#point-set-topology

1 messages · Page 163 of 1

real notch
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Trancendentals are nice
Tldr is they basically satisfy linear independence with their powers

gritty widget
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transcendetal + irrational may make rational

heavy gull
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Well... if you think of an irrational as having an infinite chain of decimals without any repeating pattern (lest it'd be rational), then there'd be another such irrational that complements the first to make an even and nice integer right?
You know, like 4-pi would still be such an irrational number, and that one plus pi would be... 4

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So I really can't wrap my head around an irrational for which we couldn't do that

real notch
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$\sum_{i=0}^N c_i x^i = 0$ implies all $c_i$ are zero for any $N$ and trancendental x

gentle ospreyBOT
real notch
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*c_i being rational

gritty widget
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OK YEAH I KNOW but tahhts irrelevant to the problerm actually

real notch
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True, just a small digression since you mentioned not knowing them

gritty widget
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I said I dont know their properties

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but its ok

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anyways, this problem is probably solvable using actual topology, fuck you have no idead how much shit I need to learn before my exams

real notch
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So
What closed boundary sets are there?

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Like, do you know any nice characterization?

gritty widget
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for sure those are sets of finitely many points

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can't say more

real notch
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So there’s finitely many points in each Cn?

gritty widget
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no

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I mean

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would irrationals of [0,1] work

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?

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clearly empty interior

real notch
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Is that closed?

gritty widget
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would it be closed tho

real notch
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Well, are the rationals closed?

gritty widget
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ok yeah thats closed I think

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I mean we are talking topology of all R not just interval, but yeah its closed

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so C_n are sets of points

real notch
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So that’d be R\Q

gritty widget
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C_n doesnt have interval

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thats the characterization

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Im lowkey thinking induction might work

real notch
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Wait is Q closed in R?

gritty widget
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yes

real notch
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So irrationals are open

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Right?

gritty widget
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wait sorry

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Q is neither open or closed

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obviously actually

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so same with R\Q

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ok I think C_n are made of finitely many points

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anyways, im spamming, I'll play with it

real notch
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If it’s finite, then induction works

gritty widget
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no thats not the argument at all

wanton marsh
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what's the problem question ?

gritty widget
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For all $n \in \mathbb Z $ let $Cn$ be a closed boundary set in the interval $\left[n,n+1\right]$. Let $D= \bigcup_{n \in \mathbb Z} C_n$. Show there exists $t \in \mathbb R$ such that $t + D \subset \mathbb R \setminus \mathbb Q$

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closed, boundary meaning closed and empty interior

gentle ospreyBOT
wanton marsh
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so for example, the cantor set is a closed boundary set ?

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trying to think of the biggest closed boundary set i can find

gritty widget
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Im not sure if Cantors set is boundary, probably is

real notch
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Cantor has empty interior, unsure on closed

gritty widget
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it is closed

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cause its compact

real notch
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Ah ye

heavy gull
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Since there's been a pause here again I wanna ask again, anyone know literally anything about arcpolygons or any place that is talking about them? Aside from the incredible overspecific and not so useful german wiki page I found literally nothing: https://de.wikipedia.org/wiki/Bogenvieleck

Bogenvieleck

Bogenvielecke oder auch Bogenpolygone sind eine Art von Gleichdicken. Ihnen liegt jeweils ein Vieleck (Polygon) zugrunde, dessen Seiten durch Kreisbögen zwischen jeweils zwei benachbarten Eckpunkten ersetzt werden, deren Mittelpunkt der gegenüberliegende Eckpunkt ist. Das z...

gritty widget
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@gritty widget: BCT is good for this type of things. Enumerate the rationals $\mathbb{Q} = {q_1, q_2, \ldots }$ and note that the translates $C_i - q_j = {x - q_j : x \in C_i }$ are closed sets with empty interiors, so $\mathbb{R} \setminus (C_i - q_j)$ is open and dense, and the index set $(i,j)$ with $i \in \mathbb{Z}$ and $j \geq 1$ is countable, so $\cap_{i,j} ( \mathbb{R} \setminus (C_i - q_j) )$ is nonempty. Take $t \in \cap_{i,j} ( \mathbb{R} \setminus (C_i - q_j) )$, then $D - t \subseteq \mathbb{R} \setminus \mathbb{Q}$.

gentle ospreyBOT
gritty widget
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Thanks for the answer, I'll try to understand it.

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the version of BCT that I'm using here is: Countable intersection of open dense subsets of a complete metric space (in this case the reals) is nonempty (in fact dense, so you get in addition that the set of t with D + t contained in the irrationals is in fact dense in R)

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I think Im getting it now... thanks, I have never really used BCT and I REALLY need to see how to use it so glad I could see your solution.

little robin
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Hey guys I'm pretty new to topology and I have to get through this paper and summarize it to my boss. Can anyone help me past this hurdle here?

Given a finite covering U = {Uα}_α∈A of a space X, we define the nerve of the covering U to be the simplicial complex N(U) whose vertex set is the indexing set A, and where a family {α_0,α_1,...,α_k} spans a k-simplex in N(U) if and only if U_α_0 ∩U_α_1 ∩...∩U_α_k= ∅.

So I get that we have a covering of X called U with an index set A. "Family" seems to just mean some subset of U, but what does it mean to "span" a k-simplex?

marsh forge
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Hm

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Is this for topological data analysis

dim meadow
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Are you reading Topology and data?

little robin
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Yeah, this is the Gunnar Carlsson paper that outlines TDA. XD Glad to see you folks caught that.

marsh forge
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The idea here

dim meadow
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I read that paper like 2 years ago lol

marsh forge
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Well, its weird that they are doing this in terms of open covers

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Im gonna give a more explicit tda example

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So we have a bunch of points

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And we want to talk about how ‘connected’ they are

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But like, they are points right

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They dont touch

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We want to ‘thicken them’

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So we take some thickness r

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We will be varying r but just fix it for now

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Take balls of radius r around each point

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Then we have something called a ‘k-simplex’ for every set of k r-balls that intersect

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So if two intersect, we add a 2 simplex

little robin
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A line.

marsh forge
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If three, we add a three simplex

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Etc

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Yes

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Wait no

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Sorry im tored

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If two intersect we add a 1 simplex

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If three we add a two simplex

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Etc

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One lower

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So the ‘line’ connect the two data points

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And the triangle connects 3

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Etc

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Then we gather all these simplexes

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And we remember how the edges of each triangle

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Are three 1-simplicies

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And so on

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Then this allows us to compute homology

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And gives us betti numbers

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So replace ‘balls’ woth ‘elements of the open cover’

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Thats the Nerve

little robin
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Okay. Now this is something I've been kinda unclear on, when everything pops out of Mapper we have something that appears to functionally be a graph. So essentially TDA is just a clustering algorithm of sorts? And isn't the degenerate "ball" that we can create with varying thickness just going to cover the whole space and so it'll glue all the points together? What's the stopping criterion?

marsh forge
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Ok for a second

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let's forget we care about persistence

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Then TDA can be thought of like this

little robin
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Okay. As long as we get there I'm good. XD

marsh forge
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We take data. We make it 'topological'. We turn this topological data into simplicial data. We compute homology.

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Okay sorry I wanted to be more explicit

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You can blackbox homology but not the simplicial part

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Ok so what I described

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is called the Vietoris Rips Complex

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it takes data

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makes it topological by thickening it

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and then makes it simplicial by looking at interesections

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A simplical complex looks kinda like a graph

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except we can fill stuff in

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Okay so let's talk about persistence

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this is normally applied in the 'make it topological' step

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in particular, we choose some random thickness right?

