A obviously

But Cl(A) contains Cl(A)' contains A'

so Cl(A) contains A u A' contains A

As A u A' is closed we see it is the closure of A

Is that clear

Idk why they mentioned contradiction lol

[real granite]

On the line with the highlighted bit, there are two relations that relate three sets. If you just look at the middle and RHS of the first one, we're saying $\bar{A} \subset (A\cup A') $. Using $X\subseteq Y \implies X' \subseteq Y'$, we can say $(\bar{A})'\subseteq (A\cup A')'$.

However, what they have actually said is that $(\bar{A})' \subseteq \bar{A}$

[gentle osprey]

Douglas

[unreal stratus]

I mean they did what I just said right, idk why you're doing your thing

Or are you just saying this is an alternative way or smth

Then sure

Just I guess note that A u A' being contained in Cl(A) can be easily proven without knowing the former is closed yet

[real granite]

You're using a different method I think to the notes. The notes are using contradiction

[unreal stratus]

Mine is the same as the notes just I didn't phrase as a contradiction

We both said that A is contained in A u A' is contained in Cl(A)

They said that contradicts an assumption that they arent equal but you can just rephrase that as saying they are equal lol

[real granite]

So the reasoning is

1. Cl(A) contains A
2. Cl(A) contains A'
3. Therefore Cl(A) contains A cup A'
4. Cl(A) is the minimal closed set wrt inclusion
5. A cup A' is closed
6. Therefore Cl(A)=A cup A'
   Is that right?

[unreal stratus]

Yes

[real granite]

Okie, that makes more sense

[unreal stratus]

They phrase it basically the same except say like

[real granite]

Sometime contradiction do be a bit confusing because it's a little like contrapositive but subtly different

[unreal stratus]

Let's assume that they aren't equal so A u A' strictly contains Cl(A), well then we have a conotradiction to the fact (proven as I did) that Cl(A) contains A u A'

But I think it is cleaner just to write it as I did

up to your taste

[real granite]

yeah

if you can prove smth by contradiction, does that mean there is necessarily a "direct" proof?

[red yoke]

Contrapositive?

[real granite]

p implies q
contrapositive is not q implies not p

[real granite]

like i've heard that wiles proved fermat's last theorem by contradiction and apparently sm ppl think that's rather messy, so i wonder if you could rephrase things to turn it into a direct one

[red yoke]

You can rewrite any proof by contradiction using contrapositives right

[umbral panther]

No
https://en.m.wikipedia.org/wiki/Constructivism_(philosophy_of_mathematics)

[red yoke]

It depends what you mean by "direct"

[red yoke]

But if you try to do this you may just end up writing your proof in the reverse order of motivation

[real granite]

deductively, i think

[real granite]

you go p=>q=>r=>...=>fermat's last theorem

but ig fermat's last theorem is prhased in the negative so maybe contradiction makes more sense

[red yoke]

Proof by contradiction is also a "deduction"

[real granite]

yh like idrk how to phrase it. but like, contraidction is basically reductio ad absurdum right, so when i mean direct i mean without doing reductio

mayube im not making much sense

[gaunt linden]

A simple propositional example of something that can only be proved either by some sort of "indirect" method or case analysis is Peirce's law: ((A -> B) -> A) -> A.

(Wait, why was this in the topology channel?)

[red yoke]

(A → B) ↔ (~A ∨ B) go brrr

[unreal stratus]

because I was explaining a proof and changed it very slightly so it wasn't by conotradiction in the explanation

then got asked about that

[gaunt linden]

Ah, I see.

[opaque zodiac]

So for FLT in particular it has a special form that means that any nonconstructive proof of it can be converted to a constructive proof. Specifically it's equivalent to a statement of the form 'for all natural numbers n [something computable happens]'. This is called a Pi^0_1 sentence. Any PA proof of such a statement can be converted to an HA proof and any ZFC proof can be converted to a proof in IZF or MLTT + some universes

[opaque zodiac]

So this should be possible in some technical sense, although some people might not consider the resulting thing to be a 'direct' proof

[real granite]

Why have they written the proof as "either... or..." instead of this?

If $U\cap A\neq\emptyset$, then $x\in U\cap A$ (as all neighbourhoods of $x$ contain $x$). This means that $x\in A$ and hence $x\in \bar{A}$.

Also it skips the other direction for some reason but it looks doable.

[gentle osprey]

Douglas

[tiny obsidian]

What's your definition of closure

[real granite]

Intersection of all closed supersets of A.

Alternatively, A cup A' (this has been proven earlier)
(A' being the set of limit points)

[tiny obsidian]

More importantly, $U \cap A$ being non empty does not necessarily mean $x$ is in there

[gentle osprey]

Edward II

[tiny obsidian]

since we aren't assuming $x\in A$ for the $\impliedby$ direction

[gentle osprey]

Edward II

[real granite]

Huh? If x is in U cap A, then it is in U AND A

x is arbitrary

[unreal stratus]

Non empty doesn't mean it contains x

[tiny obsidian]

Yes, but we've only assumed $x\in U$

[gentle osprey]

Edward II

[real granite]

But x means "literally any element" not "some special element"

[tiny obsidian]

not $x\in U\cap A$

[gentle osprey]

Edward II

[unreal stratus]

but x is fixed for the duration of the proof

[real granite]

Right I see what you're saying. We know U cap A is non-empty, and we know x in U, but it could be that U cap A = {y}, y≠x

[tiny obsidian]

yes (up to not great notation at the end but I get what you mean)

[real granite]

fixed

[real granite]

When proving the other direction, i.e. x in Cl(A) implies ..., can you just say if $x \in A$ then $x \in U \cap A$. Alternatively, if $x \in A' \setminus A$, then $\forall r>0, (B_r (x) \cap A)\setminus {x} \neq\emptyset$, and because $x\not\in A$, it must be that $B_r(x) \cap A\neq\emptyset$ also. This applies to general neighbourhoods because they can be written as the union of open balls.

[gentle osprey]

Douglas

[real granite]

Not sure about the last bit but I assume that's what's happening

[tiny obsidian]

Yes. A nicer way to say it would be that in any neighbourhood U is some ball Br(x) (I assume you're working in a metric space so this is a valid argument), and then because x in A' then ...

[real granite]

Yes we're assuming this is a metric space

[real granite]

So when working in metric spaces are there ever circumstances where you want to prove a general result about open neighbourhoods that you can't by saying "WLOG consider open balls...."

[tiny obsidian]

there probably are

I can't think of any ones now though

[tacit basin]

? I don't follow, I'm p sure (0, 1) is covered by this?

Wait nvm im dumb

[prime elbow]

Here stated that every countable topology is Souslin space, I don't understand how I can show the countable space has the Hausdorff property?

[opaque zodiac]

Not every finite topological space is Hausorff

So there must be another assumption in there

[tender halo]

most certainly they meant countable hausdorff space

[prime elbow]

Yes indiscrete topology with atleast two element

[tender halo]

is it true that a existence of path homotopy between f and g is the same as the question whether a function from S1 where one arc is f and the other is g can be extended to a function on a disk?

[umbral panther]

What’s a path homotopy?

Oh, the domain of f and g is an interval?

