point-set-topology

[quartz horizon]

you should just call closed sets coopen, cause you’re cooped up in them

[fringe thorn]

is this related to the famous .coose command?

[quartz horizon]

the what

this is a cat theory joke

[fringe thorn]

don't worry about it, just a silly mathcord joke :p

[tribal palm]

where does the word "oposed" come from

[alpine nest]

Someone's deranged imagination

[opaque zodiac]

Oposed is pretty clever

Anyway yeah like we all know that 'open isn't the negation of closed' is confusing but given (a) it's pretty fixed terminology and (b) 'clopen' is a really good portmanteau phonetically and doesn't sound like any existing words, I feel like there's a strong case that clopen is a reasonable word

[alpine nest]

Absolutely, I use it a lot

I usually work in the Cantor space, so clopen sets abound

[opaque zodiac]

It is kind of funny though how different the reaction to it is in different fields. In some places people seem to think that nobody ever says it seriously but in other places (like logic) it's used all the time

[alpine nest]

Yeah, in my area no one bats an eyelid at the word

[opaque zodiac]

Wait what is your field again?

[alpine nest]

Intersection of topological dynamics and ergodic theory, with particular focus on symbolic dynamics

Hence the Cantor space

[opaque zodiac]

Ah nice

[alpine nest]

I mean, technically it was my field, since I'm on indefinite hiatus research-wise

[opaque zodiac]

ah

[heady skiff]

how does one show that $\overline{S \cup T} \subseteq \overline{S} \cup \overline{T}$? assuming x is a limit point of $\overline{S \cup T}$, every neighborhood $U$ of $x$ contains a point of $\overline{S \cup T}$, but this point doesn't always have to be in either $S$ or $T$, right?

[gentle osprey]

okeyokay

[gritty widget]

What definition of closure are you using?

[heady skiff]

Union of limit points with original set

[gritty widget]

There is another definition that would make this extremely easy if you know about it

Closure of X is smallest closed subset containing X

But it's okay if you don't know about it

Why are you talking about limit points of the closure of (SuT) instead of limit points of (SuT)?

[heady skiff]

oh i meant limit points of (SuT)

[gritty widget]

So, x is a limit point of SuT

That means every nhood of x intersects with either S or with T

[heady skiff]

ye

but how does that mean that x is a limit point of S

or, if it's not a limit point of S

that it's a limit point of T

right because we can take some neighborhood U of x and it can intersect with S U T

but say it only intersects with an element in S

similarly take another neighborhood U' and say it only intersects with an element in T

then we can't say it's a limit point of S or T

[gritty widget]

Yep

[heady skiff]

yeah so I'm confused as to why a limit point of (S U T) is either a limit point of S or a limit point of T

[gritty widget]

The only way it could fail

Is if there is a neighbourhood V_1 of x not intersecting S

And a neighbourhood V_2 of x not intersecting T

Right?

[heady skiff]

well what if all neighborhoods U of x intersect either S or T, but never intersect both

because then technically x would be a limit point of (S U T)

but not a limit point of S or T

[gritty widget]

It could still be a limit point of S (or of T)

For example if they all intersect S, but never T

[heady skiff]

right, but say such a point exists

then it wouldn't be a limit point of either right

because if we consider some limit point x of (S U T) then it could certainly meet that condition right

that's the edge case i'm worried about

oh wait

yeah it's not even true

wait nvm

ignore the last three messages

[gritty widget]

What could a priori happen

Is that there are some nhoods that only intersect T

And others that only intersect S

[heady skiff]

right

so is my logic a priori?

[gritty widget]

You weren't asking about this case

[heady skiff]

oh i was, sorry if i wasn't clear

[gritty widget]

.

Let V_1 and V_2 in the above conditions

What can you say about their intersection

[heady skiff]

ah right my fault

well it doesn't contain any points of S or T

i'll come back to this, wasting too much time on this small detail in my reading lol

rusty on point set it seems

thanks for the help

[heady skiff]

could i get a hint for 1)? it's the intersection condition for a basis

[thorny agate]

Stuck on step (4) and (5) for this this hint for (1)

Ok so I have a maximal filter F for {x_lambda}

I know that a filter F is an ultrafilter if and only if for all subsets A of X that either A or X \ A is in F

I have shown that if {x_\lambda} is not frequently in A then it must be frequently in X \ A.

But this does not immediately say that for such a maximal filter F that X \ A must be in F

so I'm not quite sure how to do this

I want to say that towards contradiction, suppose that we also do not have that A^c is in F

then I should be able to somehow construct a filter G such that G is a filter for {x_lambda} and G contains both F and A^c

but idk how to construct such a filter.

[thorny agate]

ahhhh sets with FIP generate a filter

I think that gets me (4)

not sure about (5) still 💀

[thorny agate]

nvm it does not get me (4)

[low flame]

does infinity here mean inf in cardinality

or does it mean a set that goes to infinity? for example (-inf,0)

because (0,4) is infinite cardianlity in R as well

[fringe thorn]

I would presume so

[low flame]

okay. i could have something like (0,4) intersect (4,10) which have inf cardinality, but their intersection gives only 4, thus is a singleton right?

[fringe thorn]

the intersection is empty

if x is in (0,4) \cap (4,10), then 0 < x < 4 and 4 < x < 10

but there's no real number which satisfies this

so the intersection is empty

[low flame]

intersection of (-inf,0) u {1} and (0, inf)

[fringe thorn]

that intersection would be {1}, yes, but what would you do with that?

oh, I see

[low flame]

to show its not a topology

[fringe thorn]

yes, I misread the question

I think that would work then

unless, of course, by "infinite subset", they mean smth like the open rays (a, inf) and (-inf, b)

[low flame]

yeah the question has two ways to look at it

[fringe thorn]

but I would probably interpret infinite subset to mean infinite cardinality

[low flame]

if thats the case with rays its simple since you can just go (-inf, 5) cap (-3, inf) and its not an inf ray

oh wow infinite subsets of R, even numbers and prime numbers. they intersect at 2 which is finite

[heady skiff]

here is the map described on the last line continuous since if we have $\prod_{i \in I} U_i$ such that $U_i = Y_i$ for all but a finite number of the $i$, then it's preimage is just going to be $\bigcap_{i = 1}^n f_i^{-1}(U_i)$ which is open in the weak topology

[gentle osprey]

okeyokay

[craggy cedar]

Yes

[heady skiff]

here are we assuming that U and V are disjoint and nonempty?

[opaque zodiac]

Yes

[prime elbow]

In normed vector space X, if x_n converges to x then || x_n || converges to ||x||. Because, | d(x_n,0) - d(x,0) | ≤ d(x_n,x), since d(x_n,x) converges to 0 so d(x_n,0) converges to d(x,0).

Is it correct?

[uncut tartan]

oh you meant to say ||x_n|| converges to ||x||

[heady skiff]

is it true that if X = U u V is a separation of X then cl(U) and V are disjoint since otherwise, there would exist a limit point of U contained in V, and since V is an open set, this would intersect U

[prime elbow]

If U and V are separation that means U and V are disjoint clopen so cl(U) = U and cl(V) = V

[prime elbow]

Yes

[slim raft]

I don't know if this is the right place to ask the question but I wanted to know what will the centre of a sphere with infinite volume . Will every point inside the sphere will become a center or like regular sphere it's center will remain fixed

[gritty widget]

What do you mean by a sphere with infinite volume?

[slim raft]

The same that we mean by infinity

Or lemme reframe the question

In our culture it's said that

The consciousness or god whatever u like to call it

Is present everywhere

If I go with this analogy

Then each point in our universe space time

Will be identical to each other

[haughty jungle]

[slim raft]

Nd if that happens then the only solution that I can come up with is this universe converges to a single point

But that is not so in our experience

So either what we r experiencing is a lie

Or my assumption is false

Nd I am pretty sure my assumption is not false

[haughty jungle]

This is like some Terence Howard stuff fr

Ppl try and use math to make like grand statements about the universe but lowk I rarely find them to be meaningful

[sonic crane]

Let them have their fun hehe

[alpine nest]

In R^n a ball with infinite volume would have to be the whole space, have infinite radius, and the notion "centre" ceases to be well-defined.

(and a sphere in R^n has volume 0 whatever the radius)

Unless of course you're not using the Lebesgue measure, but then you would need to specify what measure you are using.

So yes, the centre of a ball of infinite volume can be said to be everywhere, if that helps

[gritty widget]

I guess you could say it is a ball of infinite diameter, in which case any point works as a center

[alpine nest]

Yeah, that's what I meant

[heady skiff]

where am I misinterpreting things? if we take X = (-1, 1/2) U [1, 6) and p = 1/2 this is not true, right?

oh they said point of the closure of A in X

so 1/2 wouldn't be in X so it wouldn't count

[quick delta]

Yes

[civic verge]

Hi guys could someone tell me that what I have done makes correct logical sense I just did (=>) I'm missing the opposite one

Let $f: U \subset \mathbb{R}^{n} \to \mathbb{R}^{m}$ be a bijection onto its image. Then, $f$ is a homeomorphism if and only if $f(\overline{X}) = \overline{f(X)}$ for every $X \subset U$.\

$(\Rightarrow)$ Assume $f$ is a homeomorphism. If $X \subset U$, we have that if $\overline{X} = X$, then $X \subset \overline{X}$, where $\overline{X}$ is the smallest closed set containing $X$. Since $f$ is a homeomorphism, we have:
$$f(X) \subset f(\overline{X})$$
where $f(\overline{X})$ is a closed set in $\mathbb{R}^{m}$. Then, we have:
$$f(X) \subset f(\overline{X}) \Rightarrow \overline{f(X)} \subset \overline{f(\overline{X})}$$
By the property $\overline{\overline{Y}} = \overline{Y}$, we obtain:
$$\overline{f(X)} \subset f(\overline{X})$$

$(\Leftarrow)$ Assume $f$ is a homeomorphism and $X \subset U$. Let $\mathbb{B}{\varepsilon}(f(x))$. Then, $f^{-1}(\mathbb{B}{\varepsilon}(f(x)))$ is an open set in $U$ that contains $x$. Hence, $V = X \cap f^{-1}(\mathbb{B}{\varepsilon}(x)) \neq \emptyset$, meaning that $V$ intersects at some point $y$. Similarly, $\mathbb{B}{\varepsilon}(f(x))$ intersects $f(X)$ at point $f(y)$, so $f(x) \in \overline{f(X)}$.

