point-set-topology

[tidal cedar]

\partial_0 here is the zero map by definition of our chain complex

[trail charm]

mhm

[tidal cedar]

the image here can be thought of as points w/ paths between them

so once we quotient out by the 1-simplices or paths

we get only the isolated ones

[trail charm]

isolated points?

[tidal cedar]

like, the connected components here

it can help to see this by actually drawing some simple examples

like just a small finite discrete set

[trail charm]

i tried drawing and potato said it works for H1

but why does it not work for H0?

[tidal cedar]

do like

[trail charm]

the goal was the top one represents H1 and bottom represents H0

[tidal cedar]

X = {a,b,c} discrete

your 0-chains are Z[a,b,c]

so Ker \partial0 is also that

The image is trivial

Because you have no higher simplices

[trail charm]

mhm

[tidal cedar]

Now what happens if we add a 1-simplex connecting a and b

[trail charm]

im partial 1 is b - a?

[tidal cedar]

yeah

so what does this quotient give us

[trail charm]

{a, c}?

[tidal cedar]

kinda yeah

remember it's free abelian so we are just getting representatives

but you now only have Z^2

[trail charm]

oh yea

[tidal cedar]

instead of Z^3

[trail charm]

hmm okay

that makes sense

[tidal cedar]

so now you just have Z^2

and it's generated by the connected components there

similarly if we add a 1-simplex

between b and c

we "kill off" another generator

and just get Z

[trail charm]

gotcha

[tidal cedar]

does that help answer your question?

[trail charm]

yea ok yea that makes sense

[tidal cedar]

cool!

[trail charm]

so in S1 there’s the 0-simplex v and the 1-simplex e

\partial 0 is zero so ker \partial 0 is Z[v] itself?

oh wait partial 1 is zero

wait but partial 0 is 0 by defn

so u just have Z/0?

≈Z

[tidal cedar]

yeah

[trail charm]

sick 😎😎 tysm

[high hill]

what does "colaxly closed" mean here

(do i need to understand what a filter is, to understand that)

[feral copper]

At first I thought this was a typo, but then no, googling it proved me wrong x') what a weird name...

[high hill]

eh? i couldnt find anything when googling colaxly

[feral copper]

https://www.google.com/search?q=colaxly+closed&rlz=1C1ONGR_enFR1014FR1014&oq=colaxly+closed&aqs=chrome..69i57j33i160l3.1808j0j7&sourceid=chrome&ie=UTF-8

[high hill]

bruh

theres nothing

[feral copper]

But I mean, you're reading the nLab, so of course it's gonna be ugly and not digestable

[high hill]

was there supposed to be something here

[high hill]

thats not what i get at all

even in an incog tab

anyways... it appears I can skim over this term?

[feral copper]

In incog I still get the same, weird

Yes ofc you can x')

[high hill]

morbid curiosity getting to me tho

[feral copper]

You can skim over the entirety of nLab

[high hill]

Well Im looking to intuit 1st countable and have not liked anything at all found.

so Im trying nlab's

[feral copper]

This was not satisfactory?

[high hill]

Nope.

This is like the 5th time I'm trying to do this btw

Like sure, I can understand the given definition, but not why we define it this way

[feral copper]

It's really nothing fancy!

[high hill]

How it relates to 2nd countable

and why a local basis is a "basis"

So I'm trying again from scratch

[feral copper]

The nuance is:
(1) 1st countable, you have a countably many things at each point, but you may have a shitload globally
(2) 2nd countable has globally countably many things

Have you tried proving that R² is 2nd countable for instance?

[high hill]

(2) is easy to understand for me, but I just do not get at all how (1) like... works?

Like why does the definition need to be exactly like this.

[feral copper]

(1) is ugly, the typical spaces you live with are 2nd countable

[high hill]

For reference - I've black boxed 1st countable every time til now, and think its about time I come to terms with it

[feral copper]

Where exactly do you need to handle the details of the definition?

[high hill]

I like - definitely could not reconstruct the definition of 1st countable from 1st principles

it doesn't "make sense" to me

I could do that with basis, and 2nd countable, for example

[feral copper]

1st countable you have a basis for the topology

It's just that this basis is extra fancy because at each point you only need countably many of the basis elements to envelop all neighborhoods

[high hill]

but why is it envelop

[feral copper]

Take the union of all the countably many thingies at x for all x in X

[high hill]

as opposed to having a union

[feral copper]

Wdym?

[high hill]

I would 'expect' for "local basis" to mean "we have a basis, but locally"

but that does not appear to the case

as far as I can see

[feral copper]

Isn't this what this is?

[high hill]

I definitely cannot see it

from the given statement

[high hill]

like if I were to define this it would be:

"take a neighbourhood of x. A local basis is a basis for this subspace topology"

But a problem with this is we apparently want this to apply to all neighbourhoods of x

[feral copper]

Taking an open neighborhood of x, it's the same

A basis for the induced topology on an open neighborhood is a local basis at x

[high hill]

yeah well, if i can prove this given the usual defns of 1st countable, im convinced id be happy with the defn

(what im trying now)

my intuition for point set isnt strong enough to see these are the same

[feral copper]

Do this in two steps:
(1) Assume I have a local basis (B_n) at x. I need a basis for the induced topology on N. Take the basis N intersect B_n
(2) Assume you have a basis for the induced topology on N. This gives directly a local basis

N is an open neighborhood of x in

Again, don't bother too much with this stuff, most spaces you'll work with are metrizable anyways. (Smooth) manifolds are metrizable, bundles are, CW spaces are working almost as if they were metrizable

(I think compact CW spaces might be metrizable too? I'm unsure, I don't want to make a bold statement)

[high hill]

Consider this topology on the natural numbers:
t = {{}, N, {0}, {0, 1}, {0, 1, 2}, ...}

Then {{0}} is a neighbourhood basis for every point, right?

This "basis" with 1 basis element, {0}, doesn't coincide with what I understand a basis to be.

[feral copper]

You mean N with the discrete topology? Then for any n, {n} is a neighborhood of n yes

This "basis" with 1 basis element, {0}, doesn't coincide with what I understand a basis to be.
The discrete topology is not a good example to visualize what a basis should be, it's not even a great example to see what a topology is... It's like learning what topology is through the Zariski topology: it's not even separable, so it's just messy. You wouldn't want to think that singletons are open in a typical space

[high hill]

this isnt the discrete topology?

how do you make {1}

[feral copper]

{0} is closed and open, so {1} is {0,1} intersection the complement of {0}

[high hill]

eh?

how is {0} closed

[feral copper]

I'm unsure what I said is correct, but I'm pretty sure the topology you described is the discrete one

[feral copper]

And in any case, this comment still stands: if you pick a random ugly topology, intuition is going to fail you

[high hill]

I quite frankly don't like that - if an intuition fails for edge cases, Id consider there to be a problem with it

[feral copper]

Welcome to topology

And I'm pretty serious, topology is usually intuitive, until it comes to writing it all down. If you want to make sure edge cases are fine, you'll end up writing 400 pages long proofs. This is why we do pictures

[high hill]

I honestly still have no idea what 1st countable means rn.

Picture or no picture

[opaque cloud]

pictures as in category theory?

[feral copper]

There are pictures in cat theory?

[trail charm]

this is kind of confusing me. is the motivation for choosing the "maximal collection of cancelation pairs" just so that the new space doesn't have 'unnecessary' terms?

im also stuck on intuitively understanding the construction of the triangle complex via disjoint unions

i.e., i dont know what "identifying the pairs... corresponding to chosen canceling pairs" looks like

i also dont even know what to look up to read about this :(

[tidal cedar]

this is literally the opposite of what you want to do for intuition

[high hill]

well idc, i just want some source somewhere

And what I can see rn is the confusion stems from me assuming the word "base" in local/neighbourhood basis has anything to do with a topological "base"

[feral copper]

nLab isn't going to help your confusion more than Wikipedia... The definition is what it is, it's not very complicated, just sit down and write it down

[high hill]

I just do not understand why a "neighbourhood basis" or a "local basis" is termed such

These aren't topological bases - they are some other sort of basis

[feral copper]

In topology, you always have things like 'up to taking a smaller neighborhood, we can assume [...] wlog'
A basis makes sure you can still do that

[tidal cedar]

https://nforum.ncatlab.org/discussion/4188/laxness-of-bases-of-topologies/

[high hill]

Unless I also have this wrong. But the confusion should be that

[tidal cedar]

this has discussion of the verbiage used

but this is not what you should be looking at for intuition

especially if you're working through 1st countability

[high hill]

Ok forget nlab

But even wikipedia mentions local and neighborhood bases

and I just have no idea at this point what to make of these

To me, a topological basis is really a generating set under unions for a topology

But those are not this

[feral copper]

In case you're confused because of this: nothing to do with vector space bases

[high hill]

yes i am totally aware topological bases are not those

they are more like generating sets of groups say

[tidal cedar]

so there's a poset of all the neighborhoods of a point x right

[high hill]

sure

X less than Y iff X contained in Y

[tidal cedar]

yeah

this is also a structure known as a filter, though that's not super important for this

a basis of this is some subset of that poset

so that for every neighborhood U of x

there's some neighborhood in the basis

contained in U

and first countable is asking that this must be countable

[high hill]

So this is a filter basis

yh, i think I understood what a filter is

[tidal cedar]

yeah

so for something like R

we can take neighborhood basis to be open balls of radius 1/n, for example

this is countable

and any neighborhood of a point fully contains one of these

[high hill]

Yh, I follow that (and have seen examples for non-1st)

[tidal cedar]

I don't remember the usual non first countable ones

[high hill]

cocountable works

[tidal cedar]

oh

cocountable on R?

[high hill]

yh or any uncountable

[tidal cedar]

ok yeah that makes sense

[high hill]

i now see 2nd countable easily implies 1st. Just consider all the open sets containing that point

Still cant see why 1st is an important characterisation say, but getting there...