The obvious definition of a homotopy between f and g is a map from a square where opposite sides are f and g. But the two remaining sides are unspecified. If you fix the whole boundary to be f and g, it is more constrained

[red yoke]

A path homotopy fixed endpoints

So that would be a quotient of a square by collapsing the two sides

[red yoke]

Which is this I believe

[tame arrow]

Is "surger" the correct spelling for the verb meaning to perform surgery on a manifold as in Surgery Theory or is this a word that should be avoided writing papers and is just used casually?

[red yoke]

operate

[opaque zodiac]

Surger really sounds like an antiquated word to me

But it seems like that isn't the case

I dunno mathematicians are awful with words. 'Homotope' is not a verb

[alpine nest]

butcher

[rancid umbra]

Suppose that point sets in $B$ are open and that $B$ is regular. Let $p:E\to B$ be a covering map. I want to show that $E$ is regular. $\newline$

Let $e_0\in E$ and set $b_0=p(e_0)$. It suffices to show that for any open nbhd $U$ of $e_0$, there is a nbhd $V$ of $e_0$ such that $\overline{V}\subseteq U$ (this is an equivalent formulation of regular). $\newline$

So let $U\ni e_0$ be open and let ${V_{\alpha}}{\alpha\in I}$ be a partition of $U_0\ni b_0$ into slices evenly covering $U_0$, with $e_0\in V{\alpha_0}$ $\newline$

As $p$ is open, then there is an open nbhd $V_0\subseteq U_0\cap p(U\cap V_{\alpha_0})$ such that $\overline{V_0}\subseteq U_0\cap p(U\cap V_{\alpha_0})$. $\newline$

How do i finish this proof off?

[gentle osprey]

c squared

[red yoke]

What's a point set

[opaque zodiac]

Set of points

[tender halo]

pointy sets

[rancid umbra]

meant singleton sets

[tender halo]

Kiki as opposed to Bouba

[tender halo]

so like... a discrete space?

[red yoke]

Then E is discrete too

Completely metrizable confirmed

[coarse night]

wait point sets in B are open?

[rancid umbra]

closed

sorry

[tender halo]

its called a T1 space

[rancid umbra]

bruh

Suppose that point sets in $B$ are \textbf{closed} and that $B$ is regular. Let $p:E\to B$ be a covering map. I want to show that $E$ is regular. $\newline$

Let $e_0\in E$ and set $b_0=p(e_0)$. It suffices to show that for any open nbhd $U$ of $e_0$, there is a nbhd $V$ of $e_0$ such that $\overline{V}\subseteq U$ (this is an equivalent formulation of regular). $\newline$

So let $U\ni e_0$ be open and let ${V_{\alpha}}{\alpha\in I}$ be a partition of $U_0\ni b_0$ into slices evenly covering $U_0$, with $e_0\in V{\alpha_0}$ $\newline$

As $p$ is open, then there is an open nbhd $V_0\subseteq p(U\cap V_{\alpha_0})$ such that $\overline{V_0}\subseteq p(U\cap V_{\alpha_0})$. $\newline$

How do i finish this proof off?

[gentle osprey]

c squared

[rancid umbra]

i think it’s p^-1(V_0) intersect U intersect V_{a_0} that i want to look at

[red yoke]

Yes

What's its closure?

[rancid umbra]

there’s no nice form for its closure. it’s just contained in p^-1(cl(V_0)) cap cl(U) cap cl(V_{a_0}})

[red yoke]

It's actually an equality

[rancid umbra]

am i just missing something with the local homeomorphism when restricted to V_{a_0}?

[red yoke]

Wait

It should be p^-1(cl(V0)) ∩ U ∩ V_a0

[rancid umbra]

in general the closure of the intersection is not the intersection of the closure

[rancid umbra]

how?

[red yoke]

"x is in the closure of S" is local w.r.t. x

And locally E is homeomorphic to B

[rancid umbra]

you mean the closure in U cap V_{a_0}?

no, that doesn’t make sense to me

[red yoke]

No

What I mean is, for any point x in E, you can take an open nbhd on which p is homeo, and hence x will be in the closure of p^-1(V0) iff p(x) is in the closure of V0

[rancid umbra]

this nbhd is my V_{a_0}

[red yoke]

No x is another point

[rancid umbra]

confused

x is in what set

[red yoke]

We're showing cl(p^-1(V0)) = p^-1(cl(V0))

[rancid umbra]

one direction is by continuity

[red yoke]

This is the proof

[rancid umbra]

okay. i need to work through that part as it’s not obv to me atm. thanks!

[red yoke]

(By this I mean if U ∋ x is open, then x ∈ cl_X(S) iff x ∈ cl_U(S))

[rancid umbra]

this is the closure in the covering space tho, right?

[red yoke]

Yea

[tacit basin]

Does this work? We know for all $\varepsilon>0$, $(B_\varepsilon(p)\setminus{p})\cap S\neq\varnothing$. Let $\varepsilon_1=1$, and choose $x_1\in B_{\varepsilon_1}(p)\setminus{p}\cap S$. Let $\varepsilon_n=||p-x_{n-1}||$ and choose $x_n\in B_{\varepsilon_n}(p)\setminus{p}\cap S$ for $n\geq 2$. Clearly, $x_m\neq x_n$ whenever $m\neq n$. For all $\varepsilon>0$, choose $N\in\N$ such that $||p-x_N||<\varepsilon$. Then $x_n\in B_\varepsilon(p)$ for all $n\geq N$ and $\lim_{n\to\infty}x_n=p$

[gentle osprey]

Sara

[rancid umbra]

that looks fine

[tacit basin]

ty

[gentle osprey]

c squared

[rancid umbra]

i can't find any resources online confirming or denying this

[white oxide]

I think so, yeah

You could make it cleaner by considering a open set W in U and see that f(W) is open in f(V) and so open in general

Which has non trivial intersection, so pulling back you still get a non-trivial intersection

[rancid umbra]

are you just saying, since im intersecting with U already, wlog assume U' is a subset of U?

[white oxide]

Sure, I guess

[rancid umbra]

@red yoke would you mind seeing if this is enough to conclude the covering space problem we were working on earlier?

[white oxide]

Hm, I feel like there was a different route you could've taken.

||For x take a homeomorphically mapped U and select V in f(U) with cl V in f(U). Pulling back you get a closed set C in the subspace topology of U. Ideally we want to make sure that C is already closed in the general topology:||

||Let y be a point not contained in C. It either maps into cl V or to a different part. In the latter case use regularity to find a disjoint open set.||

||In the former case we know that y lies in one of the disjoint U_i which compose the pre-image of f(U). But by assumption y can't be in U, so it already lies in a disjoint open set.||

[rancid umbra]

isnt C closed in E because U is open

[white oxide]

Nah, [0,1) is closed in (-1,1) after all

[rancid umbra]

fwoop

[white oxide]

Open + open = open and closed + closed = closed but the other ones don't necessarily hold

[rancid umbra]

how do you know that f(U) can be evenly covered

[white oxide]

Evenly covered?

[rancid umbra]

that the preimage is a disjoint union of open subsets, each mapping homeomorphically to f(U)

[white oxide]

I just choose the open set you get from the covering space definition

Calling it f(U) is a bit awkward, yeah

[rancid umbra]

yea but you say, "In the former case we know that y lies in one of the disjoint U_i which compose the pre-image of f(U)."

how do you know that this can happen?