[gentle osprey]

Homology

[quick crane]

aren't you missing the proof for $f(\overline{X}) \subset \overline{f(X)}$?

[gentle osprey]

L

[prime elbow]

I don't know if the statement is true or not, if E is normed vector space, let y in E and fixed y then x -> (x | y ), where ( x | y ) denotes inner product, is it mapping continuous?

If yes, then I tried but I don't know how to show that | (x|y) - (z | y ) | < e when d(x,z) < delta.

[ionic bay]

Are you allowed to use bounded linear maps --> continuity?

If so, then the result is obvious from Cauchy-schwarz

Hmm, well bounded linear maps is an overkill.
note that (x|y)-(z|y)=(x-z|y) and use Cauchy-Schwarz

[prime elbow]

I don't know this stuff

[prime elbow]

I see

Yes I know if L( | x | ) < M( | x | ), then L is continuous and if L is continuous then it is bounded linear map

[ionic bay]

bounded linear maps and continuity are equivalent

[prime elbow]

x-> ( x | y ) is linear mapping

I need to show this is bounded, right?

[ionic bay]

bounded in the sense that for all v in E with norm leq 1, | (v|y) | leq M for some M>0.

[prime elbow]

Yes

[ionic bay]

Yeah you need to show it's bounded then

[prime elbow]

Isn't | ( x|y ) | ≤ M | x |

[prime elbow]

I see

| v | < 1 so | (v|y) | < M

Both are equivalent, right?

[ionic bay]

You need to show this first, it can be done with Cauchy-Schwarz

[ionic bay]

yep

[prime elbow]

Because Cauchy- Schwarz says that | ( x | y ) | ≤ | x | | y | so here we can take M = | y |

[ionic bay]

or alternatively you can use epsilon-delta by noting that |(x|y)-(z|y)| leq |x-z|*|y|

[prime elbow]

I see

So by using this, if F is subspace of E, then complement of F means set of all x which are orthogonal with every element of F is closed subspace.

For closed, let x_n converge to x, since that mapping is continuous so ( x_n | y ) converges to ( x | y) and (x_n | y ) = 0 for all n in N.

Hence ( x | y ) = 0.

Is it correct?

[prime elbow]

I am not sure because they used | | this operator for both real and complex so | | denotes Euclidean metric, right ?

[ionic bay]

Well the notation || || is used to denote a norm

[prime elbow]

Yes

But our co-domain of mapping is K, K can be real or complex

[ionic bay]

It can be any norm since all norms are equivalent on finite dimensional vector spaces

[prime elbow]

So we have | (x|y) - (z|y) | ≤ | (x-z | y) |, so now here we used Cauchy Schwarz,

| (x|y) -(z|y) | ≤ | x - z ||y|.

[prime elbow]

I have to prove it

[ionic bay]

haha it's a fun exercise. I just did it a few days ago.

[prime elbow]

Okay thank you ❤️

[prime elbow]

Banach space is closed, right?

Let X be non-empty set and E be Banach space and B(X,E) denotes the set of all bounded functions from X to E. Then B(X,E) is Banach space.

If it is Banach space, so when I say B(X,E) is closed so in which relative I say it is closed?

[ionic bay]

Every convergent sequence is cauchy and so Banach space is always closed

[prime elbow]

But it is a metric space so it is always closed, right?

[ionic bay]

I don't quite get your question sorry

I'm not really sure B(X,E) is Banach tho. It is Banach when X is a normed vector space, E is banach and B(X,E) is the set of bounded linear map from X to E

[prime elbow]

No I think X not need to be a normed vector space, E needs to be normed vector space

And Banach space

[ionic bay]

Hmm what's the norm on B(X,E)?

[prime elbow]

| f | = sup{ |f(x)|_E | x in X }

|x|_E denotes norm of E

[ionic bay]

Hmm I guess so yeah

[granite crane]

do i need to show that G(y) is well defined? (this is an example in Lee's ITM book)

for continuity of $F$, I showed $\lim_{x\to p} F(x)=F(p)$

[gentle osprey]

Afzal

[granite crane]

is it fine or there is some other way?

[low flame]

so im doing 3

and i was thinking isnt the empty set considered open

and then the compliment of the empty set is X

thus X is closed

so X - Ø = X

[tender halo]

yea

its both open and closed

[low flame]

same with X too right? i can do like X-X

which is the empty set

then X is open, and the empty set is closed haha. yeah. okay thanks their clopen

[low flame]

for one right, X and the empty set are clopen. Also, wouldnt the discrete topology also be clopen?

[quick delta]

Every open set in the discrete topology is clopen, yes

[low flame]

okay so the discrete space is where each element contains itself ie T = ({Ø}, {0}, {1}, {0,1}) i didnt feel like doing up to two but it would be bigger. its almost like its the power set. Then essentially every set is the compliment of the empty set, and the empty set is open then every other set is closed. Then you can reverse this and say that the empty set is the compliment of every set, and since its open, then the rest are closed

this good for a?

[gritty widget]

What do you mean by each element contains itself?
It is the power set.
The only set whose complement is the empty set is the whole space, so that is not correct

[low flame]

okay so its only a compliemnt with the entire space, rgr

[tender halo]

sets only have one complement

[low flame]

Okay empty set is open by being in T. But then its compliment is X which is {0,1}. then Ø is open, and {0,1} is closed

then the singleton {0} has the complment of {1}, {0} is open, then by def {1} is closed. this works the other way around too.

Then if {0,1} is open, its compliment Ø is closed.

[low flame]

alright ill keep that in mind

[low flame]

can anyone explain to me what question D is asking for?

[raw stag]

What space do we get if we take a mobius strip and identify it's boundary circle to a point

Like uhh i tried to draw fundamental polygons and stuff

But like i can't see it lol

Is there a nice space this is homeomorphic to

[quartz horizon]

I think the real projective plane…?

[raw stag]

RP^2?

[quartz horizon]

Yeah

[raw stag]

Uhhhh

Elaborate

[quartz horizon]

So you draw the fundamental polygon of the mobius strip

A square with the left edge identified appropriately with the right edge

Then you shrink the top edge to a point, and the bottom edge to a point

[raw stag]

Oh right i see

And what happens to the interior

[quartz horizon]

You get a circle with the left half identified appropriately with the right half

That gets ya the real projective plane

[raw stag]

But the fundamental polygon has an interior right

What happens to that

[quartz horizon]

Yeah that gets affected by the squishing

Goes to the interior of the circle

[raw stag]

Confused

[quartz horizon]

I can tell you how the homeomorphism works if you want

[raw stag]

Sure

[quartz horizon]

Let’s start with a solid square

[0, 1] x [0, 1]

We want to collapse the top edge to a point, and the bottom edge to a point

Different points though

And we want to show the resulting space is homeomorphic to a solid disk, right?

[raw stag]

Agreed

[quartz horizon]

Now, do you know about the universal property of the quotient?

[raw stag]

Yea

[quartz horizon]

Cool cool, that’ll make this easier

[raw stag]

But it has to be same on the identifications

[quartz horizon]

Let’s choose our solid disk to be the one centred at (1/2, 1/2) with radius 1/2

So it’s “inscribed” in the square

[raw stag]

Okay

[quartz horizon]

What we’re going to do is horizontally squish the square into the circle

[raw stag]

Okay radial projection stuff?

[quartz horizon]

Not quite

The points will keep their y coordinates

Only the x coordinate will shift

[raw stag]

Okay not radial projection then

[quartz horizon]

Hmm I’m realising it will be more convenient to be centred at the origin

Ok so let’s use [-1, 1] x [-1, 1], and a disk centred at the origin of radius 1

Given a point (x_0, y_0) in the square

The places where the line y = y_0 intersects the circle are (+- sqrt(1 - y_0^2), 0)

So what we do is send (x_0, y_0) to (sqrt(1 - y_0^2) x_0, y_0)

Do you agree this is a continuous map from [-1, 1] x [-1, 1] to the closed disk of radius 1 centred at the origin?

[raw stag]

Perhaps

Okay say it's cts now what

[quartz horizon]

This map is constant on the top edge, and on the bottom edge

[raw stag]

Agreed

[quartz horizon]

So it induces a continuous map from the quotient space to the closed disk

[raw stag]

Agreed

[quartz horizon]

You can verify this is a bijection too

And then you have a continuous bijection from a compact space to a Hausdorff space, so it’s a homeomorphism

[raw stag]

Okay okay

But what about the Rp2

[quartz horizon]

So, now you start with the fundamental polygon

With the identifications of the left edge with the right edge

Apply this homeomorphism

This gets you a disk with the identifications of the left edge with the right edge

And that’s RP^2

[raw stag]

I lost you

There is no identifications on the interior

How is it RP2

[quartz horizon]

I’m not saying there are

It is true that if you have a closed disk, and you identify antipodal points, you get the real projective plane

[raw stag]

Antipodal points just on the boundary?