From what I read up its about sequential arguments

a lot like analysis-1 arguments you can do by using that countable sequence

[tidal cedar]

the closure properties of first countability are prolly the biggest things

[high hill]

ah ok, ill check those, havent seen

thanks 👍

[tidal cedar]

in a 1st countable space nets / filters can be replaced with sequences

the closure of a set U is the set of all limits of sequences in U

in spaces that aren't first countable that isn't true

[high hill]

a limit is a point where every neighbourhood contains a tail of the sequence?

right that immediately falls out - a sequence is a neighborhood basis for any of its limits. I think. will think on it to double check

[fickle elm]

Forcing the coefficients to be 1 or -1 in a singular chain means you will very likely end up with a simplex appearing several times in the chain. After applying the boundary map to the singular chain, you have a collection of degree-1 simplices and if you find 2 degree-1 simplices with opposite signs, you glue them up (identify them). The singular chain being a cycle means it becomes zero after applying the boundary map, i.e. every degree-1 simplices can be identified with another one.
Think of triangles viewed as 2-simplices. The sign before them can be thought as an orientation. After applying the boundary map, you have a collection of lines (1-simplex) with directions (orientations). Now glue up 2 lines (boundary of the triangles) if they have the opposite direction. If the singular chain is a cycle, every boundary can be glued with another, so you get a manifold.
I am not very sure about the (n-3)-skeletons because it is really hard for me to imagine things beyond 3 dimension. If they are only points, I think everything is fine. But if they are of higher dimension, something weird may happen as the book said, i guess.

[tidal cedar]

yeah I believe so

might be some specific verbiage around filters vs. pre filters when it comes to this

[fickle elm]

I suppose not many people like to write explicitly their intuition like Hatcher, so it is hard to read about this. I recall that Hatcher is a geometric topologist, he loves these kind of "visualization".

His book becomes extremely long because of this particular reason, and consequently, many people hate Hatcher's book. But still some may find his book enlightening, especially those who like to "imagine" spaces.

Btw, personally I dont like it.

Seriously, people's intuition about dimension 4 or larger is a very big mystery to me.

[feral copper]

In dimension 4, it's only an intuition, because nothing works like you'd expect. Example: exotic R^4's

[trail charm]

ok yeah this makes sense thanks

[abstract saffron]

this is why everyone should study knot theory 😄

[feral copper]

You mean knotted surfaces, right?

[heady skiff]

yo i'm trying to construct a homeomorphism from the unit cube to the unit ball, and i'm thinking about $x_i \mapsto \frac{x_i}{n}$ where we are working in $n$ dimensions and $x_i$ is the ith component in $(x_1, x_2, \dots, x_n)$. am i on the right track? or am i missin something here

[gentle osprey]

okeyokay

[heady skiff]

i'm trying to figure out surjectivity right now which is a bit tricky

it's equivalent to showing that for $s_i$ in $(s_1, s_2, \dots, s_n)$ in the unit sphere that $ns_i$ a coordinate of some point in the unit cube ig

[gentle osprey]

okeyokay

[gritty widget]

draw a picture of what the map you proposed does

[heady skiff]

yep

ur right

doesn't work

fuckkk

can i have a hint pls lol

[gritty widget]

draw a picture

[heady skiff]

thanks

i did

[gritty widget]

draw the unit ball and draw the unit cube

[heady skiff]

yea i did

my first thought

was identity for points of the unit cube inside the unit circle and dsquishing everything outside the unit circle into the unit cube but that obviously wouldn't be injective

[unreal stratus]

wdym everything outside the unit [psphere]

when that is the boundary

[heady skiff]

wdym

kinda like

let me draw it

[gritty widget]

so you're fixing the stuff in the ball and scaling the stuff outside the ball down to it?

[unreal stratus]

like you seem to be defining a map from the cube to the cube unless i am confusion

[gritty widget]

i like the idea actually. what if you also scaled down the stuff inside the ball?

could you make it injective?

[heady skiff]

ye exactly

[heady skiff]

oh nah

sorry maybe a picture will clarify

[heady skiff]

yea lemme try that

[gritty widget]

it is fixable

[heady skiff]

just looking at 2 dimensions for simplicity lol

[unreal stratus]

But uh

[heady skiff]

uh oh

[unreal stratus]

From experience it is easiest if you just do 2D and imitate the same formula

[heady skiff]

wym

[unreal stratus]

for 2D you can write out a pretty explicit formula

If I remember correctly

[heady skiff]

wait i'm confused how could you even define a bijection from the unit sphere to the unit cube if the unit sphere is contained in the unit cube

[gritty widget]

...

[heady skiff]

they're not of the same order

or am i trippin

[gritty widget]

you are

[unreal stratus]

How can you define a bijection from [0,1] to [-1,2]

[heady skiff]

OHHHH

shitttt maneeee

[unreal stratus]

They are both uncountable subsets anyway

[heady skiff]

actual factuals

[unreal stratus]

assuming by order you meant like cardinality ig

[heady skiff]

yea my b

we do a little abusing language

[heady skiff]

form for what

[unreal stratus]

for a homeomorphism

[heady skiff]

oh that's crazy

[unreal stratus]

by thinking about the difference between the two metrics giving rise to the ball vs cube basically

[heady skiff]

so something something involving maximum functions

max functions

🤯

ok i give up

time to consider enlarging the circle to a square

[heady skiff]

what if i divided by sqrt(2)

🤯

[gritty widget]

just doing that isn't going to work, for the same reason your first idea didn't work

[heady skiff]

damn

[unreal stratus]

i mean think geometrically

dividing by a constant is just gonna give smth of the same shape

[heady skiff]

oh lmfao my bad

ur right

damn bro

i don't know how to map those damn regions outside the circle but contained in the circle injectivelyu

bc if i let it be the identity inside the circle then those points are already taken

so maybe i shouldn't let it be the identity inside the circle...

does this require some high school geometry

not sure if i wanna do allat

[unreal stratus]

basically it's enough to do the boundary

because then you can scale stuff to get the whole thing

so try that

[heady skiff]

like idk the formula for this

[heady skiff]

huh ok

[gritty widget]

when you have an element of R^n what is an easy way to get a new element with norm 1

using the properties of norms

[heady skiff]

oh shit ur right

that will lead to my answer huh

damn

[gritty widget]

it might help

[heady skiff]

well in this case we're looking for elements that fit the criteria x^2 + y^2 = 1.... so we want x = sqrt(1/2) and y = sqrt(1/2)

the problem is is that this won't be defined for negative coordinate sof the unit cube lol

RIP

anyways i'll come back this problem's a bitch

well prolly trivial for y'all

[tawdry widget]

They both have “layers”. Like layer r for the former is {(x1,…,xn): max{|x1|,…,|xn|}=r}
Layer r for the latter is {(x1,…,xn): Σxi^2=r^2}. Their layer r are homeomorphic. (Cubic surface homeomorphic to sphere, construct the homeomorphism)
I think not just cubes, it works for some other bounded contractable set too,like simplex. Barycenter as center easily we can see what “layers” of it are.

[heady skiff]

Ooh ok imma ignore this until I get stuck on it tmrw for an hour, thank you tho!!

[hidden crag]

@tawdry widget this is what we mean

Several people gave hints and they made progress on the problem

And you decide to paste the entire solution

[tawdry widget]

Sorry…
I just wanted to tell him cube is layers of cubic surfaces, ball is layers of spheres. Cubic surface homeomorphic to sphere. Is it better… I will edit

[rustic haven]

can anyone help with this plz??

my weekly questions set i cant get this one

[tawdry widget]

Consider these two facts:
1, any locally-path connected space, its connected components are its path-connected components.
2, any open and closed subset, is union of connected components.

You can exam whether your space here is locally path-connected.
If it does, what would its path-connected components be

[rustic haven]

Thanks I’ll have a think about that, I tried showing that the max point in the closed set of points in U intersect Ln is also in the cloven set Z\U which is a contradiction. Just to prove the hint. My professor said correct track but I’m not sure where to go from here

[tawdry widget]

So you proved the hint?

[tawdry widget]

What I said is also to prove the hint. But since you proved the hint already, the rest is clear: it contains one of them, say (0,0), so there is a ball B of radius r center at (0,0), such that intersection of B and Z is in U. This ball will intersect any L_n for n large enough. Consider what happens for the other end of each L_n for n great enough, since U is also closed

(Closed subset contains limit point of points in it)

[rustic haven]

(0,1/n) for increasing n and this point is inside our ball. Since (1/n)^2 < r^2. So for other side: we need ((1/n), 1). For the same n?

Then I show U is closed so it must have both points?

[tawdry widget]

What I wanted to say was (1/n,1) is in U for n large enough, and (0,1) is their limit

[rustic haven]

Okay so for other side it (0,1/n)

Or y coordinate is max value/n

[tawdry widget]

No in my text I said assume it contains (0,0)

The other side is (0,1)

[rustic haven]

Oh okay right

I see see

[tawdry widget]

This disk B, it contains (1/n,0) for n large enough, and you already showed that by the hint, L_n is in U, so (1/n,1) is in U

[rustic haven]

👌

Thank you 🙏

[abstract quail]

Let $a$ and $b$ be two points of a metric space $X$ with the metric $d$, let $q$ be continuous parameterised path from $a$ to $b$, $q \colon [0,1] \rightarrow X$ by $t \mapsto q(t) \in X$, such that $q(0) = a$ and $q(1) = b$. A metric is defined on the space of these paths $q$ by $$d_1(q,q') = \sup_{0\leq t \leq 1} d(q(t), q'(t)).$$

The geometric paths, equivalence classes of parameterised paths under changes of parametrisation ($[0,1] \rightarrow [0,1]$, monotone and continuous) also form a metric space under the metric $$D(q,q') = \inf_{P} \sup_{0\leq t \leq 1} d(q(t), q'(t)),$$ where $P$ is the set of all possible parametrisations.