[white oxide]

I actually choose an open set W in Y with disjoint union of open sets each homeomorphic to W

I was just a bit sloppy

[rancid umbra]

oh okay, yea that is a bit awkward

[white oxide]

Pretend it's all hidden inside "homeomorphically mapped U"

[rancid umbra]

why are we considering points not in C?

[white oxide]

To check if it still closed

[rancid umbra]

i don't think i understand why we would do that

wouldn't you want to look at limit points of C?

[white oxide]

Basically we choose a closed set in (0,1) but we want to make sure you haven't hit the edge and there are limit points missing

[rancid umbra]

oh oh i see

you are showing that all other points are isolated from C

[white oxide]

Yeah

That C is cl V pulled back is relevant in the case where points outside C are mapped to a point outside cl V

[rancid umbra]

okay, cool. thank you for the alternative proof

[white oxide]

Isn't that sorta what you did in your original writeup now that I look at it?

Ah right. You showed equality of the pre-image of the closure and the closure of the pre-image. The former is a disjoint collection of closed sets since they live in different V_i. By the latter you know the disjoint union of those sets is closed, can you see why each one of them individually is closed?

[rancid umbra]

ig im not seeing it. is it the same argument you have laid out previously?

[white oxide]

Nope

Hint: try to intersect it with an appropriate closed set which isolates one of the slices

It's enough that all other copies of cl V_0 are disjoint from said closed set

[novel plank]

Let $X$ be a compact (and T2) space, $A$ a dense subspace and $f:X \to X$ continuous. What can I conclude about $f(A)$ or about $A$? I know that this is vague, but I'm working on a problem where this is all I have and I'm not very well versed on topology, so any facts you can deduce starting with this setting would be appreciated

[gentle osprey]

Eduude

[novel plank]

I'm just trying to gather as much facts about this as I can

[rancid umbra]

f(A) is dense in f(X)

[white oxide]

Classic

The usual stuff is also there. The map is closed (closed maps to closed) and f(X) is compact.

[novel plank]

so f(A) is dense in f(X) which is compact. Does this allow us to conclude anything else about f(A)?

[lethal oxide]

dense subspace wdym? meaning A a dense set and T topology on A, (A,T) is dense subspace of X (X a topological space) right?

[white oxide]

Just take the closure of the V_i containing x or take the complement of the union of all other V_i, in either case you'd show that cl V_0 considered as as a subset of X is closed

[novel plank]

What I'm asking is, does being dense in a compact space imply something stronger then dense?

[rancid umbra]

the graph of f is closed, if it’s a cts bijection then the map is a homeo

[novel plank]

Yes, I think that's what I mean 😅

[lethal oxide]

can't compare?

i think

[rancid umbra]

hmm. that seems to make sense. i think i am going to type these arguments out later. a bit tired of them

[white oxide]

I mean not sure what interesting thing you can say about the rational points of a compact set in R^n

[novel plank]

That's true...

[white oxide]

So in general I don't expect much. Being dense is mostly a cardinal property, i.e. they are interesting in the cases when they are countable and also for some larger cardinals

I however, luckily, forgot all the cursed higher cardinality topo stuff

[rancid umbra]

Kerr, just to make it clear, we have three different arguments for showing that the covering space of a regular space is regular, right?

[novel plank]

well, thank you guys 🙏 I'll see what I can do in the problem I'm working in with all you've said

[white oxide]

Well, two? The one I gave independently and then finishing up yours?

[lethal oxide]

i think ı dont know something it is possible or not possible this change, I can't comment now because it is change according to defined subspaces

[rancid umbra]

then what was this? just a different way to finish up mine?

[white oxide]

Well, you stopped when you showed that cl f^-1(V) = f^-1( cl V) no?

[rancid umbra]

right

[white oxide]

Drawing a picture that's actually infinitely many copies of cl V_0

You can only have one without ruining the condition that your closure fits inside U. So you need some trick to pick one out

Well, sorta. Finishing up is a one liner, it just takes some time to figure out

[rancid umbra]

but mine only did this for local homeos, not covering maps

this is in reference to my first attempted proof, right? before i stopped to prove stuff about pre-images and closures

[white oxide]

You stopped?

Also, I mean, covering maps are local homeos wdym

[rancid umbra]

uhh.

so i tried to prove that for a local homeomorphism, cl f^{-1}(V) = f^{-1}( cl (V) )

and then i think you picked up where i left off trying to show that E was regular, after giving your independent proof

[rancid umbra]

if im following the convo right, then you first gave your independent proof that the covering space of a regular space was regular here.

[rancid umbra]

and then you made a comment on my attempted proof showing that the preimage of the closure is the closure of the preimage for a covering map

im just trying to see where everything is fitting in;
we have the first proof i gave, your independent one, and then the comment you made?

im confused about where the comment you made fits in

[white oxide]

Yes, that's entirely self-contained

And the your second reply correctly marks when I moved onto finishing up your attempt

[rancid umbra]

https://mathoverflow.net/q/74415/167473
this confirms my conjecture here, since local homeos are cts open maps

okay, and thank you Kerr, i will try to fit these together and type them up some other day. these kind of gave me a headache

[rancid umbra]

i hate to keep bugging you. i want to understand what you are saying here.

i see the picture of the infinite sheet of the preimage of the closure and how this is equal to the closure of the preimage.

i see how p^{-1}(cl(V0)) cap cl(V_{a0}) = cl(p^{-1}(V0) cap V_{a0}), so each copy of cl V0 is closed in E

what i dont see is how this shows regularity of E. i need to show that this is contained in U some how

[white oxide]

f-1(cl V_0) is contained in f-1(U)

The latter are the infinite sheet

[rancid umbra]

but i thought i needed to look at the set

p^{-1}(V0) cap V_{a0} cap U (this is the U that im referencing) and find the closure of this set using the fact that the preimage and closure commute.

[white oxide]

Um

What purpose does the U serve in your original message, actually?

You just need U_0 and the V_i making up it's pre-image

[rancid umbra]

this was the formulation of regularity i was trying to use

U was my open set around e0 from this lemma

[white oxide]

Oh, sure.

[merry geode]

O separating points from opens

[white oxide]

I feel like you are mostly getting hung up on the notation. You have your U, your U_0 and the V_i. Then you can select a W in U_0 cap f(U), whose closure is still contained in it

And then forget about writing out the intersection

[merry geode]

How is regularity defined here? The lemma seems how regularity was defined for me.

[white oxide]

The fact that the closure commutes is relevant since the left side tells you it is closed (cl first). The right side that the pre-image of the closure is contained in the union of the slices

[merry geode]

I presume you are trying to show the lemma?
I guess it has slightly different statement from the definition

[rancid umbra]

i am trying to show that a covering space of a regular space (whose points are closed) is regular (and the the points in the covering space are also closed)

[merry geode]

Oh, I see. That makes more sense

[white oxide]

The equivalent definition using nested open sets and their closures is used here

[merry geode]

You can directly use E -> X being locally homeomorphic, right

[rancid umbra]

Kerr has provided a proof, but he also provided a patch for mine

im trying to understand the patch

[merry geode]

Ah, sorry that I was lost on context

[unreal gull]

why is ${0}$ open in $A$? It's complement $A - {0} = C$ is open in the lexicographic ordering of $\mathbb{R} \times \mathbb{R}$

[gentle osprey]

Henry

[rancid umbra]

its the intersection of A with a small open ball/interval around the origin

[merry geode]

I think delineating what each symbol means (how they are constructed) would help.