[quartz horizon]

Yeah

[raw stag]

Or everywhere?

[quartz horizon]

Just on the boundary

[raw stag]

Interesting

I don't quite believe that ig

Non trivial to me

[quartz horizon]

What’s your definition of the real projective plane?

[raw stag]

R^2-{0} under equivalence relationship x~y if x=zy

[quartz horizon]

I see

[raw stag]

Which can be seen as a quotient of S^1

But I don't see how it's the closed disk one

[quartz horizon]

Hmm, hang on

[quartz horizon]

Is z arbitrary here

[raw stag]

z is a real number

Non zero

[quartz horizon]

Ok then I don’t understand this definition

[raw stag]

It's just lines identified as points

[quartz horizon]

RP^2 should be 2-dimensional

[raw stag]

x,y are elements of R^2-0

[quartz horizon]

Your construction just seems to give S^1

[raw stag]

No

[quartz horizon]

Right but by making z arbitrary, you collapse it to 1D

[quartz horizon]

As you say here

[raw stag]

No i said x~y if x=z.y

[quartz horizon]

This sounds like RP^1

[raw stag]

Okay

[quartz horizon]

Which is just S^1 again

[raw stag]

Then perhaps I have notation problems

What is your RP^2

Don't say the closed disk one ;-;

[quartz horizon]

I’m not sure what you mean by the last line defining it to be zero

[quartz horizon]

The definition my geometry course gave was a quotient of S^2 by identifying antipodal points

[raw stag]

I see

Then i think I can agree that it is RP^2

I must have thought RP^2 as RP^1

(i always mess up the dimension of S^n fyi) 😭

Okay thanks @quartz horizon

[raw stag]

It would just be R^3-{0} and identifying the lines ig

[quartz horizon]

That works too

[unreal stratus]

The important thing is S^n, RP^n look like R^n "locally" (more specifically they are dim. n manifolds)

[raw stag]

,rotate

[raw stag]

Can't think of any counterexamples

Wouldn't everything workout the way it is i mean i don't see any problem to as why it should not be true

[robust drum]

My favorite way to think of it is as the space of lines, with the angular metric

In fact this def works for any RP^n

Any two lines determine a plane. Then the distance between them is the angle between them in that plane

This I feel like captures morally what’s going on with RP^n

[quartz horizon]

Why not just take the angle between the lines?

[robust drum]

Yeah that’s the same I think?

It’s just to me a little unclear how to do it in super high dimensions

Unless you just look at the angle between them in the plane they determine

[quartz horizon]

Dot product I would think

[raw stag]

Bump

[robust drum]

Here’s how I would do this problem: find an identification map X-> Y, and take a subspace A that maps surjectively onto Y via f via this map. Do it such that A/{Relation determined by images of points} is not homeomorphic to Y (easiest way to do this is to just make it also map injectively)

There’s a theorem that says that were it an identification map, A/{relation determined by images of points} would be homeomorphic to the image, in this case Y.

I can think of one specific example RN involving RP^n

[raw stag]

Surjectivity here must mean that I must have atleast one representative from each equivalence class

Uhh

Idk

What example did u have in mind

[robust drum]

Well I’m thinking that if you take the top hemisphere of S^2, and delete points on the boundary circle in angles [pi, 2pi)

This contains one point from every equivalence class

But it’s basically just a disc (with part of the boundary circle deleted)

So the projection to RP^2

Can’t be an identification map

A way simpler example: consider the interval [0, 1] and identify 0 to 1.

Then let A = [0, 1)

A/~ = A

And also, the map restricted to A is surjective

Yet clearly A is not a circle

[last marlin]

Given A to be a subset of X, if g and h are two continuous functions from closure of A to some space Y such that g=h on A, then prove that g=h on X.

Here is my solution:

Please check if it is even correct.

If it is correct, then this seems like a very "ugly" way of going around it.

Is there a better way as well?

[tender halo]

what if A is empty?

then your statement means that any two continuous functions on X are equal

[tiny obsidian]

This statement is very much not true as you've written it

You're also using Y hausdorff in your proof? You haven't told us this is part of the problem

Can you write down the exact statement you're trying to prove

I would assume it's to prove that f = g on A closure and not X, with the extra assumption that Y is Hausdorff

The basic idea of taking separating neighbourhoods of f(x) and g(x), and using that to lead to a contradiction back in A closure works, although I think you've overcomplicated somewhat in the middle

[tender halo]

easy proof: take cartesian product of f and g, the diagonal of Y x Y is closed since Y is Hausdorff, and its preimage is exactly the set where f(x) = g(x)

[prime elbow]

Any hint to show that any linear function from R^m to R^n is lipschitz?

[quartz horizon]

Try to show continuity first

[wispy veldt]

its enough to verify continuity at 0

from there you can consider contradiction to show that it has to be lipchitz

altho other paths lead to the same conclusion too

note that R^m and R^n are vector spaces of finite dimension

[quartz horizon]

Do you need contradiction?

I was thinking doing a direct proof using epsilon-delta

[wispy veldt]

you don't need it but its how i did it

[prime elbow]

Yes if it is continuous at one point then it will be bounded

[wispy veldt]

.

[prime elbow]

Yes I want to generalize this but I don't know how to show it

[prime elbow]

I tried sequence criteria but it didn't help me

[quick crane]

My hint is $$|Ax - Ay| = |A(x - y)| = |x - y|A(\frac{x - y}{|x - y|}).$$

[gentle osprey]

L

[wispy veldt]

so here is the contradiction i was using (which works for the general case too) to show continuity -> lipchitz

[gentle osprey]

James Banach*-alg
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[wispy veldt]

i think you can do this directly too tho

[wispy veldt]

so you are struggling with showing continuity at 0?

[prime elbow]

Should I know show that supT(|z|) over |z| = 1 ?

[wispy veldt]

try to write down the epsilon delta definition of what it means to be continous at 0, and remember how the basis plays a role representing each x

[prime elbow]

Yes

[prime elbow]

We are taking the orthonormal basis, right?

[wispy veldt]

yes, but any basis works

[prime elbow]

Yes

So | T(x) | ≤ (| a_1 | +...+ | a_n |) max { T(e_i )}

[quick crane]

Yeah I use $|Ax - Ay| \leq |x - y|\sup_{|v| = 1}|Av|$. The remaining step is to show $\sup_{|v| = 1}|Av| < \infty$.

[gentle osprey]

L

[wispy veldt]

that works too

[prime elbow]

So max { T(e_i) } exists

Thus we have |T(x) | < c |x |

[wispy veldt]

yeah, most proofs will rely on using a representation of x by a basis

rest follows from there

L's proof and proving continuity at 0 can be proved with a similar reasoning

[prime elbow]

I see

So this can be generalized to finite dimensional vector space

[wispy veldt]

do note that you dont need finite dimensionality to show that continuity -> lipchitz for a linear operator

[prime elbow]

Yes

[wispy veldt]

the finite dimensional just ensures continuity

[wispy veldt]

i made a argument here by contradiction

[prime elbow]

I showed that in normed vector space continuity is equivalent to lipschitz

Right?

[prime elbow]

Yes

[prime elbow]

Linear mapping

[wispy veldt]

well you only showed it in finite dimensional case

but thats what was required from you

[prime elbow]

No I did that other day

Is it for only linear mapping, right?

[wispy veldt]

lipchitz is generally stronger than continuity

[prime elbow]

Because uniformly?

[wispy veldt]

uniformly is weaker than lipchitz

[prime elbow]

Yes

[quick crane]

Yes $\sup_{|v| = 1}|Av| < \infty$ if and only if $A$ is continuous at $0$. An $\varepsilon$-$\delta$ argument shows this.

[gentle osprey]

L

[prime elbow]

Then why is lipschitz stronger than continuity? Means how it helps me to prove better results?

[quick crane]

finite-dimensionality isn't needed for this

I guess that's why they say "bounded" linear map instead of continuous sometimes

[wispy veldt]

many theorems require lipchitz continuity, see for example picard lindelof theorem

[prime elbow]

Still I am thinking about how to show sup | Av | < ∞

[prime elbow]

Okay

Thank you ❤️

[quick crane]

The quantity $|A| = \sup_{|v| = 1}|Av|$ is a norm on the space of continuous linear maps. It's very useful, and appears in concentration inequalities and other quantitative bounds involving matrices.

[gentle osprey]

L

[wispy veldt]

also known as the operator norm

[prime elbow]

To confirm, let v = a_1e_1 + ... + a_ne_n, where e_i are orthonormal basis.

So |T(x) |≤ (| a_1| +...+ |a_n | ) max { T(e_i ) | i= 1,2,..,n. }.

But | x | < | a_1| +...+ |a_n |, so how do I say that if | x | < delta then |T(x) | < e ?

Because then I have to choose |a_1 | +...+|a_n | < e/max{T(e_i) }, T ≠ 0

[wispy veldt]

you can choose them as small as you want by making ||x|| small enough

[prime elbow]

You mean we can take | x | < delta ≤ | a_1 | +...+ |a_n |

[wispy veldt]

you can take sum |an| < c||x|| for any norm, because all norms are equivalent on R^n and n(x)=sum |an| is a norm

so whatever norm you are using for x, if its small enough

the summation can be as small as you want

and if you are using a specific norm, you can directly show that sum|an| is less than that norm times a constant c

either way is fine

[prime elbow]

I see

[last marlin]

Oh yes, sorry it is this.

[prime elbow]

Is it correct?