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well, we can change our units when recording data

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and make it arbitrarily far or close together

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So it doesn't make sense to just randomly choose a thickness

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what if instead we considered

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all possible thicknesses (up to some maximum)

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That's persistence

little robin
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You gradually increase, right?

marsh forge
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so for every number in say [0,100]

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we get a topological space and collect its simplicial data

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and do homology to it

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Let's think about what happening at the topological stage

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at r=0

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we have our points

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and we add a 0-simplex (a point) for each

little robin
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Nothing is connected. Stars in a void.

marsh forge
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at maybe r=50, maybe some stuff connects, let's assume that only pairs of things connect

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this is exactly a graph!

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So we don't have any 2-simplexes (triangles)

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but some of the points have lines between them

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Homology at this stage

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can tell us if we have any 'cycles' in our graph

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if you know what that means

little robin
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Yeah, I know graph theory. I've had literally three hours of topology. XD

marsh forge
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Now, let's thicken, r=75

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Now we have triplets connecting!

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So we have points that are connected to nothing

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points that are connected by lines

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and then the closest points are connected by triangles

little robin
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SO it's analagous to a hypergraph, edges can have three things that they connect to.

marsh forge
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imagine now that these triangles piece together

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and make a sphere

little robin
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Tetrahedron, right?

marsh forge
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(to picture this, take a sphere, and think of like a low resolution video game)

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tetrahedron is 3 simplex

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sphere is empty

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on the inside

little robin
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Oh, yes, go on.

marsh forge
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Ok so, we learn two new things at this stage

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some of our cycles from before

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are now filled in

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Homology now 'ignores' these filled in cycles

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and only counts cycles that aren't filled in

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we also have potential 'spheres' in our data made by a tiling of triangles

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homology sees these, and records them separatley

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If you know the terminology, this counting is given by the betti numbers

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(if you dont know the terminology I just told you it so lol)

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Okay so like

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we can keeo doing this

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at the next stage maybe some of our data connects in sets of 4

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so we fill in tetrahedra

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and this makes homology ignore some of our spheres from before

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and eventually everything is connected to everything else (just set r to be the max distance between any two points)

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This is what persistent homology measures

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how connected our data is at different resolutions, and how many 'holes' and 'connected components' are in the data at each stage

little robin
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So when you say "ignore" could you rigorously define what you mean? Does it literally identify the points and say "you're all the same" now or what?

marsh forge
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This requires us to get into homology

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the slogan

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is that we literally just dont count them

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the idea is

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Homology is like, all of the n-dimension cycles that do not bound a larger n+1 simplex

little robin
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Oh, cuz we only care about holes. That's now "a sheet".

marsh forge
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so we ignore a filled in triangle because the sides of the triangle, which are a cycle

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now bound a 2-simplex

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(and we also ignore things that bound like, sums of 2-simplicies)

little robin
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"bound"? In this context I'm not quite sure how you're using that. (My boss is SUPER precise, he won't let me get away with this. XD)

marsh forge
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Hm

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Bound as in, are the boundary of

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like if oyu take a triangle

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remove the interior

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you get three lines

little robin
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Okay. Carry on.

marsh forge
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connected

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This is more or less the full story

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But we can generalize

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What I just gave you wasn an assignment

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for each resolution 'r'

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a topological space + an open cover of that space that depends on 'r'

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But the simplicial stuff

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doesn't really care about how we build the open cover

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or what the underlying space was

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it just takes in cover data

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and spits out simplicial data

little robin
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SO this is the "filter" function thing.

marsh forge
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Yeah I think thats the terminology the TDA people use

little robin
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Any function that roughly maintains the notion of connectedness/distance.

marsh forge
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It depends how much you want to generalize

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if you want

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we can get away with any machine

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that takes in data

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and spits out a simplicial complex

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But that kinda is too general

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what we want is a machine that does

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data -> Topological Space + Associated Cover at 'r' -> Simplicial Complex -> Homology

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The only steps here that we can't mess with

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are homology

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and data

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we can do each intermediate step in any way we want as long as the output has the right 'type'

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For example

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Data -> Send everything to a single point -> the simplicial complex with 1 point -> trivial homology

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is a perfectly valid pipeline

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if useless

little robin
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And the other degenerate one, stars in a void, nothing is connected. XD

marsh forge
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To be honest, I only have a very vague idea of pipelines other than the one I just gave you

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because the Nerve is the one mathematicians care about

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outside of TDA

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I think theres a Cech comlex mathematicians care about too but idk it

little robin
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Yeah I'm actually a math grad student working for a math professor on Data Science, so he wants me to "grok" it from both sides.

marsh forge
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Ah I see

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One example I saw recently

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thats worth keeping in mind

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is that you can do like

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as many intermediate 'data transformations' as you want

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so you don't need to topologize your original data

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to interpret it topologically

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Like someone used TDA for financial research

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but they did a few things to the data itself to make sense of 'how is the stock market topological'

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I guess this has been kinda a rant

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@little robin feel free to ask any specific questions

little robin
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Eh, I kinda interjected where I needed to for clarification/confirmation of understanding. The rest is just implementaiton, which is the easy part right? XD

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I guess most of my questions are about how to interpret results which are really outside the domain of mathematics and really back in the world of domain knowledge.

marsh forge
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tbh

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that's probably the most nontrivial part of tda

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its hard to go back to the real world from the math

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and you need to have a good hold on the steps to really have any intuition for the output

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Here's a cute example: someone took a bunch of cancer data

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and realized that at almost all resolutions

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there was a chunk of the data split off from everything else

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they studied these patients and their particular cancers

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and found a new sub-type of cancer that had important distinctive qualities

little robin
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So from an applied perspective, this is just mathematically rigorous and flexible clustering that's invariant to some traditional Euclidean "weirdness".

marsh forge
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I like to think about it like

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what properties of your data are maintained

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if you don't think about the numbers themselves

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but instead think about how they are positioned relative to eachother

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in a much looser sense than you can do without topology

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So like, you lose a lot of the info in your data

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but if you get anything out of it

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you really are seeing something intrinsic about the data

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that's the theory I think

little robin
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Okay. Thank you Max this has been really enlightening. If I think of anything else I'll pop back on but I gotta finish getting a writeup done. You have a great day. 🙂

marsh forge
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Feel free to @ me

little robin
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Thanks. 🙂

sick elm
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My answer is: Yes, No, No, ??, No

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could anyone correct me here ?

marsh forge
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I agree

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Including on ??

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I think it means

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The 42 sphere

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Whoch is false

tacit stratus
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Hi, I had a question about algebras of functions.
I know you can completely reconstruct a smooth manifold from its algebra of smooth real valued functions (and vice versa). From what I can tell, to do the same on topological spaces requires the algebra of complex valued continuous functions, instead of real.