[white oxide]

Drawing a picture too, honestly

Like you can reduce a lot of stuff just making sure that the right choice of open sets mean you don't have to keep carrying several intersections around

[merry geode]

Yeah, pictures help a lot

[unreal gull]

but doesn't R x R use the lexicographic ordering? So a ball is not open in this topology

I just assumed that R x R use the lexicographic ordering because the book, Munkres, says to assume that R always uses the order order topology, unless states otherwise

[rancid umbra]

this is the picture as i understand it

the V_{alpha}'s are homeo to U_0

[merry geode]

So are you trying to put inside U, right

[rancid umbra]

yes, i am trying to show that p^{-1}(V0) cap U cap V_{a0} has closure contained in U

[white oxide]

Don't sully a man (person) for asking 😔

[merry geode]

V_0 is already inside p(U \cap V_a0), and as p|_U is bijective, p^(-1)(V_0) is inside U \cap V_a0. Do you get this?

[rancid umbra]

yes

[white oxide]

But yeah, once you've drawn out the V_i keep in mind they are homeomorphisms

So the closure stays inside them

[merry geode]

The same holds for the closure as well

And.. it's done!

[rancid umbra]

okay, so p^{-1}(cl(V0)) is inside of p(U cap Va0). and then you just pull back with the local homeo

[merry geode]

(Just matter of replacing V0 with cl(V0))

[rancid umbra]

well wait actually, i thought p^{-1}(V0) was also an infinite sheet and not inside of a single sheet Va0. p^{-1}(V0) are the pink blobs

[white oxide]

Well, the inclusion stays intact for any subset of U_0 actually, even without the homeomorphism hypothesis

[white oxide]

It is

[merry geode]

Oh, I intentionally restricted p|_U to captire the bijective portion.

[white oxide]

That's what the whole bit here was about

[merry geode]

I guess this is a different proof then? Hmm

[white oxide]

You do stuff in U_0, consider pre-image. Use that closure and pre-image commute

[merry geode]

Yeah if we talk about general p, you need that other copies of p^-1(U0)'s are disjoint.

[white oxide]

Then you get your infinitely many copies of cl W. Then intersecting with a closed set excluding all other V_i you get a closed set which is the closure of some open set in U

Which is exactly what you were after

[rancid umbra]

uh. its just not clicking for me

thank you guys for all the help

ill just keep banging my head against a wall till it clicks

[white oxide]

Well, pro tip

Sleeping over a problem is pretty op

[merry geode]

Well, honestly I dunno if this thing will work well

[white oxide]

Wdym work well

[merry geode]

Ah nvm, it works

[white oxide]

Btw here is the initial independent one

[merry geode]

Yeah each sheet of V_a contains its own copy of \barV_0, which are all disjoint.

So by intersecting with V_a0, we are left with the "correct copy".

[white oxide]

Yippee

[gentle osprey]

c squared

wait a second. I think i got it:

we know that since the perimage and closure commutes under $p$, then $$\texttt{cl}_{U\cap V_{\alpha_0}}(p^{-1}(V_0)\cap U\cap V_{\alpha_0})=p^{-1}(\texttt{cl}V_0)\cap U\cap V_{\alpha_0}$$ is closed in $U\cap V_{\alpha_0}$. \newline

But as Kerr said, we have 
$$\begin{align*}
p^{-1}(\texttt{cl}V_0)\cap U\cap V_{\alpha}
&=p^{-1}(\texttt{cl}V_0)-\bigsqcup_{\alpha\in I-\{\alpha_0\}}(p^{-1}(\texttt{cl}V_0)\cap U\cap V_{\alpha_0})\\
&=p^{-1}(\texttt{cl}V_0)-\bigsqcup_{\alpha\in I-\{\alpha_0\}}(U\cap V_{\alpha})
\end{align*}$$
which is closed, since $p^{-1}(\texttt{cl}V_0)$ is closed and a closed set cut an open set is closed\newline

