Mathematics - 2024-10-25

Obliv

00:00

How come $\mathbb{Q}\cap [0,1]$ isn't compact? I feel like if we pick the countably infinite rational numbers in this interval, we get an infinitely discontinuous set similar to the cantor set, which is compact.

Thorgott

@BenSteffan does that include CW-approximation at least?

@psie yes, I'm not saying the norms are the same, just that they induce the same topology

@Obliv the Cantor set is uncountable, it has a lot more points

Obliv

like isn't it just $\{0,q_1,q_2,\dots,1\}$

why wouldn't that set be compact

Sine of the Time

is it closed?

Thorgott

cause you can find a sequence of rational numbers without a convergent subsequence

Obliv

that doesn't converge to something in the set.. right

@sineofthetime i.e., it doesn't contain all of its limit points (every irrational between 0,1)

Ben Steffan

00:09

@Thorgott why would you want CW-approximation?

it does include cellular approximation

(but not CW-approximation)

Thorgott

oh, cellular is what I mean

Ben Steffan

yeah, that we have :)

I guess I should also say that you may assume $G$ is finitely generated

for reference, Hatcher's proof goes as follows: You show that $\pi_n$ of a wedge sum of $n$-spheres is a direct sum of copies of $\mathbb{Z}$ (Hatcher uses homotopy excision for this, but you can do it using cellular approximation). Then you look at the tail end of the l.e.s. $\ldots \to \pi_{n + 1}(X, X_n) \to \pi_n(X_n) \to \pi_n(X) \to 0$, and finally you use homotopy excision to see that $\pi_{n + 1}(X, X_n) \cong \pi_n(X / X_n)$ and the structure of the connecting map to conclude.

Put another way, if $G = \mathbb{Z} / n$ for some $n > 1$ and $f\colon S^n \to S^n$ is of degree $n$, then the question essentially reduces to showing that a map $g\colon S^n \to S^n \hookrightarrow S^n \cup_f D^{n + 1}$ is nullhomotopic iff its degree (as a map $S^n \to S^n$) is a multiple of $n$

There is, in fact, a way to do this """elementarily""", according to Fomenko-Fuchs, but it pretty much sucks

It's essentially a PL-approximation argument

on the other hand this is an exercise in a class I'm tutoring (obviously; why else all these restrictions? :) in which moreover the exercises are usually on the easy side, so there should be an easy solution

in a nutshell, I'm trying to figure out whether somebody messed up writing this exercise or whether I'm just stupid

Obliv

00:27

Let $K$ and $L$ be nonempty compact sets of $\mathbb{R}$ and define $$d = \inf\{|x-y|:x\in K \text{ and } L\}.$$ I've shown that if $K$ and $L$ are disjoint, $d>0$ but what if $K,L$ aren't guaranteed to be closed? I said no, but idk how I'd prove this.

I wanted to say consider an open set $A$ and its set of limit points $L$. by definition $L = \{x: \forall \varepsilon>0, A\cap V_{\varepsilon}(x)\setminus \{x\}\neq \varnothing\}$

Ben Steffan

well $L \cap A \neq \emptyset$ :)

consider intersecting with the complement of $A$ and this will work

but you could also give a concrete counterexample

hint: intervals

well, it will usually work: of course $A$ could also be closed, and then the other set is empty

anyways I'm converging to bed

Obliv

Thank you

Good night

It's because in the underlying logical foundations

by convention, the law of excluded middle tells us that $A\cap A^c=\varnothing$ which is probably an axiom or something

Thorgott