What different information does the algebra of real or complex valued (continuous/smooth) functions provide about the underlying space? More fundamentally, what's the difference between the actual structure of the algebra of real and complex valued functions?

tacit stratus
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oh interesting

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i guess i'm wondering why you use complex in some cases and real in others

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in my intuition the algebra of real functions contains less information than that of complex functions

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so for the case of smooth manifolds, you can recover the manifold from the real algebra, and the complex algebra from the manifold

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so in a sense its like they contain the same infromation? which is fine i suppose but just feels weird to me

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yeah that was my thinking

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in the case above, you can construct the complex algebra from the real algebra, and vice versa

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but I can't imagine theyre isomorphic or anything

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yeah if you look pointwise

dire warren
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Not even pointwise

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The constant function C = i satisfies C^4 = 1

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Like everywhere

tacit stratus
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yeah i mean if C^4 = 1 everywhere then pointwise C(x)^4 = 1

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which can't happen for real functions

marsh forge
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Well

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It cant happen of@order 4

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It can happen if say C(x)=1

dire warren
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Jan is this summary of banach alaoglu correct

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The unit ball of X* is homeomorphic to a closed subset of product D_x. Since D_x is compact the product is compact by Tychonoff, the unit ball is compact and done

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Here D_x is all complex numbers z such that |z| <= ||x|

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Seems pretty intuitive now haha

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I didn’t get it before

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but weak* topology is just pointwise convergence

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With that detail it makes sense

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Eh?

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I mean If L is in X* then L_n -> (weak *) L iff L_n (x) -> L(x) for all x in X right?

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Oh

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Shit

marsh forge
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If the norms converge?

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Could you expand on that

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Nw

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Ah i see

dire warren
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But I mean weak* topology on X*

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Not X

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Ah ok

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Then it’s fine haha

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Cool, now just need to understand why Tychonoff theorem is true

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I am so bad with aoc constructions..

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Tychonoff, stone cech, Hahn Banach

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All of them no make sense

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Ye gH

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Haha

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But I can’t intuit it

marsh forge
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Can you explain how we pull weak * on X* back yo X?

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To

dire warren
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U can do that ?

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I mean pulling back

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I know y can embed

marsh forge
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I was using the term loosely

dire warren
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Oh lol

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Ergodic theory ya

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Comes up all the time

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Haha why

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It seems like the kind of thing u would like

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Functional analysis everywhere

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And it’s soft analysis

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Oh haha

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I feel like Z-systems are pretty geometrically intuitive

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But when they talk about general group actions

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My intuition flies out the window

marsh forge
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Im doing bb ergodic rn

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W amie Wilkinson

dire warren
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Group actions? @marsh forge

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I’ve.. never heard of powers factors before hah

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Is it an oa thing?

marsh forge
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Rn just like, endos f such that f*mu = mu

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For a prosbility space

dire warren
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Oh so Z actions

marsh forge
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Oh I suppose

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Never thought abt itnlike that

dire warren
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Wait, not invertible?

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Endos

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Well I guess it’s invertible ae

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So doesn’t matter

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they’re very common iirc cause of the ergodic decomposition

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Every pmp action can be broken down into ergodic components

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It’s so spooky lol

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Defining measures on null sets

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Oh I’ve seen a crossed products book

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For c* algebras

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Apparently it’s god stuff but I need the basics of c* algs first

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It’s kinda like semi direct product in group theory?

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I haven’t gone through the proper defintion

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God I hate those

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So hard to figure out wtf the action is doing

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Yeah

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Especially in the fully general case of semi direct products

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Where it’s just any homomorphism from X To aut Y

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Eric?

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What’s his surname

tacit stratus
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oh i have

dire warren
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Ah ok

tacit stratus
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i'm sorry i didnt realize

dire warren
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Ur doing crossed product shit?

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No I mean elliot

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Oh LOL

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Eric hahaha

bitter yoke
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Did you say Eric

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Thx

minor stag
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For this, am i supposed to show $d_1(a,b) < \delta$ \xrightarrow{} d(a,b) < $\epsilon$?

gentle ospreyBOT
minor stag
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I'm confused cause isn't $d_1:X \cross X \cross X \cross X \xrightarrow{} \mathbb{R}$ but $d:X \cross X \xrightarrow{} \mathbb{R}$

gentle ospreyBOT
minor stag
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i.e show d_1 is continuous?

west spindle
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what?

minor stag
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Idk i'm confused

west spindle
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you're supposed to prove $(\forall (x, y) \in X^2)(\forall \ep > 0)(\exists \delta > 0)(\forall (x',y') \in X^2)[d_1((x,y),(x',y')) < \delta \to |d(x,y) - d(x',y')| < \ep]$

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fuck

gentle ospreyBOT
minor stag
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oh ok that makes more sense

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thanks!

little hemlock
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When can you (nontrivially) have $x \in A$ but $x$ is not a limit point of $A$? Clearly, if $A = {x}$ then this is the case, but are there any more interesting examples?

gentle ospreyBOT
real notch
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Isolated points are interesting kinda

little hemlock
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oh hm wait does that mean any subset of a set X with the discrete topology has no limit points? Since {x} is a open set for each x in X?

real notch
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Every set in discrete is open and closed yes?

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Closed contain all their limit points

little hemlock
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oh yea when u think of it that way...

real notch
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Wait I’m an imbecile but ignore me

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Limit points need neighborhoods to hit more than one other point right?

little hemlock
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more than just the point itself

real notch
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hence “other”

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if {x} is an open subset of A (in subspace topology)...

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it’d be a bit hard for every nbhd to intersect A\x

dim meadow
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In a discrete Topology all convergent sequences are eventually constant

real notch
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A slightly less trivial example (I.e. not discrete) might be ${0}\cup [1,2]$

gentle ospreyBOT
dim meadow
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Yeah a isolated point

real notch
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0 isn’t a limit point there

little hemlock
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ah yep

dim meadow
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That's a good example, in fact that's kind of the only example

real notch
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The only good one anyway

dim meadow
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No

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All the other examples are kind of the same

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But yeah, a point x is a limit point of a set S if there is a sequence of points in S-{x} which converges to x

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Or equivalently if x \in \bar S-{x}

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In the discrete case you have that every convergent sequence is eventually constant, so the sequence must be eventually x

marsh forge
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i think that only works in locally second countable

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you'd want nets otherwise

little hemlock
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I showed that $\overline{A\cap B} \subseteq \overline A \cap \overline B$ and I am trying to explain why the other inclusion doesn't hold. \ \ When $x \in \overline A \cap \overline B$ is a limit point of $A$ and $B$, it could happen that there is an open set $U \ni x$ which intersects $A$ at different points than $B$, such that $U\cap A$ is nonempty and $U\cap B$ is nonempty but $U \cap A$ and $U \cap B$ are disjoint.

gentle ospreyBOT
little hemlock
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that's correct, right?

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*the bar notation is for closure, in case that notation is not standard or somethin

marsh forge
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are you just asked to explain

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if you want to disprove

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you would need a counterexample

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The intuition is that collecting the limit points and then intersecting should capture more points than intersecting and then taking limit points

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You're just intersecting more stuff in the first one

little hemlock
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yea, ill see if i can think of a counterexample...

marsh forge
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lmk if you want a hint

little hemlock
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open intervals A = (a, b) and B = (b, c). b is an element of the intersection of their closures but the closure of their intersection is empty.

marsh forge
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Yes

little hemlock
little hemlock
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lol, yeah I guess. I think the result in the first sentence somehow is used to show the second result, but its not quite clear what to do hmm

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the closure of A is closed, so I know that the closure of A in Y is the intersection of some closed set in X with Y (from the previous result), but I'm not sure how it follows that that closed set must be the closure of A in X (if that makes any sense).