@Kerr @Absta does this look right?
```Compilation error:```! Package amsmath Error: Erroneous nesting of equation structures;
(amsmath)                trying to recover with `aligned'.

See the amsmath package documentation for explanation.
Type  H <return>  for immediate help.
 ...                                              
                                                  
l.65 \end{align*}
                 $$
Try typing  <return>  to proceed.
If that doesn't work, type  X <return>  to quit.```

[merry geode]

Yeah this seems right (modulo the latex overflow)

[rancid umbra]

okay. it was going to bug me until i was either bone tired and passed out or figured it out

[merry geode]

Btw, another way to see this is that p^(-1)(cl V_0) is disconnected with one component per each sheet.

[rancid umbra]

wdym one component? there could be several components per sheet

[merry geode]

Ah that's assuming (cl)V_0 is connected, which we did not assume

[rancid umbra]

right

[merry geode]

But you can still use the same argument even if multiple components lie on each sheet

[rancid umbra]

i dont think so. even though components are closed, there is no way to guarantee that you have finitely many per sheet

or at least i don't see how you woud argue that the union of all the components in a single sheet is closed

im happy with the two proofs i saw and understood today. thank you both for being there for my struggle. i hope my brain absorbed some of the advice you guys gave haha

[merry geode]

Good that you resolved it!

Btw about the component bit:
Consider a closed set C = A \cup B where A and B are disjoint and clopen in C (Basically how C can be disconnected)
A and B are both closed in C, so by some kind of transitivity, you have A, B closed in the whole space.

[rancid umbra]

that only works when you have finitely many components though, right?

[merry geode]

Nah, it works with any amount of components.

[rancid umbra]

i see

[merry geode]

It only requires the subset of C to be clopen in C, in fact.

[rancid umbra]

whats the general statement here?
connected components of a subspace of an open subset are closed in the whole space?

maybe im not seeing what C is here...

[merry geode]

(Any union of) connected components of a closed set C is closed in whole space.

You can consider C = p^(-1)(clV0) in this problem.

[rancid umbra]

so C is closed and it’s connected components are closed in C, hence each connected component is closed in the whole space

now the union is closed in the whole space since they union to C

[merry geode]

Any union of connected components of C is clopen in C.

Like, if C = A1 \cup A2 \cup A3, A1 \cup A2 is clopen in C.

[prime elbow]

Prove that a set U is open in X iff each point in U has a neighborhood V in X such that U ∩ V is open in V.
i think first direction is trivial because since U is open we will take V = U, right?

to converse, i need to find open set N such that for each u in U there exist a N_x open set containing x such that N_x contained in U.

[opaque zodiac]

Yes you can take V = U

[prime elbow]

some Authors define neighborhood as superset of open set or just as open set, which one should i consider?

i think first one is better

[red yoke]

The one that your school uses

For no ambiguity the first one is "x is in the interior of S" and the second one is "S is an open neighbourhood of x"

[opaque zodiac]

Yeah but the problem is that you get tired of writing 'open neighborhood' over and over again

[red yoke]

open nbhd

[opaque zodiac]

No joke I have seen nbd in published topology papers

[red yoke]

S is a U

[prime elbow]

How can I show a set of interior points of non-empty set A is open? I think I did in metric space but I don't have any idea how to show in any arbitrary topological space

[opaque zodiac]

Show that it's a union of open sets

One way would be to use this definition (or show that this is equivalent): the interior of A is the union of all open sets contained in A

[prime elbow]

I want to prove that the interior of A is the union of all open sets contained in A

So I need the interior of A to be open set

[limpid fern]

what is your def of interior of A

x in int A if there exists a nbhd of x contained in A?

[prime elbow]

Yes

[limpid fern]

can you express int A as the union of some open sets?

[prime elbow]

I think no

[limpid fern]

why not

"some" could be arbitrarily many btw

[prime elbow]

Yes but how?

[limpid fern]

well what open sets do you even have to work with

given a point in the interior there is some open set that contains that point

so really there's only one decision that makes sense

[prime elbow]

Yes

Any hint?

[limpid fern]

well

i can't think of a hint that wouldnt give away the solution immediately

[prime elbow]

got it, thank you

[prime elbow]

Find an example in which a sphere is disjoint with the closure of the open ball with the same center and radius.

i think ultrametric will work, right?

[prime elbow]

Formulate a necessary and sufficient condition on the topology of a space which has an everywhere-dense point.

I think necessary one is it should not be T_1 space

[final grotto]

is every point in a 0 dimensional Hausdorff space an isolated point? if we can find arbitrary small points?

[red yoke]

{1/n : n} ∪ {0} has covering dimension 0

[final grotto]

i dont understand

[red yoke]

0 is not isolated

[tender halo]

that would mean that your space is discrete

but there are many 0 dimensional spaces

[alpine nest]

The Cantor set has no isolated points and is zero dimensional

[final grotto]

ahh okk

okk thank you guyyss

^^

[wet crow]

I'm have a small question about (co)homology of simplicial sets, most sources I see, the cochains C*(X,A) are defined as the dual Hom(C_*(X,A),A)

but in the book "methods of homological algebra" it seems they only talk about "functions on simplices with values on A" which to me would correspond to the group

Hom(C_*(X,Z),A)

is there a significant difference ?

[tame arrow]

Are there results for exotic structures on manifolds separating not just topological, PL or smooth manifolds? For instance, do $C^k$ manifolds have exotic $C^{k+1}$ structures on them for large enough dimension d (so a finer form of exoticness)? Can something similar be done for Lipschitz manifolds or Hölder manifolds?

[gentle osprey]

Light Yagami

[umbral panther]

C^1 manifolds have essentially unique C^oo smoothings. In fact, they have essentially unique real analytic structures

Topological manifolds have essentially unique (locally) Lipchitz structures (except in dimension 4?)

[real granite]

For the converse, how do we know if the correct implication is $x\in A' \implies x\in A$ or $x\in A\implies x\in A'$? (I.e. $A'\subseteq A$ or $A\subseteq A'$.)

[gentle osprey]

Douglas

[alpine nest]

You're showing that the set of limit points of A is a subset of A

I.e. that if x is a limit point of A, then x is an element of A

(as a side note the characterization in this lemma is a very useful characterization of closedness, worth remembering and very preferable to "a closed set is one that contains all its limit points", in my opinion)

[real granite]

Well yes, I realise that you can show A' is a subset of A, but which implication (you only want the former) comes "naturally" seems to depend on the order of the argument. I.e. "consider x in A'. By assumption, x in A too, therefore x in A' implies x in A", or "consider a sequence x_n in A that converges to x in A (x_n ≠ x). x is clearly a limit point, so x in A implies x in A' ".

[alpine nest]

Well, the "conversely" part is about showing that if every limit of a convergent sequence of elements of A lies in A, and from this assumption you want to deduce that A is closed.

[real granite]

I realise the second statement is not something we're interested in, but is it actually true?

[alpine nest]

Presumably closed is defined as "contains all its limit points"

So to show that A is closed, you need to show that the set of limit points of A is a subset of A

[real granite]

Yes, I'm convinced that this is true, but what about the other way around, i.e. all elements of A are also limit points

[alpine nest]

So you start with two known facts: "every limit of a convergent sequence of elements of A lies in A" and "x is a limit point of A"

And from that you need to deduce that x is an element of A

[alpine nest]

That isn't necessarily true, whether or not A is closed

You can have closed sets some of whose elements aren't limit points, and you can have non-closed sets whose all elements are their limit points.