I want to ask if my understanding of these two metrics is correct.

In the first case, i.e. the metric $d_1$, if I imagine two particles travelling along two paths, given their paths and velocities, $d_1(q,q')$ is simply the farthest distance the particles will ever be from each other.

I don't understand the meaning of the metric $D$. I have been thinking about it for quite some time, but can't figure it out.

[gentle osprey]

Apoorv Potnis

[abstract quail]

Given two geometric paths, if I choose my parametrisations (velocities) such that one particle travels very slowly initially and speeds up at last, and the other particles speeds up initially and slows down at the end, I can get the distance between $a$ and $b$.

[gentle osprey]

Apoorv Potnis

[abstract quail]

I can also choose the parametrisations such that the particles come closest together such that the distance between them is the shortest distance between the two geometric curves.

[abstract saffron]

this is true, but it doesn't matter, D is the infimum over all the possible parametrisations. In some sense, it means something more intrinsically geometric.

[abstract quail]

the curves intersect at end points, so the distance is zero🤔

[abstract saffron]

I.e. given the paths traced out, you shoule be able to determine D, without needing the parametrisation

Of course you can always find some weird parametrisation to screw d(. , . ) up, but the point of inf is to now allow that not to make any difference

[abstract quail]

then what is the geometric meaning of $D$?

[gentle osprey]

Apoorv Potnis

[abstract saffron]

it's very hard to define, I am also struggling to make an intuition about it

you can go from simple cases, e.g. q and q' are two line segment AB and CD, then D is the min(max(AC, BD), max(AD, BC)).

likewise you can think about q and q' more discretely: choose some n points on q, and see what n points on q' will minimise the maximum distance between pairs of points

[rustic haven]

How can I use this to prove the hint? I’m not confident with my proof 😢

[abstract saffron]

but the fucked up thing is the continuity of the paths creates an order for points. Otherwise it'd have been easy

without continuity, something like maximal distance between any two points of two paths would have sufficed

[tawdry widget]

Oh because path-connected components of Z is L_n or {(0,0)} or {(0,1)}. So U is union of some of them. Naturally either L_n doesn’t intersect U, or is contained in U

[rustic haven]

Oh this very quickly shows Ln is subset of U

[abstract quail]

do you mean a quadrilateral with points A, B, C and D?

[abstract saffron]

yes

[abstract quail]

lemme think for a bit

[tawdry widget]

Nvm I thought something wrong…

[abstract quail]

I can think of two cases:

1. Let q and q' be paths corresponing to semi-circles of a circle. So that a and b lie on the endpoints of a diameter. Then the line joining particles must cross the center of the circle at some time, in all parametrisations. Thus, the infinum would be equal to the diameter, as the superemum shall be the diameter for all paths.

[abstract saffron]

😄

usually, for problems like this, what I do is: line segments -> polygonal chain -> rectifiable paths -> continuous paths

[tawdry widget]

Oh, isn’t D(q,q’) simply maximal distance d(x,y) for x in im(q) and y in im(q’)

[abstract saffron]

no

you see, it would be, have it not been because of the continuity of the path

[abstract quail]

2. I can think of this way in the case of a rectangle, perhaps this is too vague. The value of D is equal to the length of the line segment joining the two particles, for any given velocities.

[abstract quail]

sorry, i am confused again...

[abstract saffron]

wait, actually...

it's not that far off

suppose the maximal is attained for some x = q(t) and y = q'(t') for some 0 < t, t' < 1, then this is correct

it's just when the maximum is attained at the boundary that's fucked up.

ofc D(q, q') <= sup (q(t), q'(t')). The trouble is to show the equality sign, which is troublesome exactly in the case above

[rustic haven]

assume there exists at least one point in common between U + Ln

Consider the path-connected components of U in the subspace topology of Z. Partition U into its path-connected components:
U1, U2,…,Uk

Assume, for contradiction, that Ln is not entirely contained in any of the path-connected components.

If Ln is not entirely contained in any Ui​, then there must exist some points on Ln that belong to different path-connected components, Ui and Uj where i does not = j.

This implies two points, one in each of these path-connected components connected by a continuous path within U, as they are in the same path-connected component.

this is a contradiction.

Is this satisfactory proof for the hint?

sorry for interrupting with your other questions guys

[abstract saffron]

but suppose the case above, then we are back to the case where q and q' are two line segments, that we solved above. Again, it's only my intuition. To be proven here.

[tawdry widget]

I think so. Except one small part: you shouldn’t have assumed that there are finitely many path-connected components. Other than that I think it’s fine

[abstract saffron]

oh wait, lmao, it doesn't matter, it's just me being foolish

[abstract saffron]

this is true actually. Fun exercise indeed

hint: ||you may not get to the boundary, but you can get arbitrarily close. Then the argument above works.||

[rustic haven]

will this still be fine if i make this statement? i mean the properties and proof still holds

[abstract saffron]

ohhhh, wait, hold on, I'm mistaken again, it's the infimum over all parametrisations, not supremum

[tawdry widget]

Yeah I wasn’t thinking clearly. inf

[abstract saffron]

But I'm surprised actually. Inf makes a metric?

pretty sure the triangle inequality will fail. We almost always define metrics with sup for this reason

[abstract quail]

the book says that it does define a metric, on equivalence classes of paths

[tawdry widget]

Well I proved it’s a metric indeed. But can’t answer your original question though… finding a geometric explanation for D…

Choose p,q,r: I=[0,1] to the space respectively
Then A=D(p,q)=inf{max{d(p(f(t)), q(t)): t from I}: f monotonic homeomorphism from I to I}
B=D(q,r)=inf{max{(q(t), r(g(t))): t from I}: g monotonic homeomorphism from I to I}
So any ε we can find A<=d(p(f(x)), q(x))=max d(p(f(t)), q(t))<Α+ε
B<=d(q(y), r(g(y)))=max d(q(t), r(g(t)))<B+ε
Then d(p(f(t)),r(g(t))<=d(p(f(t)),q(t))+d(q(t),r(g(t))<= d(p(f(x)),q(x))+d(q(y),r(g(y))<Α+B+2ε
So indeed we have D(p,r)<=D(p,q)+D(q,r)

But I can’t find geometric explanation. You have the two line case by Tadokoro anyway, I tried polygonal chains but went nowhere

Polygonal chains case p, q , im(p), im(q) has m+2 or n+2 vertices respectively
max appears at d(x,y) where at least one of x and y are vertex of im(p) or im(q). But I need to take inf for choices of n points on im(p) and choices of m points on im(q). It becomes too complicated…

[ebon galleon]

[abstract quail]

@tawdry widget thanks. will try to think about it

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

[median sand]

If G is a topological group and G/N is a quotient by a subgroup, then the quotient map is open, right? Since if U is any set, then p'(p(U))=\bigcup_{x\in U}xN=UN=\bigcup_{y\in N}Uy, so if U is open, this set is open, and p(U) is open in the quotient topology.

[gritty widget]

so true

Haha

Me who doesn't understand shi about topology

[ebon galleon]

wut

[quiet thorn]

#chill ?

[tiny ridge]

What did I watch

[feral copper]

💩 music, 💩 meme format and 💩 relation to maths

[tawdry widget]

Yeah, p: G->G/N. Open U, p^-1p(U)=union of hU for h in N. And multiply by h is homeomorphism

[umbral panther]

I think this argument is circular. What is your definition of the quotient topology? I think you’re assuming that it’s an open map

Indeed, the statement is false if N is not closed, eg, if N is dense

[median sand]

I don't see how it's circular. My definition is that given the set G/N imbue it with the finest topology such that p:G->G/N is continuous (the standard definition of a QT afaik). In this case a subset of G/N is open if and only if its preimage is open in G.

[median sand]

What's the joke here, is it just the incongruity of the superimposition? Why does his mood change on a dime, you'd expect the background to change too for it to make sense.

[umbral panther]

I gave you a counter example, so your proof is wrong. I haven’t identified the error, so maybe it isn’t circular

[opaque scroll]

What is the counter example, i.e. which open set will not be mapped to an open one?

[median sand]

IDK what you're on about, mon ami, here is the exact same statement and proof from Stroppel's "Locally Compact Groups".

[opaque scroll]

Seems to work fine for R/Q at least

[umbral panther]

Ok, I don’t have a counter example

[limber wren]

I think a counter example would just be taking [0, 1] and identifying the endpoints, giving S^1. Then [0, a) is open in [0, 1] for any a < 1 but it's image in S^1 isn't open

[opaque scroll]

You'd be hard pressed turning [0, 1] into a topological group I think

Especially if {0, 1} has to be a normal subgroup

[limber wren]

ahh true, I was thinking general quotients

never mind lol

[tawdry widget]

At least I didn’t find mistake in my context… multiply by h is a homeomorphism , hU open, so union of them is still open…

[median sand]

For conversation's sake, can't you just use some bijection with S^1 to transfer the topology and group law? Not particularly useful, but it would work, no? Have the bijection identify 1 with 1 and 0 with -1 and you get a normal subgroup (since it's abelian).

[opaque scroll]

If you put a different topology on [0, 1] you can do whatever you want I guess

[median sand]

No, sure, I didn't mean it for the sake of disproving the "quotient map of subgroup is open" statement, since it's patently true.

[opaque scroll]

And I guess every topological group is homogenous, so [0, 1] has no hope of becoming a topological group

[narrow cairn]

ive realised im really bad at actually showing that 2 spaces are homeomorphic

like im stuck on this problem:

4.5

i have an idea for what the homeomorphism is but its such a drag to actually show continuity both ways

[tawdry widget]

But the homeomorphism I can think of is very well expressed by elementary functions… so naturally continuous

polynomials actually…

[narrow cairn]

huh

[fickle elm]

Use the following model for the infinity cylinder (cosx,sinx,y) where 0<x<=2pi, y is any real number. Consider the map (cosx,sinx,y)-->(ycosx,ysinx,y). It is continous and its image is C. And at y=0, it collapse into a point, so it must factor through M/A. At other points it is a homeomorphism.