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yes

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yes

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hmm okay, I'll think about this

little hemlock
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So the closure of $A$ in $X$ is the intersection of all closed sets in $X$ containing $A$. When we intersect this with $Y$....

$$Y \cap \operatorname{cl}X (A) = Y\cap \left(\bigcap{X \supset C_i \supset A} C_i \right) = \bigcap_{X \supset C_i \supset A} (Y\cap C_i) $$

and by the previous result, $C_i \cap Y$ are closed sets in $Y$ containing $A$. So hmm, now I somehow have to conclude that gives the closure of $A$ in $Y$.

gentle ospreyBOT
little hemlock
#

Y n C_i are closed sets in Y containing A, but I'm not sure that all closed sets in Y containing A have the form Y n C_i.

#

yes

little hemlock
#

not completely sure about this question pandaOhNo

#

I thought maybe: call $A$ the subset in question (inside a topological space $X$). Suppose $A$ contains one element from each basis element. $A$ is clearly countable. We want to show that the closure of $A$ is $X$. So let $x \in X$, and let $U$ be an open set containing $x$. Then $U$ is a union of basis elements, each of which contain elements of $A$, so the intersection $(U \setminus { x} )\cap A$ is nonempty.

gentle ospreyBOT
little hemlock
#

but I don't see why you couldn't have that U only contains x. In that case, the intersection would be empty.

little hemlock
#

<@&286206848099549185> 🙏

#

I am guessing that X just can't be the discrete topology? I'm so confused rn megathink

#

oh I am dumb. We don't care about when x is in A 🙃

sweet wing
#

consider a random point x, does every neighbourhood of x contain A? if so x is in the closure of A

midnight jewel
#

if a certain one-point set {x} is open in X, then it must itself also be a basis element

#

@little hemlock

#

(in particular x in A and therefore x in cl(A))

little hemlock
#

right, and it doesn't matter whether points of A are limit points for A. Rather, every point that is not in A should be a limit point. 👍

pseudo crane
#

Prove me wrong: the set E = {1/n | n \in Z} isn't compact; Let e be the distance between 1/n and 1/(n+1). Then let an open set G_n be the neighborhood of radius e/2 of 1/n. Therefore, every point of E has its own point of G = \Cup_{n=1}^\infty and G is an open cover of E. If we consider only finitely many subsets of G, then we are missing infinitely many points of E, and so there does not exist a finite subcover for the open cover G.

bitter yoke
#

This is right

pseudo crane
#

I know it's flawed because E is both closed and bounded, i.e. compact

#

Also the exercise says to prove that E is compact

#

Wait

#

Nvm

#

The E in the problem contained 0

#

So the proof is correct but it's not a counterexample to the exercise

bitter yoke
#

Yes

#

This E you've written down is not closed

pseudo crane
#

Correct

#

Ok, well thanks anyways ig

#

Should be much easier to prove now

cobalt crescent
#

so I've been doing a project where I want a square grid to wrap around in various ways, so I've been doing some reading. Wikipedia says this

📎 2020-01-29-005744-ss.png
#

are they somehow equivalent to some of the ones in the wikipedia article?

#

If i do the group theory looking stuff down here (which I kinda get but not really), I get BB for the first one, which is equal to the projective plane, and AABB for the second one, which is equal to the klein bottle, but I really don't see how

📎 2020-01-29-010451-ss.png
#

basically I don't get how this works

midnight jewel
#

@cobalt crescent

cobalt crescent
#

ah ok, thanks

wooden scarab
#

are there any refs that cover geodesic sprays?

gentle ospreyBOT
sleek thicket
#

Can someone help me with a Hartshorne problem? I'm working on showing that if you have a morphism f : X -> Y and there is an affine open cover {Vi} of Y such that f^(-1)(Vi) is compact for each i, then f^(-1)(V) is compact for each affine open subset V of Y

#

I tried writing f^(-1)(Vi) as a finite union of affine opens {Uij}, writing V intersect Vi as a union of simultaneous distinguished opens D(sik) = D(tik) for tik in Γ(Vi, O_Y) and sik in Γ(V, O_Y), and then pulling everything back and using the fact that pulling a distinguished open D(tik) of Vi back along Uij -> Vi is the distinguished open D(φij(tik)) where φij : Γ(Vi, O_Y) -> Γ(Uij, O_X) is the map corresponding to Uij -> Vi

#

But now I just have a big messy union

#

Is this union somehow finite? I don't see why that would possibly be the case

#

I guess I'm stuck on the fact that the cover of Y by affines could be really big and the cover of V intersect Vi by distinguished opens could be really big

#

might just look up how to do this in vakil, feeling pretty dumb

white tusk
#

hi guys I am working through Topology, Geometry, and Physics by Nakahara. can someone help me with exercise 5.8?

west spindle
#

spose this might go here

#

define a graph to be k-embeddable if it is possible to embed it in R^k such that no edge passes through the (k-1)-simplex spanned by any k nodes in a k-clique

#

one sees that 2-embeddability under this definition is the same as planarity

#

are there any graphs which are not 3-embeddable?

#

to clarify: i'm considering undirected, finite graphs

#

and an embedding is required to keep edges straight

rugged swan
#

Yeah I think so, if it is embedable you can compute euler's characteristic and deduce an inegality on number of edges and vertices

#

Like planar graphs

west spindle
#

hm

#

okay so

#

define the embedding dimension dim(G) to be the lowest k such that G is k-embeddable

#

dim G = 0 iff G is a single point
dim G = 1 iff G is a path of at least two points
dim G = 2 iff G is planar but not a single path

#

is it possible to characterize all graphs of embedding dimension 3 in this way?

#

is it possible to construct explicitly a graph with embedding dimension, say, 4?

#

hmm

#

i think K_n may have embedding dimension n-2 always but i'm not sure

rugged swan
#

Doesn't it works by defining CW complexes ?

west spindle
#

hm?

#

i'm just playing around here. i came up with these definitions on my own.

rugged swan
#

K6 seems to be 3-embeddable

west spindle
#

can you give me an example of a 3-embedding of K6

rugged swan
#

Nope

#

It doesn't work

#

Mb

floral gust
#

@west spindle There is a very nice similarity to your definitions and the definitions of n-skeletons of a CW complex

gritty widget
#

Can someone render the formulation in 3D if x=3

little hemlock
#

let $f$ be a continuous map from $\bR$ to $\bR$. I need to show that the set of points left fixed by $f$ is closed. Any hints?

gentle ospreyBOT
little hemlock
#

there are some basic things I can see like: if A is such a set of points then f(A) = A, but that doesn't completely encode the fact that every point is fixed by f

#

It seems like the easiest thing to do would be to show that A is equal to its closure but I'm getting nowhere thonk

#

Oh, well I suppose you also have f^-1(A) = A hmm

dim meadow
#

g(x)=f(x)-x is continuous

#

The point 0 is closed

#

This is a very common trick @little hemlock

dire warren
#

If a_n in A converges to x, then f(a_n) converges to x by continuity, done

west spindle
#

${x \in \bR \mid | f(x) = x } = (f - \id)^{-1}{0}$

gentle ospreyBOT
little hemlock
#

damn all right, thank y'all.

gentle ospreyBOT
west spindle
#

@floral gust "f, g ∈ X -> Y" did you mean f, g : X -> Y

honest narwhal
#

So you can actually package this result into some others

#

A space Y is Hausdorff if and only if the diagonal in Y\times Y is closed

#

That's a special case of this result, namely let X = Y\times Y, f be the projection to the first factor, and g to the second

#

But let's say you prove that special case

#

Well, given f,g:X->Y, you get a map x->(f(x),g(x)), which is a continuous map X->Y\times Y

#

Then your set is the preimage of the diagonal

little hemlock
#

All functions are continuous' explanation is about the most I can parse atm but thank y'all. very cool catthumbsup

floral gust
#

@west spindle Soz typo. I meant f and g are two function from X to Y

gentle ospreyBOT
fleet trench
#

do you understand what the question is asking?