(but if they aren't closed, they will also have limit points that aren't their elements)

So whether or not all elements of a set are its limit points isn't directly related to closedness

[real granite]

Ah... I think I'm not being careful with my quantifiers.

I was thinking that "every sequence converges to a limit in A" means "all points in A have at least one sequence that converges to them" (i.e. are limit points), but these are actually two distinct statements

[alpine nest]

There's an important distinction between "having a sequence of elements of A converging to x" and "x being a limit point of A", because the latter specifically excludes the case of (eventually) constant sequences.

If A = {1,2}, then 1 is a limit of a sequence of elements of A (such a sequence being 1,1,1,1,....) but it is not a limit point of A

Because the definition of a limit point specifically excludes sequences that include said point as a term

So A is a closed set but none of its elements are its limit points

[real granite]

Yeah. I'm guessing whether you can find a convergent sequence of distinct points depends on the metric, because the metric defines distance and hence what is considered "arbitrarily close to the limit"

[alpine nest]

Also that's why I prefer the characterization given in the lemma, because a lot of the time it doesn't really matter whether the approximating sequence includes your particular point or not.

It mostly only does when discussing function limits and continuity

[alpine nest]

It depends on the metric and on the particular set in question.

[real granite]

okie

thanks for the help

This channel is being very helpful

And I have noticed I'm understanding metric spaces and topology better

[alpine nest]

Considering you seem to be learning metric spaces and topology, it would be very worrying if you weren't 😄

[prime elbow]

They define contraction mapping. And I want to prove that if f is contraction mapping of the metric space (X,d). Then f is a continuous mapping.

Let x in X and V be open ball which contains f(x).

Now there exists r>0 such that B( f(x), r) is contained in V.

Since f is contraction there exists s in (0,1) such that d( fx_1, fx_2 ) ≤ s•d(x_1, x_2) for all x_1, x_2 in X.

So if I take B(x, t), where t = r/s then f(B (x, t) ) contained in V, right?

[real granite]

Was thinking about this some more... does it matter if the neighbourhood(s) U are open/closed?

Earlier in the lecture notes, they said that generally neighbourhoods are assumed to be open but sometimes that convention isn't followed, so was wondering what the case was here. (Most likely they are following their own convention of course, but just wanted to check.)

[alpine nest]

My preference is definitely for "neighborhood of x" to mean "an open set containing x"

It's slightly a matter of preference, but ultimately you do want an open set containing x.

Even if you start with some more general definitiion of neighborhod, you will almost always pass to the open version early on in the argument.

[lethal oxide]

i think it is not slightly a matter of preference if he do want an open set containing is x this is definitely open neighborhood of x or itself of x

[real granite]

Well the theorem/lemma in question is $x\in \bar{A} \iff U\cap A \neq\emptyset$, for all neighbourhoods $U$ of $x$. I'm not sure if it is assumed $U$ is/are open.

[gentle osprey]

Douglas

[real granite]

Also, to show every set under the discrete metric is closed, can't you just say "let C be an arbitrary set. Since every set is open under d, in particular X\C is open, meaning C is closed."

[alpine nest]

Yes

[real granite]

The worked solution did some needlessly involved thing with sequences

[alpine nest]

No idea why

[alpine nest]

What's the alternative definition of neighborhood exactly? If it's "a neighborhood of x is a set that contains an open set containing x", then it will be equivalent

[normal umbra]

Which papers of Grothendieck/others (i.e. Calabi/Kobayashi) in Complex Projective / Vector (or Fibre) bundle theory should I reach before approaching chern1946 and yau1978?

[alpine nest]

So yeah, this will work whether you use the open neighborhoods or the more general definition in your screenshot

Because all that will matter will be that ball

An argument involving the more general neighborhood will almost always start with taking that open ball and going with that.

Which is why the more general definition doesn't really gain you much over just going with open neighborhoods

[real granite]

Right and with the second definition A isn't necessarily open, because we only know it contains a particular ball, rather than balls at all points?

[alpine nest]

Yep

You know that x is in the interior of A, but that doesn't immediately mean that A is open

Since it might have other points that are not in its interior

[lethal oxide]

yes this is true and U can be open or closed

the important thing is neighborhood

[alpine nest]

The imporant thing is the ball

[lethal oxide]

mean, if x a point of A closing, it is point of A and containing every neighborhoods of x(for every U) mean U interior A not empty set @real granite

[lethal oxide]

ı talking on about this image

[lethal oxide]

.

[alpine nest]

I'm talking about balls

[lethal oxide]

yes 🙂 but he said me on about different thing

[real granite]

If the interior of a set is $$\interior(A) = \bigcup_{\text{open}\ U\subseteq A} U$$ then is $$\interior(A\cap B)=\bigcup_{\text{open}\ U\subseteq A,\ \text{open}\ V\subseteq B} U\cap V?$$

Initially I wasn't sure because $A,B$ might be disjoint, but then I realised the intersection of open subsets would just be the emptyset, which makes sense.

[gentle osprey]

Douglas

[alpine nest]

It's true, although I don't think it's hugely useful, as you might realize if you try to prove it.

[real granite]

It's useful to prove the question I'm doing (that the intersection of the interior of A and B is equal to the interior of the intersections of A and B)

[alpine nest]

Sure, but using that fact would require you to prove that fact.

Which is just as much work as proving the fact about the interior of intersection directly

[tender halo]

i can remember only one good place where its convenient to have neighborhood with arbitrary sets instead of open ones

and its where you cover a compact space with a family closed nbhds

then its nice cuz you can just take the union of the finite subfamily

and get a closed set

[real granite]

Seems that the worked solutions assume it to be true. I could probably prove it but won't do it right now

[final grotto]

how do I prove that there exists such a c?

[alpine nest]

this doesn't require your characterization of interior, it's directly definition of interior (and of intersection)

although that's why I meant by your characterization not bringing much advantage, since it's also pretty much just that

[final grotto]

I dont know how to show the existence of a function lol

[red yoke]

Try to construct the function

[final grotto]

you mean like this?

[red yoke]

Yup

[real granite]

How not?

Oh actually I do kinda see it

Yeah he's just talking about U_x without breaking it down into open subsets of A and B

[alpine nest]

yep, or rather the point is that a subset of the intersection is a subset of all the sets being intersected

which is a very often helpful insight

[white oxide]

It's just the map in set. Since a and b are continuous the resulting map is then also continuous

(the forgetful functor has both a left and right adjoints, so any (co)limit will turn out be the (co)limit in set equipped with the right topology)

[quick bough]

how do you show that for a self map f : CP^2 to CP^2, the degree of f are just the squares n^2 for a non-negative integer n, one direction is pretty clear by naturality of the cup product, but why can we recover each square by the degree (note that the degree of f is defined in this case by the unique map Z \cong H^4(CP^2) -> H^4(CP^2) \cong Z, given by multiplication with d)

[prime elbow]

Q is not compact because we have open covering by (x, x+2), where x in Z, right? And Q is not covered by any finite collection of it

[alpine nest]

yes, you've got an open cover with no finite subcover

[prime elbow]

And if T_1 is finer than T_2 and if A is not compact in T_2 then it cannot be compact in T_1, right?

[alpine nest]

Also Q is not bounded, so it can't be compact

[alpine nest]

The finer the topology the harder it is to be compact

[prime elbow]

Right

[tender halo]

compact Hausdorff topologies are minimal in the topology order

so if you make a compact topology coarser, it stops being hausdorff