[tawdry widget]

Yeah I constructed this one too

(x, y, z) to (zx, zy, z), polynomials

Inverse being rational except one point…

[narrow cairn]

wait i dont follow, where did (cosx, sinx, y) come from?

oh wait nvm

sorry i thought you were talking abt the cone

okay i think i get it

[wind moat]

There is actually a neat way to prove this without exhibiting an explicit homeomorphism. Can you define an quotient/identification map from S^1 x R to C? Well okay you’re kinda doing the hard lift here by defining the quotient map, but you’ll get the homeomorphism part for free.

[feral copper]

That's the kind of problems you answer with a picture 😛
The space S¹×R is a cylinder, and quotienting it by S¹×0 is like tightening a belt and squishing its stomach.
The space z²=x²+y² is a cone (an actual cone, not just a half-cone).
Those are the same spaces!

It is not locally Euclidean near the apex of the two half-cones that make it

It's topologically a wedge of two disks

You could explicitly describe the homeomorphism by working out a parametrizarion of the z² thing: write it as a union of circles of radii increasing linearly.

From this, there is an obvious map from the cylinder to the z² thing, and it factors through taking the quotient by S¹×0

[somber phoenix]

He's lying bro..

V can't be contained in a neighborhood which is at most radius 1 in all dimensions, because V is infinite in almost all dimensions

He's chosen N, so 1/N > 0

[gritty widget]

[narrow cairn]

oh no i totally get why they are the same thats not the issue most of the time

[hidden crag]

stable homotopy theory

[somber phoenix]

Yeah this is just plain wrong. If y is in V, it can in some factor be anywhere on the real line, and that certainly can be outside of the at most 1x1x1x... Of the Ball of D.

Not to mention how specific this set U already has to be.

Omg he's right, that bastard

He did it

But this U has to span to infinity in almost all direction, I think

[somber phoenix]

Hm, I guess it's a metric topology on D

[heady skiff]

why couldn't we just say that since X has a finite subcover and since C is a subset of X, this finite subcover must cover all of C as well?

[queen prism]

you need to show that any cover of C has a finite subcover
a cover of C need not cover X as well
that’s basically what the “add the X - C” part is about

[heady skiff]

oh nah i meant a cover of X covers C, since X is compact we can find a finite subcover for X, and thus a finite subcover for C, since C lives in X

or am i smokin

[ebon galleon]

That's... what it does???

Except a cover of C needs not cover X

So you need it to be closed so that you can add the complement X-C to your open cover

[queen prism]

just start from the hypothesis
“let O be an open cover of C”

[heady skiff]

ok sure

ig that makes more sense then

thanks

i'll just ask in OH

tfw $\phi$ not $\emptyset$

[gentle osprey]

okeyokay

[queen prism]

(\varnothing)

[gentle osprey]

Bladewood

[heady skiff]

$\varnothing$

[gentle osprey]

okeyokay

[heady skiff]

no wy

[ebon galleon]

tfw sequentially compact

[heady skiff]

why is it we can consider C being contained in the union of these specific balls centered at the origin with integer radius, and not a union of arbitrary open sets?

nvm

[tidal lynx]

yeah the open cover in question are the balls

[heady skiff]

ye

[heady skiff]

here, what do they mean by "natural topology"? in my head i'm visualizing 2 dimensions, and if we're considering the x and y axis with the open sets being the union of open intervals then wouldn't the product topologies just be open squares? idk i'm trying to think about balls as well

[ebon galleon]

Metric topology under the usual (Euclidean) metric

Sum the squares of the coordinate (distances), and then sqrt

this give you balls which gives open sets

[heady skiff]

hmm okay thanks

[gritty widget]

Check whose paper this is 😉

[quiet thorn]

memes belong in #chill

[rustic haven]

could anyone lend a hand with this ive tried for a week and cannot get anywhere

[tawdry widget]

Say t_0 =0, and WLOG, for t<0, ψ_1 (t)=0, for t>0, ψ_2 (t)=0. So ψ_1 ‘(0)=0=ψ_2 ‘(0) since it’s smooth, you need only one side. for example you calculate ψ_1 ‘(0) it equals limit of ( ψ_1 (t)- ψ_1 (0))/t for t->0-,

[rustic haven]

just wondering why t<0, ψ_1 (t)=0, for t>0, ψ_2 (t)=0 implies that the first order derivatives = 0 at t=0, kind of understand it geometrically but how do you explain that formally

[gritty widget]

What does it mean they say "embed klein bottle in R^4" (I am taking intro to discrete mathematics class and have only taken linear algebra | ) so take it easy explaining if anyone wants to. Can this image be described using linear algebra?

[novel acorn]

It means there exists an injective continuous map taking the Klein bottle to R^4

[unreal stratus]

I guess cause the klein bottle is compact that is correct lol

[gritty widget]

Is this linear algebra or some other shit that would take me years to learn? WHere is the map from?

If you have a simple of example of this I would like to see it

[unreal stratus]

This is topology

It won't take you years though

[tawdry widget]

Because ψ_1 (t)-ψ_1 (0)=0-0=0 for t<0

limit of ( ψ_1 (t)- ψ_1 (0))/t (for t->0-) is therefore 0, similarly ( ψ_2 (t)- ψ_2 (0))/t is also constant 0 for t>0, so ψ_2 ’(0)=0

[gritty widget]

Why can't it be done in terms of linear algebra and vector spaces?

[unreal stratus]

Why should it be?

That's just one course

[queen prism]

kleine bottles don't look very linear to me but idk

[gritty widget]

A sphere doesn't look very linear to me

[hidden crag]

What's your point?

[gritty widget]

We can represent all points of the sphere in 3d vector space, so why can't we do the same from the klein bottle

Except 4d

[hidden crag]

embedding something in R^3 doesn't mean we can do linear algebra on it or that we can do so by means of lin alg

sure it's a subset but there's no linear structure on either of the objects you're embedding

[gritty widget]

So what R^4 are you talking about

[unreal stratus]

Lol

[gritty widget]

I don't understand, your embedding something into a space with 4 points, what does this even mean?

[hidden crag]

it means you have an injective continous map from the klein bottle to 4 dimensional real space that is a homeomorphism onto its image

[queen prism]

klein bottles don't have much (anything?) to do with linear algebra
they're just a subset of R^4, which is the set of ordered quadruples with real coordinates {(x, y, z, w) | x, y, z, w in R}
in much the same way a cube or a sphere is a subset of R^3
R^4 is often described as a vector space because you can put that kind of structure on R^4, but it doesn't have to be

[gritty widget]

4 dimensional real space? What dimension are you talking about? Not spacial

[gritty widget]

Right

[hidden crag]

just the space of 4- tuples of real numbers

$\mathbb{R}^4$

[gentle osprey]

Timo

[gritty widget]

So the klein bottle is not an object at all its just a mathematical thing

[hidden crag]

what's an "object" for you

[gritty widget]

Yeh thats a question which has layers... 😬

I honestly cant even answer it, I can answer it in terms of reality, but none of math is reality

[queen prism]

if you really care about "what" R^4 is, maybe watch https://www.youtube.com/watch?v=zwAD6dRSVyI

he has some good visualization tips

[gritty widget]

What is a number

[tidal lynx]

#proofs-and-logic is good for this

[gritty widget]

That videos stupid, he's talking about a "hypersphere" like its a real object that has distances and what not, 4 dimensional space simply doesn't exist, there are no objects in the space. I get what they mean about the shifting lines and increasing to higher dimensions, but actually talking about a "hypersphere"? Come on man. Will definetly not be majoring in math 😢

[queen prism]

even if you don't believe in higher-dimensional geometry, higher dimensions are extremely useful in applied math

I believe machine learning plays with thousands of coordinates

[tawdry widget]

Not funny. This is a serious channel. Are you trolling? Like you asked this in a topology channel so you clearly know you are discussing something related to topology, but you acted like “is this linear algebra?” But you didn’t ask in linear algebra channel. And say things like this is not real that is not real…

[hidden crag]

this isn't topology anymore

go to #math-discussion

[gritty widget]

This is different, I agree with him about increasing the singluar lines and allowing them to represent increasing dimensions, say 3 dimensional space and then time as the fourth one , but there are physically no "shapes" "hyperspheres" or whatever rubbish hes talking about in that video, there"s no "cutting sections" of a larger object thats just stupid

[hidden crag]

you just have no idea what you're talking about

and now move channels

[unreal stratus]

Lol

[queen prism]

potato is topologically equivalent to a ball

[ebon galleon]

"Nothing I don't understand is real"

[void tapir]

wait until you find our about infinite dimensional space

[tidal lynx]

Is it just me or is showing that (open set continuity) <=> (nbhd continuity) <=> (epsilon-delta continuity) much more natural than what is commonly done (jumping from the former to the latter)?

[limber wren]

Whats nbhd continuity?

[tidal lynx]

if N_2 is a neighborhood of f(c) then there exists a neighborhood N_1 of c such that f(N_1) is a subset of f(N_2)

[limber wren]

oooh yeah that makes sense

that seems pretty directly equivalent to the inverse open defn'

unless I'm missing some subtlety

[tidal lynx]

No yeah it is

So going from A to B and then B to C is much smoother in my head than just A to C

[fickle elm]

You mean f(N_1) is a subset of N_2, right?