#

$e_N$ is the function that tells you how the group acts on $N$. Thus, if you start at $p\in S^{2n-1}\setminus {-N}$, you want to pick $s(p)$ to be the group element that takes $N$ to $p$, so that $e_N\circ s(p) = p$

gentle ospreyBOT
viral yoke
#

Cooper's Exercise #2

#

Alexander's trick does not give us a smooth isotopy (only isotopy). To remedy this, I tried to construct bump functions that will try to "smooth" out the homotopy.

#

In particular, I did this.

#

Homotopy follows using the pasting lemma.

#

Since f, p1, p2, and the identity are all diffeomorphisms, isotopy follows.

#

However, smoothness still seems to be an issue.

viscid bolt
viral yoke
#

Hmm. Our course hasn't gone that far yet.

#

I'll look into it though.

little hemlock
#

I'm trying to prove that $f(x) = \frac{e^x}{1 + e^x}$ is a homeomorphism from $\bR$ to $(0,1)$. How does proving continuity work exactly? I know I should take some basis element in $(0,1)$ and show that its preimage is open in $\bR$, but... how exactly?

gentle ospreyBOT
honest narwhal
#

I mean you know that e^x is continuous

#

And that 1+e^x is continuous

#

So the quotient is

little hemlock
#

oh wow, okay.

#

similarly for the inverse:
ln(x/(1-x))
x is continuous from R to R. 1-x is continuous from R to R. So x/(1-x) is continuous for x != 1. ln(x/(1-x)) is continuous for x/(1-x) > 0 so x > 0 blah blah inverse is continuous on (0,1)?

honest narwhal
#

Yup. I didn't check to make sure that's the inverse exactly but on (0,1) that's a continuous function by the same reasoning

gloomy hatch
#

a) The real projective space $\mathbb{R}P^n$ is the space of lines through the origin in $\mathbb{R}^n+1$. What does $\mathbb{R}P^1$ look like?

b) The canonical line bundle over the real projective space is constructed by attaching, to each point of $\mathbb{R}P^n$, the line that it represents. What does the bundle over $\mathbb{R}P^1$ look like?

So, by drawing a diagram, it was obvious to me that $\mathbb{R}P^1$ looked like a circle, $S^1$, touching the origin once and extending however far I liked. However I am having a slight degree of trouble understanding part b intuitively. I thought that since a line looked like $\mathbb{R}$, I could understand the line bundle as looking like $S^1 \times \mathbb{R}^1$, which would just be a cylinder. However, this wasn't right

gentle ospreyBOT
bitter yoke
#

RP^1 isn't exactly a circle in the way that you're thinking about it, because points that are opposite on the circle are also "equal"

gloomy hatch
#

Oh?? What do you mean?

#

My teacher said that my response was, well, correct enough I guess

bitter yoke
#

Well, (1,0) and (-1,0) get identified to the same point of RP^1

#

Because the line going through those points and the origin are the same

gloomy hatch
#

Well, yes, of course

bitter yoke
#

Yeah, I just thought his explanation was a bit wonky, but I think I just read it wrong

civic creek
#

fix a direction on your line, above (1,0) (for the line bundle on b). once you do a half turn, notice that your direction has been reversed

civic creek
#

this is for b)

#

how you see it's a mobius strip

floral gust
#

Is the completion of a metric space the same as the compactification of the said metric space?

civic creek
#

no

#

R is complete

fervent citrus
#

not really, completeness and compactness are fairly different things

civic creek
#

compact spaces need to be complete though

rapid ridge
#

Are you talking about completion and compactification of the space by referring to the fact that compact metric spaces are complete ? @floral gust

tawdry knoll
#

Is anyone on here familiar with the theory of topological vector spaces?
I'm wondering if there's a standard way to topologize Hom sets in the category of finite dimensional vector spaces.
I'm more of an algebraic guy, so most of the theory on this subject I'm unfamiliar with.
Should I just assume the compact open topology, or is there a metric one I can invoke?
For context, I'm reading a k-theory book that just says to topologize hom(V,W) in "the standard way".

dim meadow
#

Compact metric spaces have bounded cardinality @floral gust

civic creek
#

compact open seems the usual way @tawdry knoll

tawdry knoll
#

But would that be the usual way for vector spaces already endowed with the metric space topology?

#

Where my linear maps are going to be represented by matrices?

#

Like I get using compact open for general topological spaces

civic creek
#

im pretty sure compact open agrees with, say W x V* when they are finite dimensional

#

convergence of matrices is the same as uniform convergence in compact sets as linear transformations

tawdry knoll
#

Ok, that makes sense.

#

Thank you!

gentle ospreyBOT
sick elm
#

Hi, how do I understand functions as (0,0) tensor field? I dont see how they can be regarded as such

gritty widget
rugged swan
#

This shows that the union doesb't contain any open set

#

Thus has an empty interior

#

Since the interior is an open subset

gentle ospreyBOT
civic creek
#

take a point which isn't {1,-1}. show that eventually the sequence is epsilon far from your point

#

it shows any other point is not in the closure

#

except possibly for points in A

#

more directly, it shows the complement is open because for any point you can find an epsilon neighborhood which is still in the complement

#

but you aren't trying to show the complement is open

#

you're trying to show you got the full closure

#

okay sure that works

#

yeah

ember knoll
#

I am a high school math enthusiast who is unsure where exactly this question belongs, but does it make sense to generalize the idea of an n-dimensional sphere to non-integer values of n? Like a 1.5-dimensional sphere for example?

civic creek
#

what do you want that to mean?

ember knoll
#

I don’t know, I’m just curious how it would generalize. Maybe something like a fractal sphere, but I don’t know what that would be

#

This is more so just a question being asked out of curiosity

civic creek
#

generalizing doesn't make sense if you don't know what properties you're trying to generalize, or what you want the generalization to do

ember knoll
#

Okay, that makes sense. I guess I need a more rigorously defined question, but I don’t have enough experience to know what rigorously defined aspects I would want to have

#

I was more so curious as to whether or not there was some frame that had been thought of that did try to make sense out of such an idea. Maybe something along the lines of the volume equation, and one of the properties that this object needs to have is that it’s measure is the same as that volume formula predicts it would have

civic creek
#

one way of generalizing dimension is as you say, vol(kX) = k^d vol(X)

#

but what's unclear is how you would generalize sphere

ember knoll
#

Yeah, I was curious if there existed some geometric object that would sort of encapsulate various generalizable aspects of spheres to get to a definitive shape that you could call for example, a 1.5-dimensional sphere, or does that not make too much sense in the study of fractals?

civic creek
#

idk what you mean by various generalizable aspects of spheres

ember knoll
#

Things like the volume equation. But I don’t completely know if the question makes sense either. I was just sorta wondering if it did make sense in some sort of frame work. Is there something you could integrate that had the same sort of form as integrating over an n-dimensional sphere, but was for a non integer integral. Maybe you could use a fractional integral to find a function that would give you the equivalent to finding the volume of an n-dimensional sphere given different boundaries(similar to integrating (1-x^2)^1/2), and perhaps that could be used to get a sense of what an n-dimensional sphere entails?

shrewd lagoon
#

if you look here, it has a little bit about what you're talking about https://en.wikipedia.org/wiki/Unit_sphere#Fractional_dimensions

Unit sphere

In mathematics, a unit sphere is the set of points of distance 1 from a fixed central point, where a generalized concept of distance may be used; a closed unit ball is the set of points of distance less than or equal to 1 from a fixed central point. Usually a specific point h...

ember knoll
#

Thank you I’ll try reading up on these in the near future, hopefully tomorrow if I get homework done, it’s getting kinda late and I probably have school tomorrow(tho I may have a snow day or delay), so I’m going to bed. Adios

fleet trench
#

@sick elm you can think of it in the context of differential forms if you want; for a C\infty function f, df is a one form that acts by df(X) = X(f). Since d in general increases the degree of a form, it's as if the function is a zero form that has had its degree increased to a one-form. Since forms are tensors, f is thus also a tensor -- of rank (0,0)

bitter yoke
#

Why are separable spaces called separable

gritty widget
tough hamlet
gritty widget
#

To hell with you all, scum

gloomy plover
hexed prawn
#

Any of you wanna help me with a problem? I posted it in #help-9

hollow inlet
#

is being able to specify balls everywhere equivalent to defining a metrization of your topological space?