and if you make it finer it stops being compact

[prime elbow]

Okay, thank you

[real granite]

This is closed and not open, correct?

[gaunt linden]

Correct.

[gritty widget]

Hi

I wanted to understand the definition of a neighborhood in terms of

A topological space

Also, I was doing this exercise where it started shifting the definition of limits and continuity from the epsilon delta version to this one

$$f: X \to Y$$
$$f \text{ is continuous at } a\in X \text{ iff}$$
$$\forall M \subset X \text{ where } M\text{ is a neighborhood of } $a\in X \exists P \subseteq Y \text{where } P \text{ is a neighborhood of } f(a):$$

$$f(M) \subset P$$

[gentle osprey]

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

[gritty widget]

So I was trying to see like how different general proofs convert

So the simplest exercise mentioned was just

Prove $f(x)=\begin{cases} 1, & x \geq a \ 0, & x < a\end{cases}$ is continuous everywhere except at $x=a$.

[gentle osprey]

bombastic side eye

[gritty widget]

But with the neighborhood def

[real granite]

iirc the topological definition of continuity refers to the pre image rather than the image, i.e. f is continuous iff the pre-image of every open subset of the range is an open subset of the domain, $U\subseteq Y\ \text{open}\ \implies f^{-1} (U) \subseteq X\ \text{open}$

[gentle osprey]

Douglas

[real granite]

this allows for the pre-image to be empty

[real granite]

This is not a well defined function as f(a) is apparently 1 and 0 at the same time

[gritty widget]

Sorry

Prove $f(x)=\begin{cases} 1, & x \geq a \ 0, & x < a\end{cases}$ is continuous everywhere except at $x=a$.

[gentle osprey]

bombastic side eye

[gritty widget]

Fixed it

[real granite]

what's your range?

it could be {0, 1} but it could also be a superset of that

[gritty widget]

Won't that require the existence of an inverse

[real granite]

.

[gritty widget]

We use R with metric |x-y| so

[real granite]

$f^{-1} (U) ={x\in X | f(x)\in U}$, so if there are no such points then the pre image is empty

[gentle osprey]

Douglas

[real granite]

perfectly well defined

[gritty widget]

Hm ok fair ig

[real granite]

in that case you want to consider arbitrary open neighbourhoods of R under the pre-image

[gritty widget]

It's been a while since I did some of this stuff

Do you mind elaborating a bit more

[real granite]

well every open set that doesnt contain either 0 or 1 will be sent to the emptyset under the pre image

so you want to look at open neighbourhoods around 0 or 1

and see if they get sent to open sets

if they do

then f is continuous

(on that part of the domain)

[gritty widget]

Hmmm

Yea that makes sense

But I have a question if that's ok with you

Why do you keep using the pre image

I was taught to use the image

The image of f should be a subset of a neighbourhood in Y

[gritty widget]

When I did topology

[real granite]

maybe there's some alternative defn but ive never seen continuity be defined in that way

[gritty widget]

I was taught this defn too

The one you're using

But the first one I was taught was that f maps a neighbourhood in X to a subset of a neighbourhood in Y

Which should be true from the way you defined the pre image

[gritty widget]

Got it

Can I also ask you one more thing

If it's ok

[real granite]

https://en.wikipedia.org/wiki/Continuous_function#Continuity_at_a_point i think you're right, there is a definition for continuity at a point that uses the regular image

[real granite]

okie

[gritty widget]

How exactly do you define an open set for a topological space

Where there's no metric

[gaunt linden]

You can't.

[gritty widget]

But then

[gaunt linden]

Oops, sorry.

[real granite]

que? in a non-metrizable space the open sets are just the elements of the topology

[gritty widget]

Iirc topology has elements called open sets

[gritty widget]

Yea

Exactly

[gaunt linden]

I misread "topological space" as "vector space" 🤦

[red yoke]

It comes with the defn of a topological space

[gritty widget]

Ok but I'm a bit unsure

[red yoke]

Just like vectors come with the defn of a vector space

[real granite]

but if it had an inner product (or more generally a norm) then you would be able to i think

[gritty widget]

Bcz suppose the simplest example

Like I wanna define sth like (R, tau)

[gaunt linden]

You don't have a topological space before you have chosen which subsets of it you want to be open.

[gritty widget]

How do I know [a, b] would not be an element of tau

But (a, b) would

[red yoke]

It depends what τ is

[gritty widget]

Ohh

So you essentially choose the topology set in that way

Damn that's nice

Genuine q, how would you define the topology of R then

[red yoke]

The "standard" topology is the one generated by open intervals

But there are many other topologies on R too

[gritty widget]

(I'll just add ik I come off like a dumbass rn who hasn't done any topology but I don't rlly work with non-metrizable spaces)

I'm a phy major

[gaunt linden]

Most often what we do is to choose some of the open sets explicitly and then say "... and every other set that the axioms for a topology force to be open given the decisions we've already made are also open".

[gritty widget]

N they work w Hilbert spaces all the time

At worst banach spaces

But not more general than that

[real granite]

slight aside...

this set is equal to $A={ \frac{1}{\pi m} | m\in\bZ\setminus{0} }$.

the compliment is $\bR \setminus A$, or the reals with a countable number of points removed. wouldn't this compliement be open and hence $A$ be closed?

[gentle osprey]

Douglas

[gaunt linden]

This leads to the concepts of "basis" and "subbasis" for a topology.

[gritty widget]

Can't a set be closed n open both?

[real granite]

yes

[gaunt linden]

Yes.

[gritty widget]

Then

?

What's the problem

[gaunt linden]

Such sets are (only somewhat facetiously) known as "clopen" sets.

[gritty widget]

Yes

I was taught that

[real granite]

im not claiming the set is open, im claiming it is closed. being closed is not a counter argument to it being open

i want to know if my claim that it is closed is correct

i mean intuitively it prolly isnt

[gaunt linden]

It's not true in general that R minus a countable number of elements is open.
For example R \ Q is far from open.

[real granite]

interesting

[red yoke]

Try to draw the set

You can tell what its closure is

[gaunt linden]

The standard example (other than Q) of a subset of R that is countable but not closed is { 1/n | n in N+ }.

[real granite]

my thinking was that $$\bR \setminus A = \bigcup_{ m\in \bZ\setminus{0} } \left(\frac{1}{\pi m}, \frac{1}{\pi (m+1)} \right)$$

[gentle osprey]

Douglas

[red yoke]

You missed 0

[real granite]

where?

[gritty widget]

Idt it should be big cap

[red yoke]

Is 0 in the union?

[real granite]

m=0 does not give sin(1/m)=0

[gritty widget]

Big cup is fine

[gaunt linden]

Sorry, I was speaking nonsense.

[gritty widget]

Yea should add that

[red yoke]

^

[gritty widget]

Lol happens issok

[real granite]

yh i realised my defn is bad because when m=-1 you run into problems

[gaunt linden]

According to your definition here, is 0 an element of A?

[gritty widget]

Once I called an ODE solution analytically unsolvable but then flipping through my nb I found I had solved it a week before

So happens

[gritty widget]

Wrong tag?

[gaunt linden]

Whops, wrong reply. I'll just shut up, I think.

[gritty widget]

Happens, chill

[gaunt linden]

Spent all of today's quota of stupid mistakes.

[real granite]

I think you can add in (-1/π, 1/π) and then follow my argument

[red yoke]

No that covers points of A

[real granite]

oh ye im being dumb

[gritty widget]

There's a small problem with this brw

You're taking the union of (-1/pi, 1/pi(-1+1))

That's not

[real granite]

yes ive pointed that out

[gritty widget]

A valid interval

[real granite]

yh as m increases you will get tighter and tighter intervals aorund 0

[red yoke]

Indeed

So 0 is not in the interior of R\A

Alternatively you can see that 0 is a limit point of A

[real granite]

and then 0 is an LP bc you take m=x_n-> infinity so sin(1/x_n)->0

[red yoke]

Yup

[real granite]

okie this is making more sense

[gritty widget]

$$\mbb R - A = \qty(-\frac{1}{\pi}, \frac 1 \pi ) \cup \qty(\bigcup_{n\in \mathbb Z - {-1, 0} } \qty(\frac{1}{n\pi}, \frac{1}{\pi (n+1)} ))$$

[real granite]

setminus

what does qty do?

[gritty widget]

Makes the bracket bigger

[gentle osprey]

bombastic side eye

[gritty widget]

Set difference isn't working for some reason for me

Hm

[gaunt linden]