Or do I miss something

[tidal lynx]

yeah sorry

[obtuse meteor]

This is either a top tier level troll or plain ignorance and I don’t know what’s funnier

[unreal stratus]

Lol

[gritty widget]

If I was walking in a 3d space, my location could be represented in r3 , now if wind starts hitting me, that could also be measured in r4. So when he says “hypersphere” what does he mean

[queen prism]

if you want to talk about wind ask the physicists

[obtuse meteor]

You just gave an example. When the square sum of all those variables is 1 :)

[limber wren]

https://en.wikipedia.org/wiki/N-sphere

[gritty widget]

Ok, so its not a spacial object, that is what I'm saying

[unreal stratus]

What do you mean by spatial object

I mean a sphere in R^3 isn't a physical object either

[limber wren]

Ok lets all agree hyperspheres are not spacial objects and move tf on

[obtuse meteor]

Notably this does not mean notions of distance are not meaningful or interesting to study

(Which was what you originally complained abt)

[gritty widget]

Yes I think I have a better understanding now, I hadn't eaten for 12 hours yesterday and was just annoyed that the teacher was talking about a klein bottle in a discrete math class

[obtuse meteor]

Hangry and predisposed to being annoyed. Truly a person after my own heart 😌

[hidden crag]

I do wonder how this topic came up in discrete math

[obtuse meteor]

Klein bottles are good and based that’s how 😌

[gritty widget]

They are also talking about the projective plane and euler characterstics, I think its to give the students who are interested in persuing pure maths a taste of it

topology is discrete math (combinatorics) anyways

[obtuse meteor]

Euler characteristics are extremely useful everywhere

[unreal stratus]

Utterly based

[tribal palm]

from a phil maths perspective, all modern maths is, in it’s symbolic-mechanistic nature

[ebon galleon]

wdym a topology is just a frame (infinitary distributive lattice), so it should be order theory

[obtuse meteor]

What an area of math is is purely about vibes not abt symbols

[tribal palm]

elementary geometry, honestly

[limber wren]

I find math gets more feels based the higher you go, like at some point symbolic manipulation is a tool but it's not what it's about, you need to develop deep vibes to do harder stuff lol

[queen prism]

ya
and that takes time

[gritty widget]

One thing I didn't fully vibe with in the video was the distance formula for higher dimension. Like, I get the three dimensional example, where the the distance in two dimensions become the base of a right triangle in 3 dimensions. But how do I extend this to 4d using the literal thinking they were showing in the rest of the video.

[queen prism]

you take it as a definition in 4D
then if you take a 3D subspace like R x R x R x {0}, the 4D distance of two points in this subspace agrees with their 3D distance

[limber wren]

If you look at length formulas in 2 and 3 dimensions, you get sqrt(x^2 + y^2) and sqrt(x^2 + y^2 + z^2), so in 4 dimensions it would make sense to define it as sqrt(x^2 + y^2 + z^2 + w^2), and you can go to 6,7, or any number of dimensions in the same way

it may be unsatisfactory from a physical perspective if you want to restrict yourself to spaces you can easily visualize but the generalization is straight forward

[gritty widget]

Yeh, I spoke to the course coardinator (discrete mathemetician), and they were baisically telling me they don't get how this stuff works either. Basically all of math is some kind of abstraction that we made up (with reason) and she was telling me the higher you go the more abstract things are and its basically not visual in the slightest at that point. I see how it works but I don't think I'll ever be persuing pure math 🙂

[limber wren]

lol well that's fine, you shouldn't pursue something you're not interested in. There's still a lot of visual aspects to different areas of higher math, but you need to be interested in the math itself for it to be worth thinking about

[unreal stratus]

Tbf e.g. general relativity uses a notion of 4D "distance" and stuff and quantum mechanics uses infinite dimensional analogues

they are useful abstractions even outside math

[queen prism]

with things like the klein bottle and other funky manifolds, visualization is basically like a blind guy feeling the different parts of an elephant and trying to figure out what the bloody animal looks like
or so I've heard

[fading vale]

Lots of math is extremely visual

[limber wren]

also for an interesting example, someone mentioned a while back that every compact manifold in R^n can be represented as the state-space of some planar mechanism, kindof like a double pendulum but it can be a lot more complicated

[gritty widget]

There are many useful abstractions for sure, some that we probably won't use until next century, some we won't use at all. I do not think I have the logic base and making these generalisations is not natural for me. I find it interesting but it would be selfish of me to persue the subject since I don't see myself making any useful contributitions to the field

[limber wren]

so even arbitrarily high dimensional manifolds can be understood in terms of something visual in 2D

[queen prism]

sounds horrifying

[queen prism]

me neither, doesn't mean you can't do it for fun

[gritty widget]

I have people telling me a sphere is just a figment of R^3 and telling me math is extremely visual, i feel a bit lost. If anyone has any books on how to think of mathematics and about it it would be greatly appreciated 🙂

Like I want to think how you think, I don’t want to be in a twisted mess of trying to associate physics reality and math into one thing

[queen prism]

it's just experience
by reading books, talking with other people, and thinking for an ungodly amount of hours, you collect thousands upon thousands of examples of a mathematical phenomenon, over time shaping and refining your understanding of what that thing is
you first learn to think of the world as just 3D space, but after familiarizing yourself with the number line, the cartesian plane, and space, you start to notice that a lot of their properties can be abstracted into rules and structures (vector spaces, topologies, algebras, manifolds) that apply very generally
you notice that object A shares a lot of properties with object B despite looking nothing like it
and once you make this connection, your perception of what A really is changes drastically

[obtuse meteor]

There’s a good physical reason: it’s the curvature 0 metric on real n-dimensional space

(“The” sorta in quotes here. There are a few easy transforms to do that preserve this property)

[hot locust]

okay dumb question after a month of topology, how exactly do you define an open set in a non-metrizable topological space? Our prof discussed it in the context of a metric space then just ran off somewhere. Even the book has done some hand-waving or I am just not able to convince myself. I am not able to deal with the fingers poking at the back of my head whenever I write 'Say U be an open set in X'. HELP.

some book I was using simply defines it as sets in the topology, I mean, what?

[ebon galleon]

A topology on a set X is a collection of subsets of X. The elements of the topology are then called "open sets"

They need not (and often don't, for nonmetric topologies) really have a great notion for why they're "open" is some sense, but that's just what they're called

[hot locust]

It is what it is huh

🥹

[queen prism]

they’re your primitive elements

[hot locust]

primitive elements?

like, just, axiomatically defined?

[queen prism]

they’re the basic objects of study
the rules of a topology define how they interact with each other—unions and finite intersections of different primitives will keep you inside the topology

but beyond that there is not much you can say about what open sets “are”—they are defined by their relations to one another

they are nonetheless very flexible, because the rules of a topology are so simple that a lot of things act like open sets in a topology
as long as your collection of subsets behaves well under set operations, for all intents and purposes it’s a topology

[hot locust]

makes sense, thanks

[narrow cairn]

any small hints on showing that the long line is path connected?

since you can have two points that are infinitely far apart

[gritty widget]

then just make an infinitely long path

[narrow cairn]

alright then

[ebon galleon]

:o

[narrow cairn]

but i thought if you could have an infinitely long path then the topologists sine curve would be path connected

wait but:
[0, ω] isnt compact since (a, a+2) for -1 < a ≤ ω has no finite subcover (and similarly, cant be contained in any compact set)
[0, 1] is compact
the continuous image of any compact set is compact

[red yoke]

Cuz what does "infinitely far apart" mean

You can embed any countable ordinal in R, then the distance becomes "finite"

[narrow cairn]

where does this fail?

[red yoke]

Does not cover ω

[narrow cairn]

oh fuck right

the intervals behave weirdly when they have to include limit ordinals

this shit is hard to get your head around

i didnt think topology would be this hard

[red yoke]

Ordinals are fun stuff

[narrow cairn]

does constructing ω_1 require choice?

[red yoke]

Idts

Take any uncountable well order and take the least uncountable element

[sinful cloak]

uncountable well order requires choice

[red yoke]

Oh

[ebon galleon]

no

knowing that there is a well-order of any cardinality requires choice

but there are uncountable well-orders without choice

idk if you need choice to know that the least uncountable ordinal has cardinality aleph_1 or not tbh. Wikipedia seems to indicate that you maybe don't?

https://math.stackexchange.com/questions/1833322/a-well-order-on-a-uncountable-set

[red yoke]

Pog

[sinful cloak]

I see, basically w_1 = { w | w is a countable ordinal }.

[ebon galleon]

Yeah I think that's what they said works. But also, it would be morally wrong if all ordinals were countable lmao. Since there's a proper class of them they should take (idk the exact formulation/proof off the top of my head) arbitrarily large cardinalities

[sinful cloak]

Well you can do the same thing I think. Assume W was the set of all ordinal numbers. W itself can be well ordered by ordinal comparison and is hence an ordinal. If W is itself an ordinal, it would need to be an element of W.

[ebon galleon]

Right, I think that's pretty much the proof that it forms a proper class

[red yoke]

Wait how is aleph_1 defined

[sinful cloak]

Cardinality of w_1.

[red yoke]

Sus

[sinful cloak]

oh yeah I think that is then by definition.

[ebon galleon]

Ooooh I never actually read up on how aleph numbers are defined

I thought every (infinite) set had an aleph cardinal without choice, but I guess that statement is equivalent to every set being well-orderable huh

Welp. Today I learned

[queen prism]

beware the pipeline

[narrow cairn]

we need to start using δ and then a picture of a subset for the boundary

i love that

[narrow cairn]

okay, for a path between 1 and ω, does f(x) = 1/x when x > 0 and ω when x = 0 work?

wait, is that even continous? seems terrible to check

its also weird because like, what are the neighborhoods of ω

like how does the order topology on ω1 even work, let me think abt this

okay so every neighborhood containing ω must contain some finite number? weird shit

but that number can be arbitrarily large

okay so if U is open and doesnt contain ω (or any infinite ordinal) then trivially this map is continuous since analysis

if open U does contain ω then just consider U = (a, ω + 1) in which case the preimage is [0, 1/a) which is open in [0, 1] so this is a path

wait holy fuck do you use transfinite induction here

thatd be cool

i know its probably easier than this but i have to try it

[heady skiff]

are there any general guidelines/ideas to follow when constructing homeomorphisms? for me it's very different from the notion of showing isomorphisms or even cardinalities, for ex i'm strugglint to show show that the unit cube is homeomorphic to the unit sphere

[unreal stratus]

Disk I suppose you mean, or boundary of the cube

[gaunt linden]

No, cube goes with sphere.