[prime elbow]

Sorry but I am not sure

I got it

You mean all norms are equivalent

On finite dimensional

So \sum |a_i | is a norm

So \sum |a_i | ≤ c|x|, for any norm

[wispy veldt]

indeed

[prime elbow]

Thank you ❤️

[wispy veldt]

for completion you can prove by consider (n(x)=\norm{x}) from (S={ x=\sum a_{i} e_{i}: \sum|a_{i}| =1} \to \Bbb{R} ) , notice this is a continous function from a compact set so it attains a minimum at some point (\alpha).\ meaning ( \norm{x} \geq \norm{\alpha}>0) for any (x\in S), and if we pick (u=\frac{x}{\sum|a_{i}|} ) then [ \norm{u}=\frac{\norm{x}}{\sum |a_{i}|}\geq \norm{\alpha} \implies \norm{x}\geq c \sum|a_{i}| ]

[gentle osprey]

James Banach*-alg

[civic verge]

$f(\overline{X})=\overline{f(X)}$

[gentle osprey]

Homology

[wispy veldt]

and in fact proving the norm is continuous is litearlly what you have done, N(x)<\sum|ai| max {N(ei)} [ here the norm we are working with is sum |ai| ofcourse]

since every norm is equivalent to the sum norm, they are all equivalent

the way you generalize this to any finite dimensional space X with two norms N1, N2 is by translating your work back to R^n by a certain choice of function f:R^n -> X and using the norms ni(x)=Ni(f(x))

something you'll hopefully see in introductory functional analysis

[prime elbow]

😭

[robust drum]

Doesn’t this just follow because a linear function is a polynomial in the basis vectors

Oh that’s what you guys said later ok ignore me

[quick crane]

this works in finite-dimensions, but the $\varepsilon$-$\delta$ works in infinite-dimensions (and is simpler)

[gentle osprey]

L

[civic verge]

Let ( f: U \subset \mathbb{R}^{n} \to \mathbb{R}^{m} ) be a bijection onto its image. Then, ( f ) is a homeomorphism if and only if ( f(\overline{X}) = \overline{f(X)} ), for every ( X \subset U ).\

((\Rightarrow))\
Let ( f ) be a bijective function with both ( f ) and ( f^{-1} ) continuous, i.e., a homeomorphism. If ( X \subset U ), we have that if ( \overline{X} = X ), then ( X \subset \overline{X} ), where ( \overline{X} ) is the smallest closed set containing ( X ). Since ( f ) is continuous, we have:
$$f(X) \subset f(\overline{X}).$$
Given that ( f ) is continuous, ( f(\overline{X}) ) is a closed set in ( \mathbb{R}^{m} ), hence:
$$f(X) \subset f(\overline{X}) \Rightarrow \overline{f(X)} \subset \overline{f(\overline{X})}.$$
By the property ( \overline{\overline{Y}} = \overline{Y} ) and the fact that the image of a closed set under a homeomorphism is closed, we obtain:
$$\overline{f(X)} \subset f(\overline{X}).$$
Let ( f ) be a homeomorphism and ( X \subset U ). Let ( \mathbb{B}{\varepsilon}(f(x)) ). Then ( f^{-1}(\mathbb{B}{\varepsilon}(f(x))) ) is an open set in ( U ) that contains ( x ), since ( f^{-1} ) is continuous under the homeomorphism. Then, ( V = X \cap f^{-1}(\mathbb{B}{\varepsilon}(x)) \neq \emptyset ); that is, ( V ) intersects at some point ( y ). Similarly, ( \mathbb{B}{\varepsilon}(f(x)) ) intersects ( f(X) ) at the point ( f(y) ), so ( f(x) \in \overline{f(X)}. )

Therefore, ( f(\overline{X}) = \overline{f(X)}. )\

((\Leftarrow))\
Conversely, if for every ( X \subset U ) we have ( f(\overline{X}) = \overline{f(X)} ), then the map ( f ) is continuous. If ( C ) is a closed set in ( U ), then ( f(C) = f(\overline{C}) = \overline{f(C)} ); hence, the map ( f ) is also closed.

[gentle osprey]

Homology

[civic verge]

[rancid umbra]

why do you say cl(X) = X?

in general, X is a subset of cl(X)

the fact that f is continuous has nothing to do with the fact that f(X) is a subset of f(cl(X))

there are continuous functions f which don’t take closed sets to closed sets

you mean because f^-1 is continuous

[quick crane]

The proof for $f(\overline{X}) \subset \overline{f(X)}$ doesn't make sense to me. What is $x$ and $y$?

[gentle osprey]

L

[quick crane]

why did you include a proof then if you can just cite this

[civic verge]

Because my teacher wants me to learn how to write demonstration (?)

[civic verge]

I am assuming that the closure of X is X, that is to say that X is a closed set, I should have written it better I suppose, I suppose that there will be counterexample, but the idea is that I am assuming that X is a subset cl(X), to apply f continuity, for something it is called theorem because it will not apply to everything or else it would be called definition, I would think not?

Being in the para of => I am assuming that f and its inverse is f is continuous and is bijective, then I am guaranteed that every neighborhood of Y is an open in X, by definition I am assuming that f is continuous and is bijective, then I am guaranteed that every neighborhood of Y is an open in X, by definition I am assuming that f is continuous and is bijective.

[rancid umbra]

im honestly not sure what is going on with the last part of your proof for the ==> direction

for the <== direction, you are citing the lemma that you said you can’t cite by deducing that f is continuous

[civic verge]

=>
I am arguing that cl(cl(Y))=cl(Y) since I have f(cl(cl(X))) and since this set is closed under the continuity of f then we have that f(cl(X))

and the last part I quoted munkres, clearly he wants to prove this from adherence.

the returned <= I have to assume that f(cl(X))=cl(f(X))

[ocean canyon]

This is a perhaps a basic question, but I am confused: in Lees ”Topological Manifolds”, he let τ be the collection of all unions of elements of B, where B is a collection of subsets of a space X such that:

• U_{B ∈ B} B.
• If B_1,B_2 ∈ B and x ∈ B_1 ⋂ B_2 then ∃B_3 ∈ B such that x ∈ B_3 ⊂ B_1 ⋂ B_2.

He then says that we see that any B_i ∈ B is in τ (by which I suppose he mean that B_i ∈ τ) since (paraphrasing) B_i is a union of a one-element set. But is not the one-element union UB_i and not U{B_i}? It is the latter we need to say that B_i ∈ τ.

Or should I think of it as U{B_1} = B_1 but for U{B_1,…,B_n} = B_1 U … U B_n?

[fringe thorn]

B_i is the union of {B_i}, yes

[ocean canyon]

Yes, I’m aware of that. I don’t think that is where my confusion lies.

[fringe thorn]

Where does it lie?

I think I didn't understand your question then

oh, I think I see it

[fringe thorn]

I believe this is wrong, actually

B_i is the union of itself

we're not looking at sets of sets here

maybe to clarify what I'm saying here, the elements of τ will look something like B_1 U B_5 U B_7, not {B_1} U {B_5} U {B_7}

so B_i is the union of B_i itself, which is what I believe Lee is getting at

[thin tide]

Yeah I didn’t realize what the question was at first lmao

You can think of open balls in R^n or something as an example

[ocean canyon]

Right, yes, I agree with this. I don’t see how that changes what I asked. I mean, assuming we are working in ZFC, surely U{B_i} = B_i by definition? Like, x ∈ U M ↔ ∃A ∈ M s.t. x ∈ A. Now take M = {B_i}. By extensionality UM = U{B_i} = B_i.

[thin tide]

I think you’re looking into it too deeply ngl

[ocean canyon]

I can’t parse this sentence, what are you saying (if you convert it to symbols) that is different from what you said earlier?

[fringe thorn]

I retracted my initial statement, actually
#point-set-topology message

[fringe thorn]

this is a bit beyond my pay grade, unfortunately

I'm not the one to ask about that

[thin tide]

Can you show a screenshot or picture of the passage @ocean canyon ?

[ocean canyon]

But your original statement is correct, assuming our background (meta) theory is ZFC

[fringe thorn]

ah, hmm

I don't really know much set theory or logic

[fringe thorn]

I'm pretty sure Benjamin is talking about this passage?

[thin tide]

Yeah I mean in this case we know what it’s supposed to be, the union of B_i’s in \mathcal{B} should cover the space and every open subset should be writable as the union of some elements of \mathcal{B}

So I don’t see where the notation issue he brought up is here

[ocean canyon]

Yes, sorry @thin tide I’m in bed (swedish time)

[thin tide]

No worries I was just trying to figure out what you were referring to

[ocean canyon]

I only found an earlier ed. online by quick search and it is formulated somewhat differently there

[thin tide]

Whenever you get the chance to show it @ me cause I’m interested in seeing what youre talking about

[ocean canyon]

Like generally U B_i is not equal to B_i since U B_i are the elements of the elements of B_i

[thin tide]

but that would only be if $B_i = \bigcup_{j} {B_i}_j$ right?

[gentle osprey]

chipotle

[thin tide]

Whereas in this case B_i is just one of the open subsets

[ocean canyon]

I’m not sure, I mean, x ∈ U B_i iff ∃y ∈ B_i such that x ∈ y. Now if B_i is transitive then x ∈ y ∈ B_i ⇒ x ∈ B_i.

• So if B_i is transitive then U B_i and U {B_i} coincide, AFAICT

[thin tide]

I think what’s going on here is that implicitly we have $i \in I$ for some index $I$ so the union of $B_i$’s would just be $B_1 \cup B_2 \cdots$ with subscripts ranging over $I$ meaning we’d just be left with one set containing all elements contained in some $B_i$

[gentle osprey]

chipotle

[ocean canyon]

Hmmm, I mean, it is not impossible that this is the case

[thin tide]

Well in this case that’s the definition

[ocean canyon]

It is the only reasonable way to view it so far

Atleast

[thin tide]

Look up “open cover” and see the definitions various sources give

[ocean canyon]

I don’t think we should assume that B_i is transitive since this is not true in V (the universe of all sets)

[ocean canyon]

I think I will tentatively go with this, atleast, thanks!