You want (-infty, -1/pi) cup (1/pi, infty) there instead of (-1/pi,1/pi).

[gritty widget]

Uh

What

[gaunt linden]

For example, 42 is not in A, and therefore it should be in your expression for R - A.

On the other hand 1/2pi is in A, so it shouldn't be in R - A. But 1/2pi is in (-1/pi, 1/pi).

[gritty widget]

Oh

Yea

Mb

Wait then the order of my interval isn't correct either

[gaunt linden]

So now its not a union of open sets anymore. :-)

[gritty widget]

Man

Wait

[lethal oxide]

Two loop is homotopic, it is possible right?

[cedar pebble]

what

[lethal oxide]

Meaning f and g two loop a point(let x be point and same point for loop f and loop g) f homotopic to g. This is correct?

[cedar pebble]

[white oxide]

What if, dare I say, they three loop a point?

[cedar pebble]

I have no idea what you are asking

[white oxide]

😭

[kindred cairn]

this looks like a treasure map

[white oxide]

Okay, interpretation time:
x is a point and you have two loops f and g? And you are asking if they are (always) homotopic?

[lethal oxide]

Yes

[white oxide]

The answer in general is no, there is something called the fundamental group which measures how many ways there are there are how loops of a point x can fail to be homotopic

If your space is called "simply connected", then the answer is always yes.

For example any retractable space works, like R, R^2, ...

If it has a hole, then the answer can be no.

[lethal oxide]

Oh okay thank you @white oxide

[thorny siren]

Does anyone have idea(definition etc.)on about ambient isotopy? I was working on about knot theory and I saw ambient isotopy.

[opaque zodiac]

Usually isotopy is like homotopy but the space stays homeomorphic the entire time

[thorny siren]

Can you give some more details?( On about of the spaces stays homeomorphic the entire time)

[opaque zodiac]

Uh maybe I can

Uhhh

[white oxide]

Have you checked the wiki page?

[opaque zodiac]

Oh yeah I didn't actually know what 'ambient' meant in this context
https://en.wikipedia.org/wiki/Ambient_isotopy

[white oxide]

Nlab provides the same definition

So isotopies are homotopies which are embeddings for each t, that is it is injective for each t and and open (that is no self-intersection + geometric niceness). Ambient isotopy is an isotopy that deforms the ambient space homeomorphically together with the object you are looking at

[thorny siren]

Thanks guys

[unreal gull]

https://web.math.ku.dk/~moller/e02/3gt/opg/S22.pdf the solution to this problem is the first one in the link that i sent

I don't see how the solution proves the problem statement

namely, how does showing U is open show that p is a quotient map?

additionally, in the solution, it says "Let $U$ be a subspace of $Y$ such that $p^{-1}(U) \subset X$ is open," but by def of $p$ being continuous, if we just assume that $U$ is open, then $p^{-1}(U)$ must be open.

[gentle osprey]

Henry

[paper wedge]

a surjective map p: X-->Y is a quotient map if U is open in Y <--> p^-1(U) is open in X

so u want to prove this iff statement

one direction is already given by the fact that p is continuous

[unreal gull]

yeah

oh i see

its like a contradction proof

[paper wedge]

wdym

no it's direct

[unreal gull]

if we assume that U is closed such that p^{-1}(U) is open, then we arrive at a contradction

[paper wedge]

where is the proof assuming U is closed

[unreal gull]

nowhere

they just say for any U

[paper wedge]

i thought u were asking

for the proof in the link 😄

[unreal gull]

yeah

i am

[paper wedge]

okay so you know whenever U is open in Y , p^-1(U) is open in X

this is the definition of continuity

you want to show that

[unreal gull]

yeah

no we know that

we want to show the other way right

[paper wedge]

if p^-1(U) is open in X, is U necessairly open

in Y

thats what you want to show

[unreal gull]

yeah

ig

[paper wedge]

so he chooses U to be a "subspace lmao" such that p^-1(U) is open in X

[unreal gull]

wait actually i thought we want to show that if U is open in X, then p(U) is open in Y

[paper wedge]

is U open in Y?

[unreal gull]

yeah ig thats what they proveed

[paper wedge]

yeah so no contradictions here no nothing

just directly assumed p^-1(U) is open in X and showed that U is open in Y

[unreal gull]

wait

[unreal gull]

but how does that show this

[paper wedge]

not all quotient maps are open

what u described here describes an open map

that's not what a quotient map is

you should revise your definition of aq uotient map again i think ur slightly confused

[unreal gull]

OHHHHHH

yes

yes

sorry

thank you

[paper wedge]

np 😄 hah

[unreal gull]

lmao

[paper wedge]

haha

[unreal gull]

thats so stupid of me

theres so many defiition when learning topology

that differ in like one word

[paper wedge]

no it's okay

[unreal gull]

or one direction of implication

💀

[paper wedge]

yeah ig you will get used to like the logic quickly

have fun

[unreal gull]

hopefully

thank u tho

[paper wedge]

no need

[prime elbow]

Munkres defined compactness on a space. When we say A is subset of X and A has open covering so should I take A as subspace or subset?

[fading vale]

It doesnt matter, which you can probably prove yourself

A is compact in X iff A is compact in itself (with the subspace topology)

[prime elbow]

Yes

But they proved (0,1] is not compact because there is a collection of open set { (1/n, 1] | n in N} covering (0,1] but not finite collection of it covers (0,1 ].

So I thought how (1/n, 1] is open set in R but then I saw it is open in (0,1].

[fading vale]

Yes they proved that it is not compact by working intrinsically within it, in the subspace topology on (0, 1]

This is equivalent to it not being compact within R, so they could have also taken the open sets (1/n, 2) in R, which cover (0, 1], and have no finite subcollection which still covers it

of course you can replace 2 with any real number greater than 1, so long as when you intersect the interval with (0, 1] you recover (1/n, 1]

[prime elbow]

Okay

[prime elbow]

Now I got it, thank you

To show { 1/n | n in N } is not compact.

Let the collection of open set ( 1/n - r, 1/n +r), where r<<0 so any finite collection of it cannot covers { 1/n | n in N }, right?

[fading vale]

well no single such value of r exists so this argument does not work

You want to be more careful in constructing your intervals, and make sure you know that (and understand why) there is no finite subcover

[prime elbow]

Maybe we can use that if it is compact then it will be closed in R but since it is not closed in R so it cannot be compact

But I don't know any other way to prove it

[red yoke]

Yes, if you already know that your definition of "closed in R" matches the topological definition

[red yoke]

I think you mean something like (1/(n+r), 1/(n-r))

Or (1/n - r(n), 1/n + r(n)), where r(n) depends on n

[white oxide]

Keep in mind you are allowed to keep making the r smaller

[prime elbow]

what do you mean?

[prime elbow]

no

[prime elbow]

i thought r be fixed but maybe it is not correct

[red yoke]

If r is fixed, there is a finite subcover

[prime elbow]

i think so

[red yoke]

Do you know how to show it is closed?

[prime elbow]

yes because its closure it is not {1/n | n in N}

so it is not closed

[red yoke]

And why compact subsets of R are closed?

[prime elbow]

i proved that if A is compact in hausdroff space then it will be closed

R is hausdroff space

[red yoke]

Yea that seems correct

[prime elbow]

thank you

let $(X,d)$ be metric space and $f \colon X \mapsto \mathbb{R}$ such that $f(x) = d(x,a)$, where $a$ is fixed and $a \in X$.

i want to show $f$ is continuous map.

now i use epsilon delta definition here, for any $\varepsilon > 0$ and $x \in X$ and $y \in X$, $ | f(x) - f(y) | < \varepsilon$. Since $f(x) = d(x,a) $ it implies that $ | d(x,a)- d(y,a) | < \varepsilon$.

if we take $\delta = \varepsilon$ such that d(x,y) < \delta.

then for $d(x,y) < \delta$ imply that $ | d(x,a)- d(y,a) | < \varepsilon$.

hence $f$ is continuous mapping.

is it correct?

[gentle osprey]

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

[white oxide]

Maybe name-drop the inverse triangle equation more explicitly, but yeah

[prime elbow]

okay, thank you

[prime elbow]

so is there construction argument?

[red yoke]

Yes

Try to find a cover where each open set covers only one element

[white oxide]

Consider drawing a picture

Ideally with a "exponential scale" approaching 0

[prime elbow]

maybe i can visualize geometrically

[white oxide]

As a hint:
|| You can separate all the naturals in R by disjoint open sets can't you? Can you somehow reduce your problem to this setting?||