[unreal stratus]

I guess usually in situations like this geometric intuition

To me cube has always meant [0,1]^n or similar

[gaunt linden]

Ah, I get it. He should have said "... to the unit ball".

[heady skiff]

oh yea

sorry

maybe i'll just think about it as two sets and try to construct something from that

instead of trying to see something visually

which i have no idea how to describe functionally

[gaunt linden]

If you're coming from abstract algebra, suppose the main novelty is that there's a lot more freedom when picking a homeomorphism than for algebraic isomorphisms. So you'll need to get comfortable with "just choose something that feels vaguely right in each part of the function, and it probably will be" whereas in algebra it's often been "once you choose this tiny little detail to begin with, it forces everything else to be exactly so-and-so".

[heady skiff]

right, bc like

lmfao just choose where generators go

and usually there's a very obvious candidate

here it's a lot different it seems yea

[heady skiff]

(if u are finitely generated for a small amount of generators that is lol)

[gaunt linden]

So, e.g. you could say something like "put the center of the cube at the center of the ball, and then scale each ray out from that center linearly such that the end of it ends up at the surface of the new shape you want it to have.

Or, since the shapes are convex, you can pick any interior points of each to be "center" and do similarly.

The main challenge here is not to find something that will work -- topological spaces are very squishy so almost anything will -- but to find something where proving it works doesn't drown you in uninteresting symbol bashing (or at least something that needs as little of it as you can get away with).

[heady skiff]

i see yea that makes sense thanks

i think my main trouble is actually putting these squishing/transformations into the form of a function

ok fine scaling down the cube to fit inside the ball is fair enough

but then expanding it to fit the ball? no clue how to come up with that

anyways for this one i'll probably look up the answer anyways, spent way too long trying to figure it out lol

[gaunt linden]

Another idea that may be easier to implement might be to do it in several steps, along different coordinate axes, using lemmas such as:

if f(x) is a continuous always positive function, then (x,y,z) -> (x,f(x)y,z) is a homeomorphsm of R^3 to itself (and therefore from any subset to its image).

Suppose we start by knowing a square is homeomorphic to a disk -- then we can apply this to the xy-plane of our cube to make it into a cylinder.

Then scale each horizontal layer of the cylinder differently, such that the middle latitudes get the correct shape, but the poles still end up flat (because we can't afford collapsing the whole end disk of the cylinder to a point).

Finally scale each cylindrical shell vertically by an appropriate amount to get the shape of the polar areas right.

The result of all this might not be as nice as scaling rays correctly in one step is, but proving that each step is continuous everywhere could be simpler.

[heady skiff]

I see, thanks Tropo

yeah that definitely makes more sense, breaking it up into chunks instead of tackling it all at once

i suppose it's a matter of practice as well too, so i'll try to show other stuff is homeomorphic

[heady skiff]

i don't understand how they got to this point, shouldn't it be t|y_i|??

[obtuse meteor]

Yes just make it y = tx

Same thing in practice

[obtuse meteor]

Note this fails at 0 a bit and one should thonk

[heady skiff]

ye ok thanks

i don't understand how that implies that t = 1 tho

[heady skiff]

also is the method of reasoning that they used for surjectivity normal in topology? seems kind of hand-wavy/non-precise, but then again it is a very visual subject

does anybody know of showing this is homeomorphic to this problems

[obtuse meteor]

This is definitively precise lol

It’s an intermediate value theorem argument

[heady skiff]

damn

that's crazy

[obtuse meteor]

Hm? Wdym

[obtuse meteor]

You can just pull multiplying by t out of the max and then it’s like Blah = |t| * Blah so |t| = 1. -1 can be ruled out easily

(The map preserves more than collinearity. It preserves direction)

[obtuse meteor]

I mean how do you know x^3 is surjective [0,1] -> [0,1] . It’s the same type of thought process

[heady skiff]

i suppose but that's not really my thought process lol, i would just take any x in [0, 1] and take the preimage to be the cube root of x

anyways thanks for the help tho

[obtuse meteor]

But my good sir

Why do cube roots fucking exist

(The only way to prove this easily is with continuity)

[tidal lynx]

what’s the easy way

I know that the usual way is to declare it as z = sup({y | y^3 < x}) (checking that z^n = x is the annoying part)

[queen prism]

if we're talking about continuity then maybe it's via the inverse function theorem since d/dx x^3 > 0 (except at x = 0 but that's a trivial case)

although proving the inverse function theorem is already a fair amount of work... (I'm sure IVT must be overkill here)

[queen prism]

yeah this is how I would prove it

not fun

[tidal lynx]

what does the inverse function theorem tell you here?

injectivity yea, but that’s also just true because x^3 is increasing

oh shit

[queen prism]

uhhhhh

[tidal lynx]

you can just spam IVT no?

[queen prism]

idk

[tidal lynx]

to show that the image of x^3 is all of R

so then x^3 is both injective and surjevtive, so it has an inverse function

I guess this doesn’t tell you what the inverse is… idk if that’s important

[queen prism]

e wait

what about the other IVT

lol

i kinda dismissed it because i forgot the exact statement

[tidal lynx]

Yeah maybe my “spamming IVT” is the proof of (usual statement of IVT) implies (image of interval is interval)

[queen prism]

does this count as an easy proof

[tidal lynx]

What I had in mind was: given an arbitrary real y, to show that it is in the image, apply (usual) IVT to (cbrt(a), a) and (cbrt(b), b), where a is some perfect cube less than y and b is some perfect cube greater than y

Maybe this is what Faye had in mind

[tidal lynx]

Idt it’s that important now that I think about it

once you know that x^3 has an inverse function, you can still prove the usual exponent laws using this inverse function

[tidal lynx]

Let f: R -> R be continuous, and a and b reals. Suppose image of the interval (a, b) is not an interval. Then there exists reals x, y, c such that f(x) < c < f(y) but c isn’t in the image of f. This contradicts usual IVT.

[tidal lynx]

This is funny, cuz the usual topological proof of showing the usual IVT statement uses the fact that the image of connected sets is connected, but in R the only connected sets are intervals…

[tidal lynx]

But then in a real analysis class you probably prove usual IVT first and then do this proof

[ebon galleon]

Does anyone have an example of a quotient map p: X --> Y with X and Y locally compact Hausdorff which known to not be preserved by - x A for some (necessarily not locally compact) Hausdorff space A?

A bit specific, I know, but it would give a nice counterexample for something I need.

[heady skiff]

can somebody tell me where the max in the function came from? what's the point of it being there?

topology hard

actually all of math hard lol

[queen prism]

you remember |x|_p norms from functional right

when you take p = 2, your unit ball looks like the ball we all know and love

as you increase p, the shape of your ball approaches a cube

and in the limit as p -> infty your norm happens to be "take the longest component of your point"

[heady skiff]

no i don't rip, i only studied functional for like 1 week and then school started

but that's just the norm you sent right

in the wikipedia article

or wait

[queen prism]

uuuuh not really, that one's for functions

[heady skiff]

i remember what this

i think

it was the one i was strugglinw tih about why the limit goes to it

or smt

[queen prism]

basically this is what you want

yeah it's that one

https://www.desmos.com/calculator/ufbffm8zj4 take this and try playing with the parameter p

as p -> infinity the graph approaches a square
and your norm approaches |(x, y)| = max{|x|, |y|}

here's the cube/sphere example as phrased by lee

try thinking about what it would mean to be in the set C

you can convince yourself that C is a proper cube

[heady skiff]

oh thank you, what page is this on?

[queen prism]

29

in top manifolds

[heady skiff]

yea, i'm seeing C to be the surface of a cube

idk what the terminology is

but with the inside empty

because we required that one of the coordinates has absolute value 1

what does he mean by cubical surface of side 2 centered at the origin tho lol

[queen prism]

coob has side length 2 and is centered at O
= coob of radius 1

[heady skiff]

yea makes sense

the cube is empty though right?

or the surface is the only thing filled in

[queen prism]

that cube is empty

(only surface)

[heady skiff]

ok

[queen prism]

but you can probably see how to adapt it to fill in the cube

[heady skiff]

mmm yea i'll have to think about it

how can i see this, specifically the part that this formula gives the unit vector projected inward in the direction of p

is this something related to basic physics/multivar that i forgot

that dividing by the distance metric or whatever gives a vector or smt

[queen prism]

take a point x in space
divide it by its length |x|
what do you get

[heady skiff]

a scaled down version of it closer to the origin right

[queen prism]

right
and its length is actually 1
which means it's on the unit sphere

the phrasing is a little confusing but what it means is take a point in C
the cube is bigger than the sphere so all this transformation can do is shrink the length of the vector (or leave it unchanged), never stretch
nor does it turn the vector
so that's what it means when we say it projects the point inward to the sphere

[heady skiff]

oh ok... so the presence of the norm of x in the denominator here is to make sure that it's of distance less than 1 so that it's in the unit ball, right?

[heady skiff]

ahhh okay that makes sense then

!