[thin tide]

No problem

Btw when I said open cover I was referring to the thing about unions, being a basis is a stricter condition cause it requires every open subset to be the union of some elements of \mathcal{B} which is the second condition in the definition you initially sent above

The first condition is what you were asking about but just wanted to note that

[ocean canyon]

Sure, I thought you rather meant that some source might explain how to think about unions with one elements. My interpretation of what you said is that if. |I| = 1 then we ”cut-off” the union B_1 U … U B_n before it starts, ie at B_1, ie we are just left with B_1. Was this a correct interpretation of what you meant?

[thin tide]

Yeah

And similarly if I = {1,2} or something then the union is B_1 U B_2 = {x such that x \in B_1 or x \in B_2}

And so on

[ocean canyon]

Yes

But we can then easily think of it as U {B_i : i ∈ I} since if eg I = {1,2} this is B_1 U B_2, and if I ={j} it becomes U {B_j} = B_j. So AFAICT this way agrees with what you wrote now.

Somehow the way you wrote it made it click though

[thin tide]

Yeah glad I could be of help!

[ocean canyon]

Yes. This has already been solved (as far as I see it).

I don’t think this proof for why a hausdorff space X with A ⊂ X and limit point p of A is such that every ngbh of p must contain infinitely many points of A is entirely correct. My idea would rather be V = M ⋂ (⋂_{i = 1}^{k} V_i), otherwise(AFAICT) we don’t know that the V_i does not contain other points of A in common with the other V_j

[quick crane]

In the proof it says $V_k$ is open in $M$, which makes it correct

[gentle osprey]

L

[quick crane]

In your mind, $V_k$ is only open in $X$, not in $M$.

[gentle osprey]

L

[low flame]

3.4. Riddle. What topological structures have exactly one base?

indiscrete topology?

[tender halo]

hmm

not only

left/right order topologies also have that property? i think?

for example, the Alexandroff topology on 2 - {{}, {0}, {0, 1}}

[tender halo]

for linear orders at least, for posets in general its not true

also it has to be finite

or well ordered in the right direction? idk im too lazy too think about it

[gritty widget]

Every topological space has at least 2 bases: if B is a basis of X that doesn't contain Ø, you can add it, otherwise you can remove it. So maybe you want to ignore bases that contain Ø.

Ignoring those, this should be topological spaces such that whenever you take a nonempty union, the result is one of the terms of the union

[tiny obsidian]

Which can be reformulated as any family of sets in the topology having a greatest element w.r.t. subset

[gritty widget]

The open sets must be linearly ordered by inclusion because the union of two uncomparable opens would give a different open. There must also only be finitely many open sets then, else the union of strictly increasing chain of length omega would give a new set incorrect becaues could have only -omega chains

The Kolmogorov quotient of a topological space with finitely many open sets is finite

So I think we get that these spaces are precisely the ones whose Kolmogorov quotient is the Alexandrov topology of some finite totally ordered set

[tiny obsidian]

Finite isn't required

I don't think

[gritty widget]

The space doesn't have to be finite, but I believe its Kolmogorov quotient does

[tiny obsidian]

I mean you can have infinitely many open sets

As long as you don't get infinite increasing chains you'd be fine

[tender halo]

yeah it has to be anti-well-ordered

[gritty widget]

Can you give me an example of such a space with infinitely many open sets?

[tender halo]

reverse omega

[gritty widget]

Oh, I'm dumb, of course

[tiny obsidian]

Take an arbitrary infinite set

Pick an infinite subset

Then pick an infinite subset of that

Etc. forever (after infinitely many steps you can start picking finite subsets)

With the empty set and whole space, the collection at the end is a topology

[gritty widget]

Why does it have to be anti-well-ordered? I can see why it mustn't contain a chain of opens of length omega

Nevermind, I see why

So the answer is spaces whose Kolmogorov quotient is the alexandrov topology of an anti-well-ordered set?

[tender halo]

looks like it, although i imagine the intended answer to the riddle was "the empty space"

[gritty widget]

Doubt it since op even mentioned indiscrete spaces

[gritty widget]

Did you get this from some book, or did you come up with it yourself?

[low flame]

book

[tender halo]

idk the answer is a little too high concept to be a riddle

doesnt really matter either way

[gritty widget]

Which book?

[low flame]

yeah the indiscrete works, but i think so does cofinite right?

vitro

[gritty widget]

No (unless the set is empty of with just 1 element)

[gritty widget]

Vitro is the book? What vitro?

Oleg Viro?

[low flame]

https://www.maths.ed.ac.uk/~v1ranick/papers/viro.pdf

sorry i typod

[low flame]

above

[raw stag]

I don't understand where the equations are coming from in no 12

[sonic crane]

Can we assume that for a subset B of X, a point x in X is either interior, exterior or on the boundary of B, or do we have to prove this?

[raw stag]

Yeah I think so cause like it's enough to show that the interior, boundary, exterior union to X right

[sonic crane]

Yea that statement u said was a lemma given in class

Im trying to prove it …

[raw stag]

And for a given x in X there are 3 possibilities of its neighbourhood intersect right?

Either all of its neighbourhoods intersect B

[sonic crane]

Ya like this seems like just some logic thing

Case by case thing

[raw stag]

Yeah imo

See it either belongs to its closure or it doesn't right

That's the logic behind this

[raw stag]

Closure of B

And closure is int U boundary

Soo yeah

[sonic crane]

Oki

[raw stag]

Bump

[ocean canyon]

Oh, ok, so we are looking at the subspace topology on an openset M of X, and then I suppose the induced subspace topology on M is hausdorff.

[ocean canyon]

Answering this

[heady skiff]

Do my answers look correct here?

\newline
\
a. Let (x \in \overline{A}), so that (x) is a limit point of (A). Let (\epsilon > 0). Then there exists (N \in \mathbb{N}) such that if (n \geq N), (d(a_n, x) < \epsilon) for some sequence ((a_n) \subset A). It follows that for (n \geq N), (d(a_n, x) < \epsilon), as required.
\newline
\
b. Let (x) be a limit point of (\overline{A}). Let ((x_n) \subset \overline{A}) be a sequence of points converging to (x), and (\epsilon > 0) be given. For each (n \in \mathbb{N}), let (a_n \coloneqq A \cap B_r(x_n)), where (r = \frac{\epsilon}{2}). Since ((x_n) \to x), choose (N \in \mathbb{N}) such that if (n \geq N), (d(x_n, x) < \frac{\epsilon}{2}). We claim that ((a_n) \to x). Indeed, if (n \geq N),
[
d(a_n, x) \leq d(a_n, x_n) + d(x_n, x) < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon
]
as contended.
\newline
\
c. If ((a_n) \subset A \subset Y) is Cauchy, then ((a_n) \to x) since (Y) is complete. Since (A) contains its limit points, (x \in A), and (A) is complete.

[gentle osprey]

okeyokay

[uncut tartan]

whats the difference between a hausdorff space and a frechet space

[gritty widget]

A frechet space might have two points x,y such that every nhood of x contains y

You should try to find an example of such a space with underlying set {0,1}

[quick crane]

I'm pretty sure Frechet spaces are assumed to be Hausdorff so that they are metric spaces

[gritty widget]

I believe by Frechet they meant T1-spaces

Some people call T1-spaces Frechet

It wouldn't make much sense to ask for the difference between hausdorff spaces and frechet spaces in the sense of functional analysis, it would be like comparing bananas to banana split ice cream

[quick crane]

Oh I see, it doesn't seem like he's talking about functional anal

[uncut tartan]

uh yeah i was talking about these spaces
https://en.wikipedia.org/wiki/Hausdorff_space
https://en.wikipedia.org/wiki/T1_space

[gritty widget]

Do you understand the difference now?

[uncut tartan]

ill be honest not really. The wiki page says that T1 spaces are spaces where all distinct points are seperated

but then isnt really clear on what seperated means

i assumed it was just the same as what hausdorf spaces were but with open sets instead of neighborhoods

[gritty widget]

A T1 space X is one where, given distinct x,y in X, you can find a nhood of x not containing y and a nhood of y not containing x

And a T2 space X is one where, given distinct x,y in X, you can find nhoods A of x and B of y such that A\capB=Ø

[gritty widget]

Also forget this message, I was thinking about T0 spaces for some reason, sorry

[uncut tartan]

ok so just kolmogorov space with open sets replaced with neighborgoods

[gritty widget]

No

Kolmogorov is weaker

[uncut tartan]

yeah

[gritty widget]

X is kolmogorov if given distinct x,y in X you can find either a nhood of x not containing y or a nhood of y not containing x

In everything I said, if you replace nhood by open, everything stays true

[uncut tartan]

you mean and 'an open set contain x but y or y but not x' is equivalent to the definition you gave?

[gritty widget]

yes

For kolmogorov

[uncut tartan]

oh T1 is what you gave but with and instead of or

so you can find both

ok

[alpine nest]

And Hausdorff is where you can find a pair of disjoint neighborhoods, which is even stronger.

[uncut tartan]

ok let me start by proving those two definitions of kolmogorov topologies are equivalent

well 1. => 2. is trivial since neighborhoods are just supersets containing open sets. let me see about 2. => 1.