#

Urysohn's metrization theorem. This states that every Hausdorff second-countable regular space is metrizable.

#

But then again I don’t really know of any examples spaces that aren’t second countable

sweet wing
#

the discrete space is not second countable for like R

cursive nebula
#

hey guys, i have a question in #help-7｜zen1thxyz , it's about measure theory so i don't even know if this is the right channel to ask..
could you take a look? thank you!

soft radish
#

Hey can anyone help me out with a problem i posted in #help-2

gritty widget
#

Hey, I need some help with this one (differential geometry). I'm thinking that we can use the hint and state that f(s) has a local maximum point at s_0, so f''(s_0) < 0. I understand that I have to differentiate f twice to get the formula for f'' in terms of alpha, but tbh I dunno what to do

📎 grtrabb1.png
celest flare
#

Is this proof correct?

#

What's taken as S_n doesn't match how S is described. For example the set {2,3,4,...} satisfies the condition for S but I don't see how any S_n can be equal to this set

#

In fact, why is there a need to introduce a set like that in the first place? Can't we just say that $S_k$ are all the S in $\tau$ and from the property that for each k, $N - S_k$ is finite prove that this is a topological space?

gentle ospreyBOT
gloomy plover
#

,rccw

gentle ospreyBOT
gloomy plover
#

I don't really understand your question, please elaborate

celest flare
#

ok

#

S is the subset of $\mathbb{N}$ such that $\mathbb{N}-S$ is finite

gentle ospreyBOT
celest flare
#

So there should be some $S \in \tau$ for which $\mathbb{N}-S={1}$ , right?

gentle ospreyBOT
celest flare
#

In the proof from what I can see they treated it like all $S \in \tau$ can be written in the form $S_n$

gentle ospreyBOT
celest flare
#

But for none of these $S_n$ is $\mathbb{N}-S_n = {1}$

gentle ospreyBOT
celest flare
#

So not all $S \in \tau$ can be written in the form $S_n$

gentle ospreyBOT
celest flare
#

Instead, why can't it be done like this:
I number all $S \in \tau$ as $S_1, S_2, S_3, ...$, so that I can say $S_k \in \tau$ for any $k \in \mathbb{N}$. Then I proceed the way it's done in the second part of the proof by replacing $S_n$ with $S_k$ without having to introduce that union thing that defines $S_n$

gentle ospreyBOT
celest flare
#

Was that elaborate enough? @gloomy plover

gloomy plover
#

Okay yeah I see your point

#

I don't know exactly why they introduced S_n in the way they did, it seems rather arbitrary

gentle ospreyBOT
civic creek
#

neither is maximal, and they are compatible

floral gust
#

Oh nvm,... the maximal atlas would contain both of them

gritty widget
#

Does this go here

#

i don't understahdn why some pairs of committees cannot meet during the same period

bitter yoke
#

Person 1 is on both committees 1 and 2

gritty widget
#

what does it matter if a person is on two teams

bitter yoke
#

They can't be at both meetings at the same time?

gritty widget
#

i thought they were meeting all in the same place anyway

#

i guess not

gritty widget
#

@gritty widget UWU yo

#

Let $A = {\left(0,-1\right), \left(0,1\right) }$ and $B = A \cup {\left(1,0\right)}$. Show $S^1 / A$ and $S^1/B$ are both Huasdorff and they aren't homeomorphic

gentle ospreyBOT
gritty widget
#

actually knowing this I need to do the other problem: Find a closed set $X$ in $S^1/A$ and a continuous $f : X \to D^2$ onto the boundary of disc $D^2 = {\left(x,y\right) \in \mathbb R^2 : x^2 + y^2 \leq 1$} such that \left(S^1 / A\right) \cup_f D^2$ is homotopy equivalent with $S^1$

gentle ospreyBOT
gritty widget
#

And to the part 1 whats the ideal way to show they arent homeomoprhic?

#

is there an easy fast way here? Like components or sth

gritty widget
#

you can remove a point from $S^1 / B$ and turn it into a space with 3 components, but you can't with $S^1 / A$.

gentle ospreyBOT
gritty widget
#

yeah thats what I thought

#

with S^1/A you can make it at most 2 right

#

sure

#

yeah so thats pretty mcuh what I wrote on my exam, but the 2nd part I don't have a clue

#

Also $S^1 / A$ is figure 8 so there's an obvious way to turn your identified space into something that is homotopic to $S^1$

gentle ospreyBOT
gritty widget
#

namely, take $f$ to be the map that sends a trace of an arc from $(0,1)$ to $(0,-1)$ once around the boundary of $D^2$

gentle ospreyBOT
gritty widget
#

then in the identified space $S^1 / A \cup_f D^2$ you can "collapse" into a point one of the holes in figure 8

#

which is homotopic to S^1

gentle ospreyBOT
gritty widget
#

hmm thanks, I'll think about it, I think I dont understand the \cup f thing yet

gritty widget
gritty widget
#

kevy you need a hug?

marsh forge
#

Hugs are not topology or geometry

gritty widget
#

yes they are

floral gust
#

I am a bit confused with this question. Specifically, for the forward direction, let us consider U as a manifold and i(U) as U as a subset of R.
Then a map f : U -> R is smooth iff f o i-1: i(U) -> R is smooth (calculus sense). But from this all I can show is f o i-1 is smooth (calculus sense) not f: U into R (calculus sense). Moreover, it is impossible for f:U -> R to be smooth (in calculus sense) since U is not a subset of R.
Now ofcourse over here I was treating U and i(U) as separate objects. Am I correct to think that in the question they're being considered the same thing? So f:U -> R <=> f: i(U) ->R ?

📎 unknown.png
fleet trench
#

a map between manifolds is smooth if it is smooth when expressed in coordinate functions (because for a general manifold, we define smoothness based on the normal calculus definition of smoothness for functions R \to R); the inclusion into R is a map U \to R which can be thought of as a parameterization of the mfld U

floral gust
#

So when he says show the map f:U -> R is smooth what he actually means is that f:U -> i(U) -> R where i is the inclusion map?

odd crescent
viral yoke
#

Hi again, back with another question

#

Let $X$ be a vector field on $\mathbb{R}^2$. Assume $X_p = 0$.
1.) Show that there is a well-defined integer $\mathrm{ind}_p X$ given by the degree of the self map $S^1 \to S^1$ determined by traversing a small circle about $p$.
2.) Find a vector field $X_n$ such that a.) $\mathrm{ind}_p X_n = n$, and 2.)$(X_n)_p = 0$, for each $n \in \mathbb{Z}$.

gentle ospreyBOT
viral yoke
#

I've been trying to look for a definition of degree of a map since the concept hasn't been discussed in class.