[obtuse meteor]

You need a scalar which ranges from 0 to 1

From the defn of a cube a possible scalar is that maximum

This is really the only thing you need you could fiddle with it a lot

[obtuse meteor]

The reason you need such a scalar is you want to cover all a ray of length 1 in any direction to cover a sphere

[heady skiff]

i see.... i think i 'm starting to get the idea

i guess i have to digest that it actually works

because rn all i'm seeing is just shrinking to the unit vector then possibly scaling by a real number between 0 and 1, and i'm struggling to see how that actually suffices to cover the unit ball

because rn i'm just thinking about dilating and shrinking.... it's fine tho i can go to OH

like

i can’t see how it can curve to fit the unit ball yknow

[narrow cairn]

okay so what i've been thinking: say theres a path $1 \rightarrow \alpha$. then there is certainly a path $\alpha \rightarrow \alpha + 1$, and you can just squish and glue together the paths. next suppose that $\kappa$ is a limit ordinal. since it is a countable limit ordinal, it is the limit of a sequence $({\alpha}_n)$. there is a path by the induction hypothesis between each of these, so take $(p_n)$ where each $p_n$ is a path between ${\alpha}n$ and ${\alpha}{n+1}$. take $q_n(x) = p_n(\frac{x}{2^n} + (1 - 2^{k-1}))$, and set each $q_n(1) = \kappa$. then by the gluing lemma there is a path $q$ from $1$ to $\kappa$. therefore by transfinite induction, there is a path from $1$ to all countable ordinals (trivially there is a path to 0) so $\mathscr{L}$ is path-connected.

does this work?

if this does work then this is super cool

[ebon galleon]

glueing lemma is used how? Also I feel like there is a nice closed form of what the sum^{n-1} 1/2^k looks like

[narrow cairn]

right lol

gluing lemma is used since all q_n and q_n+1 overlap at alpha_n+1

[gentle osprey]

most likely to honorable

[ebon galleon]

So q_n should be giving a path alpha --> alpha+1 or something? Also, what's the domain of q_n?

[narrow cairn]

domain of q_n is uh i dont know how to describe it

[ebon galleon]

Some part of the interval, such that the union is [0,1]?

[narrow cairn]

yes

[ebon galleon]

So like maybe you mean for the domain to be like [1-(1/2)^n, 1-(1/2)^{n+1}]?
If so, you can't apply the gluing lemma to this

[narrow cairn]

yes i did mean that

oh why not

[ebon galleon]

Glueing lemma required the cover to be locally finite, so each point has a neighborhood that intersects finitely many of the sets. No neighborhood of 1 satisfies that

[narrow cairn]

fuck

god dammit

fuck

fuck

[ebon galleon]

It's okay, I've made that mistake before too lmao

[narrow cairn]

well im not too worried about the mistake just that 'proof' took all of my brain power

and now i am very lost

i dont think i can modify it to resolve this

[ebon galleon]

yeah idk how you prove it tbh

[narrow cairn]

god dammit that was so cool too

i hate topology

ruining my transfinite induction

my book says it doesnt have to be locally finite just either open or finite closed

either way cant apply it here

wait, can you?

maybe

because you might be able to do like (1 - 2^k - eps, 1] which is open and you could maybe do some clever math to find a good eps

man now im totally lost, i felt smart writing that one

is there really not a way to construct a path from countably many component paths?

[obtuse meteor]

this is the point of the exercise right like. Corners aren't topological

It might help to make the map animated (i.e. turn it into a deformation retract to be technical).

If you add a time parameter t which specifies how far along the line from 0 to the point on your cube you go to get into the sphere

you essentially "slide" farther in if you're on the outer parts of the cube (outside the sphere).

So that's how you get rid of the corners

[ebon galleon]

It looks like you might wanna do some order topology shenanigans or something. A similar idea might work? But glueing lemma won't give you the result

[narrow cairn]

man im really gonna be stuck on this problem for another week arent i

maybe the gluing lemma holds for things like this when the domain is a compact subset of R?

R is nice so maybe it works

[ebon galleon]

You need the locally finite condition.
Define f : [0,1] --> R by mapping [1 - (1/2)^n, 1 - (1/2)^{n+1}] to the line segment between n and n+1, and then picking any real number to map 1 to. This is a cover of [0,1] for which the restriction to any of the closed sets is continuous, but clearly f is not continuous itself.

*the cover is { [1 - (1/2)^n, 1 - (1/2)^{n+1}]] } U { {1} }

So it's something special to something something ||order iso between omega x [0,1) with the dictionary order and [0,1) itself for countable ordinals omega|| or something? At least that's what it looked like from my skimming

[tidal cedar]

yeah for these sorts of problems I find myself thinking about it as a homotopy

this might not actually be a good way but idk

[tawdry widget]

This is exactly what I told you. Cube to ball, r-layer of cube (cubic surface whose distance from origin is r, under norm max(|x1|,…,|xn|}) to r-layer of ball (sphere whose distance from origin is r, under norm sqrt(Σxi^2)) , (x1,…,xn) to r(x1,…,xn)/sqrt(Σxi^2)
Therefore together (x1,…,xn) to max(|x1|,…,|xn|)(x1,…,xn)/sqrt(Σxi^2)

[onyx raft]

Are people talking about cube homeo sphere?

Just realize both are simply connected surfaces

[tawdry widget]

Solid cube to solid ball

I told him once each “layer” are homeomorphic so together he will have a homeomorphism. Strange that last time he didn’t go through

[hot locust]

If X is compact hausdorf sp., then a closed sp. F and a point x not in F can be separated using open sets.

Like, I get that if you can show that there exists, say V, open in X such that F is a proper subset of V, such that V intersection H is empty, where H is a subset of F complement then we are done. But I am stuck.

F is compact, okay, fine. so there's an open set that contains F. But how do I show that the intersection between this open set and H is empty?

[narrow cairn]

here's a fairly strong hint: ||for every point p in F, take a neighborhood U_p of p and V_p of x that are disjoint (you can do this since X hausdorff). then take the cover of F by all of these sets, and since F compact, take a finite subcover. then you have a collection of neighborhoods U_1, U_2... U_n that cover F and V_1, V_2... V_n with U_n and V_n disjoint. how can you use these to construct the desired neighborhoods?||

[hot locust]

oh got it, I think

thanks

[hot locust]

Example of a T1 space that's not hausdorf?

[quick delta]

Cofinite topology on any infinite set

[grave solstice]

Can someone help me understand this?

So what I understand: The discrete topology is the topology where the open sets are the subsets of F (the field). Then, given x in F a neighborhood of x is a subset of F containing x. Finally, F will be locally compact with this topology iff for every x in F there exists a neighborhood of x that is (contained in a) compact (subset), meaning that any cover (of the subset) admits a finite subcover.

wouldn't {x} satisfy the requirement already? or just any finite subset containing x?

[opaque scroll]

Yes, that's right. Any space with the discrete topology is locally compact.

[grave solstice]

how is this useful

like the discrete topology, unless it resembles the p-adic topology or something

[opaque scroll]

It's not, that's sort of the point

[grave solstice]

ah ok so when they say "... one adds the condition that the field should not be discrete" they mean that the topology should not be the discrete one

[opaque scroll]

Like you can equip any field with the discrete topology, and that doesn't give you anything interesting. Therefore they want to study fields that are not discrete

Yeah, they want a more interesting topology

[grave solstice]

idk why I interpreted that at first as the field not being finite lol

i c i c

[grave solstice]

I tried to prove this. This is my idea: For every rational x in [0,1] let H(x) be its denominator written in reduced form, call it the height of x. Define H(0)=1. There are exactly phi(n) numbers of height n in the interval and these can be linearly ordered, say these are H_n^1<H_n^2<...<H_n^phi(n). Now, for even n let S(2n) be the sequence (for 2n) linearly ordered and for odd n let S(2n+1) be this sequence for 2n+1 but with reversed order. Then we have two sequences S(1)S(2)S(3).... and S'(1)S'(2)S'(3)... (note that S'(1) is empty and from that point the sequences are the same) the first enumerating the rational interval [0,1] and the second enumerating the rational interval (0,1), To define the homeomorphism [0,1]_Q-->(0,1)_Q simply pair the two sequences.

This should work right?

mmh not sure, like what you are doing is moving every element in the height sequence defined above two places to the right

I'll think more about it

[grave solstice]

lol

[queen prism]

continuity is a fucking myth

[red yoke]

There's no way that's continuous

[red yoke]

What if you do
||(-1, -x)(-x, -x/2)(-x/2, -x/4)…{0}…(x/4, x/2)(x/2, x)(x, 1)
≈
{0}…(x/4, x/2)(x/2, x)(x, 1)||

And repeat

[sturdy notch]

I'm slightly confused about a definition/notation involving spectra
If E is a spectrum, then does $E^*(E)$ denote the direct sum $\bigoplus_n E^n(E)$ or does it denote $\bigoplus_n \lim_k E^{n+k}$?

Because the mod p Steenrod algebra is defined as the algebra of stable mod p operations so it can be written as $\mathbb{H}\mathbb{F}_p^*(\mathbb{H}\mathbb{F}_p)$

[gentle osprey]

dadaurs

dadaurs

[sturdy notch]

this is using the second notation

[heady skiff]

sorry boss i'm kinda new to this, will work on it tho

in a metric space, the closed sets are the closed balls right?

[amber raven]

No they're the sets F so that when a point x of the metric space verifies that all balls of center x cross F, x belongs to F

[sturdy notch]

I should say that I'm taking an inverse limit over the maps E^n(E) -> E^n(SE) -> E^n-1(E)

[limber wren]

close balls are closed yeah, but that's not all of them. In general an open set is the union of open balls, and a closed set is the complement an open set

[heady skiff]

ah ok yea that makes more sense, for some reason i thought that the open sets were exactly the open balls and same with closed sets being closed balls

got it tho

thanks!

here, I set inf${d(a, b) \mid a \in \bar{A}, b \in \bar{B}} = \delta$, and I let $U = \bigcup_{a \in \bar{A}} B(a, \frac{\delta}{2})$, $V = \bigcup_{b \in \bar{B}} B(b, \frac{\delta}{2})$. So $U$ and $V$ here are clearly open and clearly $A \subseteq U$ and $B \subseteq V$. I assumed their intersection was nonempty, so I let $z \in U \cap V$ and obtained $d(a_i, z) < \frac{\delta}{2}$, $d(b_j, z) < \frac{\delta}{2}$, to get $d(a_i, b_j) \leq d(a_i, z) + d(z, b_j) = \delta$ which contradicts our choice of $\delta$. would this argument work?