[tender halo]

my preferred nomenclature is T0, T1, Hausdorff, regular, Tychonoff, normal, hereditarily normal, T6

and by my preferred i mean "the best one everyone should use"

[alpine nest]

My preferred nomenclature is "Hausdorff, metrizable"

[tender halo]

by frechet spaces in general topology i usually understand the Frechet-Uryhson spaces, where sequential closure is the same as closure

[alpine nest]

Statements dreamed up by the utterly deranged

With the exception of the conclusion that metric spaces are perfectly normal, which they are

[tender halo]

every space is Polish

[alpine nest]

Well, all of mine are 🇵🇱

[uncut tartan]

btw when we say neighborhood of x we mean a set that is a super of an open set containing x right?

in whihc case this whole thing might just end up being trivial

[gritty widget]

Yes

[uncut tartan]

welp i just wasted a whole bunch of time unnecessarily staring at it for no reason

ok well thats proven

[alpine nest]

Although I think often it works to just think of open sets containing x

[uncut tartan]

now to prove that every T1 set is T0

[alpine nest]

Since it's the open bit that tends to be significant

[tender halo]

i know of like one proof where its specifically more convenient to have nbghds not necessarily be open

[alpine nest]

I tend to think in terms of open neighborhoods almost exclusively, to the extent that I often forget that "neighborhood" doesn't necessarily need to mean an open set.

But as established, I do fairly tame topology (genuinely hardly ever touch non-metrizable spaces)

So basically when I think topology, I think balls as I do most of the time anyway

[craggy cedar]

All the good ones anyways

[quick crane]

meanwhile, $\ell^{\infty}$.

[gentle osprey]

L

[sonic crane]

I was reading munkres, and neighbourhood of x is an open set containing x. But now i started school and my topology prof introduces neighbourhoods as not necessarily being open sets 😭

We are following Janich textbook not munkres …

A is a neighbourhood of x is there is an open set containing x inside A

[trim spade]

You see what’s the point of that tho lol, you’re basically wanting to specify that there is an open set around x anyway

[finite token]

Neighborhoods aren't open by convention outside of Munkres?

[cedar pebble]

yeah, neighborhoods can be closed if they want

[alpine nest]

Or you can be doing your topology on the Cantor set and have clopen neighborhoods

[uncut tartan]

lol

[prime elbow]

Let X be a metric space which has a dense set which is countable so X has a countable basis. Let D be that countable dense set.

Thus, take { B(x,1/n) | x in D and n in N }.

So take any open set U in metric space (X,d), let u in U so there exists e>0 such that B(u,e) \subset U.

Since D is dense so D intersects with B(u,e), say y in D intersection B(u,e), so we can get B(y,1/n) \subset U.

It is just an idea

[limpid fern]

this is a basis of X that generated the metric topology yes

[prime elbow]

Thank you ❤️

If Y is a compact subset of X and Z \subset Y then Z compact implies Z is closed.

Since X is metric space, so compactness is equivalent to sequentially compactness.

Let x_n be a sequence in Z which converges to x \in X. But Z is compact so there is a subsequence of x_n which convergent in Z, but x_n is convergent so that subsequence converges to x.

By the uniqueness of the limit, x \in Z.

But here I didn't use the compactness of Y.

Is it correct?

[limpid fern]

You actually only need Z compact

Compactness of Y is redundant. It turns out that for a compact subset of a space to be closed it suffices to assume that the space is Hausdorff. All metric spaces are Hausdorff

[prime elbow]

Yes

[gleaming umbra]

Hi, how are you today! My question is, is PST, is it too difficult to Use a Neighborhood for explaining the plane of the Graph? (in graph theory)

As if the plane is say, I'm not sure what I'm really saying atm, but as similar to a hyperbolic circle?

[rancid umbra]

maybe an example might help get your point across

[robust drum]

The theory is the same bc neighborhoods contain open sets

IE there are Munkres nbhds iff there are not necessarily open neighborhoods

[uncut tartan]

are differentials like dy and dx defined for reals? afaik they are only really defined in structures like topological manifolds

[gritty widget]

R is a topological manifold

[tender halo]

perhaps the most topological manifold there is

[unreal stratus]

And they should be smooth manifolds rly

[west brook]

Perhaps the most manifold manifold there is

[uncut tartan]

@restive agate would like to know your location (sry of the ping bud)

[prime elbow]

For the first part,

In S there cannot be two rational numbers, but if we take any open neighborhood of any point of S it will contain infinitely rational numbers so the interior set is empty.

Is it correct?

No hint

[quick crane]

seems correct

[prime elbow]

Okay thank you

[plush crescent]

Q: how useful is it to learn order topology for more advanced topics? I want to skip it and go to product topology in munkres

[gritty widget]

It's probably not a big deal to skip it. And if you ever need to use it in the future, you can always come back and learn it better

[lavish beacon]

i have a question about the 2nd property of a basis - if x is in B1 intersected w/ B2, could B3 just equal to B1 intersected w/ B2 as long as that intersection is still in the basis?

[gritty widget]

Sure, why not?

[lavish beacon]

ok js making sure cus that felt a bit 'easy' ig? cus i feel like an intersection of two elements would usually still be in the basis anyways idk 😭

[gritty widget]

It doesn't have to be

For example if you consider the basis on R² given by open balls, the intersection of two open balls will almost never be an open ball

[lavish beacon]

okioki i see thanksssss

[tender halo]

munkres doesnt really have any topologic order theory

its a somewhat useful topic to have in the counterexample toolkit?

[tender halo]

also

[alpine nest]

Entirely true

[thorny agate]

what text is this from lol

[gritty widget]

Saharon Shelah, "Classification Theory and the Number of Non-Isomorphic Models"

[tender halo]

shelah's classification theory

[heady skiff]

why is the equality pointed to by the red arrow true

is it a mistake?

ah it should be I \subseteq T'

[plush crescent]

Q: i was looking at this example; couldn't help to wonder if the set of intersection of projections is also a subbasis?

[heady skiff]

why is $\varnothing \subseteq \mathcal{I}$?

[gentle osprey]

okeyokay

[heady skiff]

is it just vacuously true?

[thorny agate]

empty set is a subset of any set

that's easy to see from vacuous truth yes

every element of the empty set is an element of I

[heady skiff]

got it thanks

[plush crescent]

I got it, turns out it is true by definition

[unreal stratus]

Not sure what you mean by "projections" here (as sets)

[plush crescent]

Pi(x,y) giving x or y, the book calls it a projection

[unreal stratus]

To me if a set js called a projection then it should be the image of smth under a projection , but here we are in a product

[unreal stratus]

Yeah but their images aren't subsets of X x Y

[red yoke]

projections preimages of opens under projections

[unreal stratus]

Sure

But then what is the difference from the set in theorem 15.2 there other than allowing for finite intersections

Or is that it

[red yoke]

I think so

[unreal stratus]

Oh okay sure

I guess "it is true by definition" referred to smth other than what I thought I.e. what was meant was it is true by definition of subbasis

[unreal stratus]

worth pointing this out anyway ig

[plush crescent]

Yes BTW cool name 1 day I wish I could learn prismatic cohom... I can't say it if I don't know it

[unreal stratus]

Aha

Thanks yeah I mean I have not worked on it as much as I'd like for sure

[plush folio]

Is this true? Isn't the topologist's sine curve a subset of euclidean space?

[alpine nest]

Yes, it's a subset of R^2, which is an Euclidean space

[alpine nest]

I think that text is missing "open"

Or at least it would be true if it talked of open subsets of an Euclidean space

Or at least of open subsets of R^n and C^n

[plush folio]

I see, thanks

[alpine nest]

But maybe the inner product lets you also establish that in a general Euclidean space

[plush folio]

but that's a bit unfortunate, because it's from a Lie theory course, where we mostly talk about closed subsets of R^n / C^n (IIRC a matrix lie group is (isomorphic to) a closed subset of GL_n(R))

I guess the more relevant fact is that connectedness implies path connectedness on manifolds

[robust drum]

For manifolds it’s fine; points are locally homeomorphic to open subsets, and therefore locally path connected

That’s actually all you need for PC <=> connected

[plush folio]

Yep, but it doesn't seem like that was what the slide what trying to convey

[robust drum]

Yeah they prob just forgot the word open

[plush folio]

Yeah, but my point is that a matrix Lie group is not an open subset of R^n. I think you have to use the fact that a Lie group is a manifold

[robust drum]

Ok yeah I agree

Mb

[fierce lily]

Question: 1, can we say a single element {x0} is both open and closed on discrete topology? 2. when f is continuous, we write f: R->trivial topology on X. How can we make sure that it is always true?

[fierce lily]

I see, because discrete space must contain all single closed element, union of closed. same for all open.

but i am wondering why this map is always continuous?

so X_tri only has open sets empty and X itself. but I don't know if I can say f^-1(empty) is empty set. f^-1(X)=R here, I have hard time imagining how a real numbers can map to a whole set X here?

how do you think of the preimage of an empty set and the set X here?

[unreal stratus]

Preimage of empty is always empty

[fringe thorn]

for the entire set X, just note that f^{-1}(X) = {x \in R : f(x) \in X} = R

[fierce lily]

do you think we need to think of how real numbers map to subsets of X?

[fringe thorn]

no, a function between sets must map every element in its domain to an element of its codomain

so the set of all real numbers in R that get mapped to X under f is all of them, i.e. R

[fierce lily]

Why there is nearly no continuous map satisfies this map?