#

Nor is it in the text, it seems (Tu).

marsh forge
#

Given a map S1->S1

#

there is an induced map on homology

#

Z->Z

#

all maps of this kind are of the form x\mapsto nx

#

n is the degree

#

You can show that the htpy classes of maps S^1->S^1 are fully determined by this n

honest narwhal
#

In fact

#

S^n to S^n

#

😛

#

But yeah no need for that now

marsh forge
#

Yeah thats why I did homology

#

bc the normaly intro is pi1

#

and it still works

#

for pin

#

but its gross

#

homology is cleaner for the Sn version

little hemlock
#

Let $d(x, A) = \inf{d(x,y) \in \bR \mid y \in A }$. Prove that $d(x, A) = 0$ iff $x \in \overline A$, the closure of $A$.

gentle ospreyBOT
little hemlock
#

So, I came up with this argument for the case when x is a limit point of A (but not an element of A). Idk... it seems correct but it might be kind of cumbersome what do y'all think?

gentle ospreyBOT
little hemlock
#

<@&286206848099549185> 🙏

gentle ospreyBOT
little hemlock
#

um ye thats what B(x, \epsilon) is by definition.

polar laurel
#

I don't think that's the best definition

gentle ospreyBOT
vocal wharf
#

this argument will work for any complete metric space, not just R

#

and yeah, open balls of radius e around x are just all points with distance strictly less than e from x

#

so you can make this more general

polar laurel
#

If you aren't told that A is a subset of the real numbers, I wouldn't assume it

#

and same with x being a real number

little hemlock
#

ok yea i see ur point, ill delete that. However, im more concerned about everything after "Let l = inf{....."

#

I don't think anything is wrong, but it seems like there should be an easier way to argue this idk thonk

polar laurel
#

It's also weird that you suppose x to be in the closure of A, but not in A. What if A is closed?

little hemlock
#

ugh true.

I mean, other than when A is closed, its not weird, its trivial because d(x,A) = d(x,x) whenever x \in A.

vocal wharf
#

cant you just take x in boundary of A, for 1/n you can find y in A with d(x,y) < 1/n, then take the limit for n -> inf

little hemlock
#

yea, i like that... I wouldn't have thought to do that.

polar laurel
#

d(x,A)=0 iff you can find a sequence of y[k] in A so that d(x,y[k]) approaches zero. This means x must be the limit of this sequence, and so x must be in the closure of A. If x is in the closure of A, any sequence of y[k] in A approaching x will show that d(x,A)=0.

#

That's how I'd show it, at least

#

(I'm not sure what bits of information you are allowed to use for this proof)

little hemlock
#

honestly, im an analysis noob, so im not always good at making those kinds of connections. quite cool tho

polar laurel
#

Can you take for granted that for a decreasing sequence, its infimum is its limit?

little hemlock
#

yea, thats monotone convergence theorem.

marsh forge
#

By contradiction, suppose it is not a limit point

#

There is some eps ball containing x that does not meet A

#

That gives a lower bound on d(x,A)

#

Contradiction

#

This works and is cleaner

floral gust
#

For the case of backward implication, it suffices to prove that d(x,A) is a continuous function. Since we have that A ⊂d⁻¹({0}) and since d⁻¹({0}) is closed (preimage of a closed set), we have that Ā ⊂d⁻¹({0}) as required.

little hemlock
#

Wow thats so clean...

polar laurel
#

The set of n×n doubly stochastic matrices is a polytope in R^n^2, right?

#

How would you find the vertices that are connected by an edge to a particular vertex?

#

As I understand, the vertices are permutation matrices, so what would two adjacent permutations look like?

gentle ospreyBOT
gloomy plover
#

@orchid forge which math font is that?

#

Thanks catpat

iron jetty
#

{0, {a,b}, {a} }? {b}' = {a}, {a}' = 0

#

so they arent always involutions like that

bitter yoke
#

empty set is a subset of {a}

iron jetty
#

oh it is empty

#

should be {a}'

#

and then b

#

but yeah

#

let me think harder

#

that's the question right?

#

if it is

bitter yoke
#

Yeah, you can come up with more counterexamples by just embedding this space into another one right?

iron jetty
#

that's what I understood

#

I don't understand what you mean

#

I interpret this question as being "is the essentially the minimal counterexample?"

#

you're asking if it's the only thing that can go wrong, right?

#

ok im confused

#

add an extra element to the set?

#

{0, {a,b,c}}, A = {a}, A' = {b,c}, A'' = {a}

gentle ospreyBOT
iron jetty
#

oh by ' you mean

#

ok

#

i see

#

{0, {a,b}, {c,d}, {a,b,c,d}}, A = {a,c}, A' = {b,d}, A'' = {a,c}

#

no you're right

#

the one before doesn't work

#

because that's not what ' means

gentle ospreyBOT
iron jetty
#

ok what does ' mean

#

are you allowing elements of the set

#

or just sequences that aren't constant

gentle ospreyBOT
iron jetty
#

oh but the point is it satisfies what you want

#

still

#

okay sure

#

right so the question is if this is the only thing that can go wrong

#

and idk im kinda ditzy apparently

#

cuz everything I write is slightly wrong

#

it's the interesting question

#

it is?

#

what exactly do you mean

gentle ospreyBOT
iron jetty
#

yeah

#

that's convincing

#

cool

gentle ospreyBOT
iron jetty
#

yeah any counterexample is going to have some U - {x} not open and you can just remove everything else and get that one

gentle ospreyBOT
floral gust
#

Notice you could do this with anything by restricting the domain enough
@gritty widget
Ah ofcourse ofcourse. If I want to work with this strategy I have to construct a function f from |R² to |R such that f⁻¹(0,1) = S

floral gust
floral gust
gentle ospreyBOT
tawny smelt
#

umm that is loterally the opposite of connected

#

there shouldnt exist such U and V if X is connected

tough hamlet
#

(assuming they mean disconnected) that is true

#

you can equivalently define disconnectedness (of a space X) as "there exists a nonempty proper clopen subset of X"

gentle ospreyBOT
fading vale
#

(I'm proving the forward direction first)

tough hamlet
#

remember how closure plays with inclusion

gentle ospreyBOT
marsh forge
#

I dont really know what my question is

#

But can anyone give me an idea

#

Of what the fuck this space is

#

And why it works

gritty widget
#

yo what is an example of relation ~ on interval [0,1] with euclidean topology such that [0,1]/~ isnt Hausdorff?

#

would collapsing rationals to one point work?

frosty sundial
#

yes

gritty widget
#

thx, yo what would be example of topological spaces (X,T) and (X,T') such that id: (X,T) -> (X,T') is continuous but id: (X,T') -> (X,T) isnt?

frosty sundial
#

just make T' a much coarser topology than T

gritty widget
#

hmm

#

ye makes sense I guess

marsh forge
#

Take one trivial one discerete

#

Such that the trivial is the domain

rancid field
gloomy plover
fading vale
#

pls

#

@marsh forge someone has a question for u dude

brittle widget
#

hi i saw the channel description

gloomy plover
#

Where's the chamber in your question

brittle widget
#

the submarine

gloomy plover
#

I wouldn't call a vehicle a chamber

brittle widget
#

plsss ineed help

gloomy plover
#

plsss use another channel

fading vale
#

its not funny to do it when u know what it is

midnight jewel
gloomy plover
#

I miss those blackboards

vocal wharf
#

they still exist

midnight jewel
#

where’re you at these days, r/j?

gloomy plover
#

Back in Delft

gritty widget
#

What do you call a process of making friends? Homieomorphism

gloomy plover
marsh forge
#

@fading vale you cant just @ me like that here

#

An @ in this channel makes me so excited

fading vale
#

ok

#

then buy me nitro

#

@marsh forge

viscid magnet
#

can anyone help me with an undergrad topology question?

coarse kestrel
viscid magnet
#

can i ask here?