[gentle osprey]

okeyokay

[heady skiff]

i'm not really sure how this rules out |t| = 1, -1, can you expand on this pls?

nvm

i'm an idiot

lol

[heady skiff]

how tho? how do we know that the maxes are equal?

[obtuse meteor]

You’re proving it’s injective so you’re assuming f(x) = f(y)

This implies y = tx for some t

[heady skiff]

mhm

[obtuse meteor]

Then max(stuff for y) * tx / |tx| = max(stuff for x) * x/|x|

[heady skiff]

oh

[obtuse meteor]

And it’s just like manipulations from here

[heady skiff]

oh wait i'm so dumb

ok thanks! i think i got it then

appreciate it

[umbral panther]

Something with a star always represents the sequence where the star takes on integer values. You can turn a sequence into a single object by taking the direct sum

In particular, E^*E represents the sequence E^nE, which are the homotopy groups of the spectrum of maps from E to E, or, equivalently, the sequence of groups of homotopy classes of maps from S^nE to E

I don’t understand your second possibility I don’t think it makes sense. There is no map from E^nE to E^n+1E. But it may be tricky to define maps of spectra, especially if they aren’t spectra, but just prespectra, and you may need to take a (different) limit. But that’s just to define a single E^nE. You still need all of them

[sturdy notch]

Yeah Im just a bit confused by wikipedia’s definition of the steenrod algebra

Because it only contains the stable cohomology operations which obviously isnt all of HF_p^*(HF_p)

Im also pretty unfamiliar woth spectra so Im extra confused

[umbral panther]

An unstable operation is a map of spaces between K(pi,n). Maps between then that are maps of spectra are stable

[sturdy notch]

Ooh and then it all makes sense ofc

Yeah ofc the maps of spectra are stable and those are precisely represented by the limit

Alright that cleared up a lot, thanks 🙂

[umbral panther]

Also, H is a spectrum, not a prespectrum, but defining a map of spectra still is an inverse limit. It is a compatible map on all spaces. That is an inverse limit

[unreal stratus]

Nice

There's a fun way to do it like consider f(x) = d(x,A)/(d(x, A) + d(x,B)l

Continuous, preimage of 1 is B, preimage of 0 is A

[heady skiff]

Holy shit I did something right

[heady skiff]

Huh interesting

[ebon galleon]

metric spaces are T_6 frfr

[ebon galleon]

hmm the standard examples for quotient maps not preserved by products is to take X = R, y = something like R/Z, and A = Q, so the map RxQ --> (R/Z)xQ is not a quotient map. But I don't think R/Z is locally compact ( since [Z] has no compact neighborhood ) (where R/Z is with Z contracted to a point)
But I can't cook up any examples

[heady skiff]

to show that any two open balls in R^n are homeomorphic, it would suffice to first use the taxicab metric to move the center of one ball onto the other center, then scale up the ball right

[gritty widget]

what does the taxicab metric have to do with just translating a ball?

just translate it

[heady skiff]

wait i think i'm confused

[gritty widget]

if you have two points a and b, then the mapping f given by f(x) = x + b - a takes a to b

[heady skiff]

so i would have to first align them one-dimensionally, then two-dimensionally, etc.?

wait

i think i might be high

oh it would just be $f(x_1, \dots, x_n) = (x + x_1, \dots, x + x_n)$ where $x$ is the center of the ball in question, at least for translation

[gentle osprey]

okeyokay

[gritty widget]

.

[heady skiff]

wait i'm confused that's only defined in terms of a right

[gritty widget]

???

x plus b minus a

[heady skiff]

yea but if you plug in something like c you're not gonna get b

[gritty widget]

okay, but who cares? all you need to do is take a to b

if anything other than a also mapped to b, we wouldn't have a homeomorphism

[heady skiff]

and here a and b are the centers of the balls right

[gritty widget]

sure

[heady skiff]

so you would have to define this function individually for each point in the ball

sorry i'm confused

bc then how would you define say f(x_1, ..., x_n) if you don't know the point you're mapping it to

for any arbitrary point in the ball

[gritty widget]

all you asked about in the first place was moving one centre to the other

i urge you to think about what this translation is going to do to the whole ball

let a be the centre of one ball and b the centre of the other. define the map f by f(x) = x + b - a. what does the image of the first ball look like?

draw a picture

the ball with centre a is being translated so that it has the same centre as the other ball. now you can scale

[uneven bronze]

Basic question. Given an open ball $B(x,r_1)$ of radius $r_1$ and choose $r_1>r_2>0$. Is the $\textit{closed}$ ball $\bar{B}(x,r_2)$ of radius $r_2$ necessarily contained in $B(x,r_1)$?

[gentle osprey]

sunside

[uneven bronze]

The question arose from the following; if X is an open set, then for every point in X there exists an open ball B(x,r) of radius r>0 contained in X. I was wondering if by this we also get a closed ball around x with radius s if we choose s small enough.

[queen prism]

define a closed ball

[uneven bronze]

Open ball: $B(r,x)={y\in X: d(x,y)<r}$ \

Closed ball: $\bar{B}(r,x)={y\in X: d(x,y)\le r}$.

[gentle osprey]

sunside

[queen prism]

alright

so let (B(x, r_1)) and (\overline{B}(x, r_2)) be two concentric balls with (r_1 > r_2)

how do you show that the second set is contained in the first set?

[gentle osprey]

Bladewood

[queen prism]

generally speaking, how do you show that one set is contained in another?

[uneven bronze]

you show that for every y in \bar{B}(x, r_2), y must also belong to B (x, r_1), right?

i.e. $y\in \bar{B}(x, r_2)\implies y\in B(x, r_1)$

[gentle osprey]

sunside

[queen prism]

alright

can you do that?

[uneven bronze]

yeah, should be no biggie 🙂

if $y\in \bar{B}(x, r_2)$, this means $d(x,y)\leq r_2$, but $r_2<r_1$, so $d(x,y)<r_1$, which implies $y\in B(x,r_1)$

[gentle osprey]

sunside

[uneven bronze]

So we can say that if a set is open, there is also a closed ball around each point

[queen prism]

yes

[pliant elk]

kind of basic, but I'm having a bit of trouble getting intuition for the proof that homotopy respects composition. If $F_0,F_1:X\to Y$, and $G_0,G_1:Y\to Z$ are homotopic, then taking $H_F,H_G$ as the respective homotopies, you can define $H':X\times [0,1]\to Z$ by $H'(x,t)=H_G(H_F(x,t),t)$. This clearly has $H'(x,0)=H_G(H_F(x,0),0)=H_G(F_0(x),0)=G_0(F_0(x))$, and identically $H'(x,1)=G_1(F_1(x))$, but I'm having issues trying to show it's continuous. Like, how do you take the preimage into $X\times [0,1]$ where $[0,1]$ "appears" twice (you're mapping $X\times [0,1]$ to $Y$, then reattaching $[0,1]$ to map it into $Z$).

[gentle osprey]

frog

[ebon galleon]

Both $H_F : X \times I \to Y$ and $\pi_I : X \times I \to I$ are continuous, where $I = [0,1]$ and $\pi_I$ is the projection map. Thus, by the universal property of products, $(H_F, \pi_I) : X \times I \to Y \times I$ given by $(H_F, \pi_I)(x, t) = (H_F(x, t), t)$ is continuous. Now compose that with $H_G$.

[gentle osprey]

Ryx (Home for flowers)

[heady skiff]

why is it that if O is any open set in Y, then f^{-1}(O) = {x} x O is an open set in {x} x Y? is it because {x} is the whole space if we're talking about {x} x Y and O is an open set in Y?

[obtuse meteor]

Yes

Essentially

[heady skiff]

ok i think that makes sense then

[obtuse meteor]

X x O is Open in X x Y so (X x O) intersect ({x} x Y) is open in {x} x Y

This is exactly {x} x O

[heady skiff]

got it

thank you then

[obtuse meteor]

You’re not trying to map everything to b this would be a bad map.

Pls leverage geometric intuition otherwise topology is pointless

[gritty widget]

for the love of god okeyokay draw pictures

[heady skiff]

sweet thanks my guy

that's a good point TTerra

[obtuse meteor]

Begging pleading and on my knees for ugs to draw (good) pictures

[ebon galleon]

this might help I wonder why people keep emphasizing to draw pictures

[heady skiff]

i do draw pictures but i'm not an artist or anything

didn't realize i was taking an art class

damn

[gritty widget]

moving a circle in a straight line does not require artistic proficiency

[heady skiff]

idk dude you're kinda overestimating my artistic capabilities

[dusk heron]

It is a classic fact that a bijective continuous function $f:X\to Y$ need not have continuous inverse (for, say, $X$ and $Y$ subsets of Euclidean space in general), and also, that if $X=Y=\mathbb{R}$, then $f^{-1}$ \textbf{is} actually necessarily continuous. Are there any simple examples of a continuous bijection $f:X\to Y$, where $X$ and $Y$ are \textbf{open} subsets of some Euclidean spaces, and where $f^{-1}$ is not continuous?

[gentle osprey]

gustavn64

[pliant elk]

right, thanks

[obtuse meteor]

Space filling curve

[opaque scroll]

Space filling curves are usually not bijective though

[obtuse meteor]

Oh oof I thought you could make them that if you were non compact

[obtuse meteor]

Open annulus and half-open square

With the wrapping function

Eek

I will think more my coffee hasn’t kicked in