[last jungle]

Ok so, I wanna check my reasoning.
I would like to show that for $S \in (\mathbb{Z} \times \mathbb{Z}){FC}$ (where FC denotes the finite complement topology), we have $S \in \mathbb{Z}{FC} \times \mathbb{Z}{FC}$. (That is, the product topology)
It is enough to show the set $\mathbb{Z}^2 \backslash \lbrace m_i,n_i \rbrace \in \mathbb{Z}{FC} \times \mathbb{Z}_{FC}$.
Specifically, we can write $S = (\mathbb{Z} \backslash \lbrace m_i \rbrace \times \mathbb{Z}) \cap (\mathbb{Z} \times \mathbb{Z} \backslash \lbrace n_i \rbrace)$, where each of these is open in the product topology. So, S is in the product topology.

compile error PLEASE

[gentle osprey]

Galstaff, Sorcerer of Light

[fierce lily]

I really don't understand why the intersection of A and B can refute that f is not continuous

so you mean the inverse image from X_dis will be disconnected, and it will make f not continuous?

[wraith bear]

It is assumed f is cont. (also "onto")and X has more than 1 element. And get contradiction cuz R cant be disconnected.
Not inverse image of X, but instead inverse image of 2 disjoint subsets of X whose union is X (or image of f).

[tender halo]

holy shit tell your lecturer to wash their rag

this is unreadable

[cedar pebble]

damn it's kinda rare that I see boards so caked in chalk that it might actually be more readable to write on them by erasing the chalk with your fingers

[fierce lily]

I guess that's his style of teaching....

[fierce lily]

really hard to read what he writes when he updates his notes on blackboards.....

[candid bronze]

Question:\\

we say $f(x)\to L$ as $x->a$ if $\forall$ neighbourhood $V$ of $L$, $\exists$ neighbourhood $U$ of a s.t. $f(U\setminus{p})\subseteq V$\
For sequences we say $x_n\to L$ if $\forall$ neighbourhood $V$ of $L$, $\exists N\in\mathbb{N}$ s.t. $\forall n\geq N$, $x_n\in V$\

So it looks like theres some topology on $\mathbb{N}$ so that the limit point $a$ is actually $\infty$ and "neighbourhoods of infinity" means something like $[m+1,\infty)\cap\mathbb{N}$. What is this topology? Is it the subspace topology from $[-\infty,\infty]$?\
Is there are metric on $\mathbb{N}$ that induces this topology? \\
Thank you ♥️

[gentle osprey]

MartinFTW

[prime elbow]

The set E = (-√2, √2) intersection Q is not compact in Q, induced from R.

Because the open covering of E {(-r,r) intersection Q | 0<r< √2, r in Q } has no finite covering, right?

[red yoke]

Yes

[gritty widget]

It has no finite subcover, yes

[red yoke]

You can think of it as a subspace of R + ∞

Or the subspace {1/n: n ∈ Z+} + 0 of R, which gives it a metric

Something interesting to note is that the order of the elements in N doesn't matter

The open sets are

1. arbitrary subsets of N and
2. subsets containing ∞ and cofinitely many naturals

And indeed if σ: N → N is any permutation then x(n) → L iff x(σn) → L

[prime elbow]

Okay thank you

[red yoke]

(this construction generalizes to something called the one-point compactificaton of a locally compact Hausdorff space)

(so you can talk about say lim[|z|→∞] for complex functions)

[candid bronze]

right i see, cuz the only thing that matters is it containing all the numbers after a point

thank you mr fox

[fierce lily]

first question is: Why empty set can satisfy openness directly while X can not? 2. To prove X is basis element here, why don't we need to check x is in B1 intersects B2 part, because in textbook basis should satisfy two properties.

for the second question, I mean part 2 in this diagram.

[gritty widget]

They probably mean that the empty set is the empty union

[gritty widget]

If x is in the intersection then {x} is in the basis

[fierce lily]

empty union? so why we can directly write it is open vacuously?

[gritty widget]

I wouldn't write write it that way

[fierce lily]

so I know x must be contained in basis element B subset of X. do you mean if x is in B1 intersects B2, then we have B3={x} subset of B1 intersects B2?

[fierce lily]

Then how will you justify this part?

[gritty widget]

The definition of open is union of elements of the basis

Empty set is the empty union

So it is open

[gritty widget]

Yes

[fierce lily]

why {x} can be the basis element here?

[gritty widget]

That's the basis the example is using

[fierce lily]

I assume this question is indenpent of example 3?

so that's the reason why I am confused about {x} can be considered as basis element

[gritty widget]

Sorry I misread. It cannot in general. And it isn't?

[fierce lily]

so do you think for X in T part, we need to examine the second condition of basis?

if no need, why we can skip that?

[gritty widget]

For every x there is a basis element containing x. So take the union of all basis elements

[fierce lily]

but how it justify the intersection part?

[fierce lily]

and also, I see that open set U1 and U2 are union, but we also need to justify the union is basis here instead of just writing it as open sets?

[gritty widget]

I have to sleep. You need to show such an union is again an union of basis elements

[fierce lily]

I think he just want to prove that union, finite intersection, X and empty are basis element

but when we aims to prove X is basis element, I really think what textbook writes seems not complete( not convinvcing to me at least

[fringe thorn]

isn't Munkres showing that T is a topology in that photo?

[fierce lily]

he wants to prove T is topology I think.

[fringe thorn]

yes, how does he define T?

is it the collection of all unions of elements of B?

[fierce lily]

I think he writes in the chapters behind that

[fringe thorn]

well, that's the definition of a topology in general

ah, is this what we're looking at?

[fringe thorn]

which part of his proof that X is in T does not satisfy you?

[fierce lily]

I am confused that why he does not write the part when x is in intersection of B1 and B2?

[fringe thorn]

he is not proving that something is a basis

he is proving that X is an element of T

[fringe thorn]

to do that, he just has to prove the condition at the bottom of this screenshot for X

which is what he did

[fierce lily]

so we can argue that for each x in X, we must find a basis element( because collection T generated by basis) that contains x and this basis element is contained in X, so X is said to be open in itself right?

[fringe thorn]

yes, for any x in X, we apply property 1) of a basis to the basis B to find a basis element that contains x

and then by definition, an element of B is a subset of X, so we have x \in B \subseteq X

thus, X is an element of T

[prime elbow]

Any hint to prove that if X is metric space and any real valued continuous function on X is bounded then X is compact space.

f:X-> R such that x->d(x,x_0) where x_0 is a fix in X.

Since X is not compact

So there exists a sequence x_n such that it has no convergent subsequence

if f is bounded then d(x_n,x_m) is bounded

And R is complete so by Bolzano Weierstrass theorem there exists subsequence d(x_n_k, x_0) which converges

Contradicts that X is not compact

Is it correct?

[prime elbow]

I think this is not complete proof

[prime elbow]

Given two compact subsets A, B of a metric space such that A intersection B is empty. Show that d(A,B) > 0. In fact, show that there exists a in A and b in B such that d(A,B) = d(a,b).

I showed the first part, but I don't know how to show the last one.

Can I say f:A×B -> R such that (a,b) maps to d(a,B) is a continuous function

Is d(A,B) = inf{d(a,b) | a in A and b in B} same as inf{d(a,B) | a in A } ?

[prime elbow]

Let X and Y be metric space and a in X. A family A of functions from X to Y is said to be equicontinous at a if for any e>0 there exists a \delta >0 such that d(x,a)< \delta implies d(f(x), f(y) ) \e for all f in A.

We say that A is uniformly equicontinous on X if for any e>0 there exists a \delta >0 such that d(x_1,x_2) < \delta implies d(f(x_1), f(x_2) ) < e for all f in A.

Now let X be a compact metric space. Let F: X×X -> Z be continuous. Let f_y(x) = F(x,y). Show that A = { f_y | y in X } is uniformly equicontinous.

X × X is compact implies F is uniformly continuous on X × X.

So for e > 0 there exists \delta > 0 such that if d_p((x_1,y_1), (x_2,y_2) ) \delta then d(F(x_1,y_1), F(x_2,y_2)) < e.

Here d_p denotes product metrics.

Now let d(x_1,x_2) < \delta then d_p((x_1,y),(x_2,y)) \delta so d( F(x_1,y) , F(x_2,y) ) < e.

Hence, for d(x_1,x_2) < \delta it will be d(f_y(x_1), f_y(x_2) ) < e for all y in X.

Is it correct?

[plush folio]

Just need a quick sanity check: uniform continuity of $f: X \to Y$ can be formulated as: for all $\epsilon > 0$ there exists a $\delta > 0$ such that $f(B_{\delta}(x)) \subset B_{\epsilon}(f(x))$ for all $x \in X$, right?

[gentle osprey]

sheddow

[prime elbow]

I think yes

[heady skiff]

can I get a hint for showing that arbitrary unions are open in b please? this is what I have so far:

Let $\mathcal{V}$ be any collection of elements of $\mathcal{T}$. For each $V \in \mathcal{V}$, there exists a closed set $C_V$ with $V = X - C_V$. It suffices to show that the union $\bigcap_{V \in \mathcal{V}} C_v$ is closed, for if this were true, we would have
[\bigcup_{V \in \mathcal{V}} V = \bigcup_{V \in \mathcal{V}} X - C_v = X - \bigcap_{V \in \mathcal{V}} C_v \in \mathcal{T}]

[gentle osprey]

okeyokay

[heady skiff]

I'm having trouble showing that the intersection over all the Cvs is closed, ie cl(n Cv) = n Cv

[opaque zodiac]

Argue that the closure of n Cv needs to contain n Cv and be contained in all of the Cv's

[heady skiff]

isn't that just extensionality?

that's what i've been trying to show yeah

we already have one direction by axiom 2

the other direction is what I need help on

i know that intuitively the closure of n Cv is a subset of every closure for each Cv but it's a bit tricky to rigorize

[opaque zodiac]

It follows from 4 that if A is a subset of B then cl(A) is a subset of cl(B)

[heady skiff]

bossmannn

you're too smart

thank you

life saver

[opaque zodiac]

Well I mean I was thinking about Kuratowski closure operators two days ago

[tender halo]

whats that

[heady skiff]

to be clear, in part c) are they asking us to show that for any given set $A$, $A \cup A'= \text{cl}(A)$ where $A'$ is the set of limit points of $A$ determined by the topology in part b?