#

@coarse kestrel

coarse kestrel
#

Omg

#

Just ask

#

It's fine

#

Idk much topology but I'm sure someone does

viscid magnet
#

@coarse kestrel

#

I know I need to take an open subset U of XxY and show i^-1(U) open in X but idk where to start

#

I tried using the projection map to get some kind of composition of functions but that doesn’t seem to be useful

uncut plume
#

whats a proof

honest narwhal
#

So the idea here is this

#

Or wait first off regarding the projection map

#

Do you know what a quotient map is?

fading vale
#

Just think abt the subspace topology?

viscid magnet
#

@honest narwhal this is assuming we only know about subspace topologies, basis, and continuous functions

honest narwhal
#

Do you know about continuity at a point?

viscid magnet
#

and projections

#

@honest narwhal yes

honest narwhal
#

And in particular, that a function is continuous iff it's continuous at every point?

fading vale
#

You don't need the projection map to do this :?

honest narwhal
#

You don't need it but it's faster lol

fading vale
#

is it

honest narwhal
#

Err I was thinking of it the other way actually

#

Like there's a universal property of quotient maps

#

If f:X->Y is a quotient, a map g:Y->Z is continuous iff f o g is continuous

#

But we can't quite use that here

#

In any event so, take x_0\in X

#

Give me a neighborhood of (x_0,y_0)

#

There's gonna be some U\times V contained in that neighborhood since those guys generate the topology

fading vale
#

i feel like its easier to just liek, think for a moment abt what the preimage of a subset under the inclusion map actually is

honest narwhal
#

Then U is a neighborhood of x_0 whose image is contained in that neighborhood N of (x_0,y_0)

#

So gg

fading vale
#

¯_(ツ)_/¯

#

idk i feel like its pretty obvious that if ur taking the inclusion map from like A to B or whatever and u take U subset B then ||i^{-1} (U) = U cap A||

#

and once u have that its like...

#

lol

#

basic property of the subspace topology

#

that seems like the most direct approach

honest narwhal
#

Not quite that simple, thing is there's the topology on X just in space

#

And there's the topology of {(x,y_0) | x\in X} as a subspace of X\times Y

#

Like there's not really much to the proof that these are homeomorphic but that does require a few words

fading vale
#

i mean i guess but

#

¯_(ツ)_/¯

#

different strokes for different folks ig

honest narwhal
#

It's not even that, in a point-set class if you just wrote that down that wouldn't suffice lol

#

Like you need to talk about the product topology on X\times Y

fading vale
#

i mean ye you would need to prove it

#

but its not hard to prove lol

honest narwhal
#

Sure, neither was what I said lol

fading vale
#

ye

honest narwhal
#

Point is you gave the part of the proof that didn't have content

fading vale
#

well yea i wasnt trying to give the full proof

#

all i said was "its pretty obvious that..."

honest narwhal
#

It's not even the full proof lol, more like

#

The actual meat of the problem isn't in that

#

It's like the computational analog of, someone gives you a square matrix and asks you to show it has full rank, and you just say "Oh just follows by rank-nullity"

#

But the point of the problem was to show that the kernel was trivial

fading vale
#

i mean yes but i wasnt giving the answer?

#

or a proof

honest narwhal
#

I mean you kinda did say "Just think about the subspace topology" as if that was pretty much the point

#

The subspace topology is not what's relevant

viscid magnet
#

well which subspace are you referring to?

marsh forge
#

wait

#

are we still doing that X->XxY is cont?

#

This is very easy if you know the basis for the topology on XxY

#

What is a preimage of UxV? Compute it directly

viscid magnet
#

How would I go about that @marsh forge

#

Sorry, I'm just really lost lol

marsh forge
#

There are two cases

#

either y_0 is an element of V

#

or it is not

#

Can you do it now?

viscid magnet
#

Not really. Sorry

marsh forge
#

Well

#

you didnt try lmao

viscid magnet
#

I've been trying for the past hr lol

marsh forge
#

Every x

#

gets mapped to (x,y_0)

#

if y_0 is not in V

#

can anything get mapped into UxV?

viscid magnet
#

I'm just not following sorry. I'm gonna just go reread some definitions in the book and see if i can restart from the top and figure it out. Thanks tho

#

@marsh forge

marsh forge
#

huh?

#

Assume

#

f(x) is in UxV

#

that means (x,y_0) is in UxV

#

if y_0 isn't in V

#

is this possible?

#

Do you know the defn of preimage?

pseudo crane
#

So apparently it is possible to construct a nonempty perfect set with no rational points

honest narwhal
#

Yeah this is a Rudin chapter 2 problem

pseudo crane
#

Yep

#

Oop

tough hamlet
#

did you see what you did

pseudo crane
#

The fuck was i doing with epsilon

#

The proof is confusingly put together, put i don’t really see the falsehood in the reasoning

#

I think I meant to define epsilon to be 2r-h

tough hamlet
#

ok it's a bit weird to work out what it's supposed to mean

#

looks like p isn't fixed and you treated it as if it was

pseudo crane
#

Does this not show that for every neighbourhood of a, there exists a rational point p which has a neighbourhood containing E and therefore x

#

Oh

#

where do i treat it as if it was?

tough hamlet
#

for p to be a limit point of E, every neighbourhood of p needs to contain a point of E (different to p)

#

(which is done for fixed p)

pseudo crane
#

Oh

#

So I could have chosen a different p for each r

#

and the proof would have failed

tough hamlet
#

yes

pseudo crane
#

Ok

#

Yeah I can now see why this approach doesn’t work

#

thanks for the help

dreamy smelt
#

Yuhhh

#

Jon I remember when you were just a noob

#

you're just like me except you're actually succeeding on your own

#

🗿

tough hamlet
#

np

pseudo crane
#

I took a break from the “flashy” stuff, so to speak

#

I never really developed proof-making skills

dreamy smelt
#

Tbh same

#

I'm just too unmotivated to actually do any math

#

so I just know bits and pieces from the few areas I've explored and that's pretty much it

#

can't do much

pseudo crane
#

I figure it will be much more rewarding to learn things like algebraic geometry etc. in a more rigorous way

dreamy smelt
#

Oh for sure.

#

My main goals right now are elementary differential geometry and algebraic topology.

#

Kinda far away from where I am right now

#

lol

pseudo crane
#

Lol yeah

dreamy smelt
#

We'll get where we wanna be one day

#

lol

honest narwhal
#

"differential geometry" stahp

tough hamlet
#

el christoffe

marsh forge
#

Wait

#

How am I supposed to read

#

The x^2+1 thing

gentle ospreyBOT
marsh forge
#

Oh ok

#

Idk why youd use an eq relation

#

And?

#

The image being oath connected doesnt help

#

You should try

#

Sure

#

I would suggest

gentle ospreyBOT
marsh forge
#

Constructing paths

#

I would suggest not doing this eq relation idea