[gentle osprey]

okeyokay

[fierce lily]

Let's call that stuff (similar to C ) C' since I don't know how to pronounce. My question is : because for every open set U in T, U contains a C, we conclude that each element of C' belongs to T, is that correct?( I am struggling about how it gets each element of C' belongs to T?" 2. is that forward direction proves that T is a topology generated by the basis C'?

[gentle osprey]

brayden

[heady skiff]

am i missing something here? isn't X - A a subset of X for every subset A of X?

[alpine nest]

It is, but I'm not sure what that has to do with anything

[heady skiff]

because they're trying to show that Sbar is not a subset of X, but if X \ B(x, e) is a subset of X and Sbar is a subset of X \ B(x, e), then Sbar is a subset of X

but they wrote X \ B(x, e) is not a subset of X

[alpine nest]

That doesn't say "is not a subset of"

It says "is a subset of X and not equal to X" (because X is its own subset, but here we want a proper subset, i.e. not equal to the entire space)

[heady skiff]

Oh

I've been using that notation wrong my entire life then

Rip

[thin tide]

Yeah you’re talking about $\nsubseteq$

[gentle osprey]

chipotle

[heady skiff]

ops

[quaint relic]

If we have a Hausdorff space Y, we take Z to be the space consistent of two disjoint copies of Y, and if we take A to be a closed subset of Z, so we consider the quotient space Z/A, how can we prove that the projection from Z to Z/A is open?

[rancid umbra]

i dont think this is true.

take Y to be the unit disk in R^2, A1 to be the closed upper right quarter disk and A2 to be the closed lower left quarter disk.
Set A to be the disjoint union of the two A1 and A2 in Z

take U to be the disjoint union of both open upper semidisks of Y in Z. the image of U under the quotient map q is not open since q^-1q(U) is U union with A1 and A2, and this set is not open in Z

[quaint relic]

I think i wrote it incorrectly
Particularly, we take $A \subseteq Y$ a closed subset and we define the relation on Z to be $(p,0) \sim (p,1)$ if $p \in A$ and $(p,x)$ just related to itself in other case

[gentle osprey]

Roscoe

[quaint relic]

Is it really false? Because I've seen that this is implicitely used in a lot of posts lol

[rancid umbra]

is my counter example wrong?

what posts?

[quaint relic]

https://math.stackexchange.com/questions/214045/what-are-the-epimorphisms-in-the-category-of-hausdorff-spaces

like that one

to prove hausdorffness its used but i dont see why this is true directly

[rancid umbra]

in Brian M. Scott’s answer, q(B0\K0) is open since q^-1q(B0\K0) is equal to B0\K0 which is open in B (by definition of the quotient topology)

it’s not because they are implicitly using the fact that the quotient map is open here

[quaint relic]

It is being used when x is distinct from y

or how do you get the neighborhoods on the quotient there?

[quartz horizon]

Yeah this is false I think

If you just take Y to be the real line

Z is two copies

And take your closed subset to be [0, 1] on one of the copies

(The other copy doesn’t really matter here)

Then (-0.5, 0.5) is open in R

Applying the quotient map, and then the preimage, gives (-0.5, 1]

Which isn’t open in R

[low flame]

cant i just take the union of the relative topology of A cap V where V is in the topology of X

or the union of all sets in the subset of an open set is open, and thus the subset being the union of these sets it he largest

[alpine nest]

The union of the relative topology would be all of A

Since A itself belongs to the relative topology

[low flame]

but woudlnt that be showing that A comes from the union of all open sets in the subset?

[alpine nest]

You would need to be more precise with the wording for me to comment

[low flame]

okay can i use the sets of the relative topology that the subspace belongs too to prove this? or can i just say okay the subspace is open, then it contains all of its open sets, which if i take the union of those open sets then i get the open subset

[gritty widget]

Openness is not preserved by subspaces

[unreal stratus]

Well that depends what you mean by this

Lol

Under inclusion into the whole (your statement is correct) vs under intersection with the subspace (then openness is preserved tautologically)

[gritty widget]

True, I should have said it is not equivalent for a set to be open in the subspace and in the original space

[unreal stratus]

You know this but I think statement is ambiguous

[quartz horizon]

The inclusion map isn’t always open

Is maybe how I’d put it

[low flame]

im confused what everyones talking about

[unreal stratus]

Indeed inclusion is open iff subspace is

[alpine nest]

Elements of the relative topology don't have to be open in the original topology. A subspace is open (and closed) in its relative topology, but it doesn't have to be open (nor closed) in the original topology.

Admittedly the interior of a set will be open both in the relative and in the original topology, but I'm not sure that's very useful.

[low flame]

so the forward direction in this isnt too bad

if U is open, then for each x in U there is a Vx which is a nbhd of x. Then U = U cap Vx, and is open in V.

but the reverse, im a bit confused

[fierce lily]

since T is a topology containing open sets, why it writes T is closed under finite intersection or arbitrary unions? 2. why it writes elements in P(x) instead of elements in T? because I think sets in topology is said to be open

[fringe thorn]

to answer your first question, this is a definition, so the author needs to say all of that

oh, I think I misread your question

by "T is closed under arbitrary unions", they mean that taking an arbitrary union of elements in T will return an element in T

[quartz horizon]

Ah yeah it’s unrelated to the idea of a closed set

[fringe thorn]

this is analogous to how you say that a (linear) subspace is closed under addition and scalar multiplication

[quartz horizon]

I can see how that’d be confusing

[fringe thorn]

from linear algebra

as for the second question, uhhh

I think that's a typo

[quartz horizon]

Yeah typo

[fierce lily]

ok, really appreciate that

[quartz horizon]

Lol

[fringe thorn]

just remember the definition of the subspace topology

[low flame]

hah ive been looking at it 😛

[low flame]

i submit, can you please help me. im not being graded or anything im doing this for meself

ive been thinking about this backwards direction most of the day

[fierce lily]

Is there anyone who is willing to answer this one?

[rancid umbra]

what is the question?

like, what is C and C'?

[fierce lily]

I think I attached the picture, can you see that diagram below that question?

[fierce lily]

C is the basis element of C' in that diagram, C' is the basis.

[vestal vine]

Let M and N be n-dimensional topological manifolds, M compact and N connected.
Let f:M->N be a continuous injection, how do I show that it is a homeomorphism?
Since M is compact and N hausdorff, f is closed, so f(M) is closed.
I want to show that f(M) is also open, and since N is connected this means f(M)=N, so f a continuous bijection from compact to hausdorff, and thus a homeomorphism.
But I don't see how to prove that f(M)/f is open.

[hidden crag]

#discussion message

Is this technically a snipe?

[opaque scroll]

By restricting to a small neighborhood in N, it's enough to show that a continuous injection R^n -> R^n is open. Not sure if that makes it easier

[vestal vine]

yur

okie dokie

thanks!

[hidden crag]

Mod btw…

[vestal vine]

mod aboose

[robust drum]

If f: X-> Y is closed and bijective, it’s also open.

Proof;
f(X-C) = f(X) - f(C) = Y - f(C) which is open because by assumption f(C) is closed. The first equality uses that f is an injection and the second uses that it’s a surjection.

[pastel linden]

I have a silly question. I'm trying to show a specific homotopy is continuous, but it's defined piecewise with the conditions on the piecewise function dependent on the homotopy variable

Is there a good way of doing this? I can't just use the gluing lemma

[robust drum]

Often you still can

Example: often in these problems the domain changes on a line with respect to the parameter of the homotopy

In this case, example if you have your boundary is like x = [5t + 3] or something random

[vestal vine]

Proving it's bijective is the hard part

[robust drum]

Then you substitute 5t + 3 into your homotopy for x and y

[vestal vine]

Anyway I already did the question

[pastel linden]

Yeah this problem is analogous to showing that gamma * c_x is homotopic to gamma

[robust drum]

If it’s injective onto its image the same thing holds; it’s an embedding, you just have to talk abt the subspace topology

[pastel linden]

I just don't remember what I did then lol

[robust drum]

Whatever your parameterization is, just replace x with it in the equations on the top and bottom and check they’re equal

[vestal vine]

Yeah it's homeomorphic to its image, that doesn't mean its homeomorphic to N as a whole

[robust drum]

The gluing lemma still applies here

Ah I misread your question

[pastel linden]

Doesn't that only imply that it's a continuous function for each fixed t

[robust drum]

No

Gouing lemma applies to any two closed sets intersecting along another closed set (doesn’t have to just be intervals)

It might help to draw the picture

If you take your homotopy rectangle and split it along a line

[pastel linden]

Oh I see what you mean

[robust drum]

And define one equation on one side of the line and another along the other

It suffices to show they agree on the line

IE, they agree when you substitute x for your defining equation for the line

[pastel linden]

The tricky thing is that I'm doing this for the action of pi_1(X,x) on pi_n(X,x) so I can't really visualize it easily but I'll think about it

Thanks!

[robust drum]

Ah yeah that problem is kinda ass (annoying)

Np

[heady skiff]

what exactly are we trying to show here? that the prime ideals of Z are a basis for Prime(Z) under the jacobson topology, ie any open set under this topology can be expressed as an arbitrary union of prime ideals?

[novel acorn]

What is Prime(Z)

The collection of prime ideals?

[heady skiff]

Yeah

[novel acorn]

Well the topology here isn't discrete