Topological Proofs of Stuff and Things

I'm writing this page to serve as a collection of proofs I come across which use topological methods to show things about non-topological-like things. I'll update this page as I come across them. I find it interesting how topological methods are useful outside of its own discipline, or, from a different perspective, how topology lives in a lot of nonobvious places.

Infinite Primes

This proof of a classic theorem was invented by Hillel Furstenberg in the 1950s.°

Theorem 1. There are infinitely many primes.

Proof. Define $S(a,b)$ to be the arithmetic sequence $\{an + b\::\: n \in \mathbb{N}\}$. Then we can define a topology on $\mathbb{Z}$ by calling a subset open if and only if it is the union of some arithmetic sequences. It is easy enough to show that this actually does define a topology on $\mathbb{Z}$. We notice two things:

The compliment of a finite set can't be closed, because this would imply that a finite set could be open. But the only finite open set in our topology is the empty set.
Any arithmetic sequence $S(a,b)$ is closed, since we can write the compliment of $S(a,b)$ as $$ \bigcup_{i=1}^{a-1} S(a, b+i), $$ which is open because it is the union of arithmetic sequences.

Now suppose for the sake of contradiction that there were finitely many primes, indexed $p_i$. Since any integer can be written as the multiples of some primes, except for $\{-1, 1\}$, we can write $$ \mathbb{Z}\setminus\{-1,1\} = \bigcup_{p_i} S(p_i,0). $$ Note that the left hand side isn't closed by our first bullet-point observation. But the right hand side, being the union of finitely many closed sets, must be closed (this isn't necessarily true for infinite unions of closed sets). This is a contradiction. $\Box$

Fundamental Theorem of Algebra

This is a classic application of the fundamental group. I'll assume that one has already calculated the fundamental group of a circle to be $ \mathbb{Z} $, even though this takes a lot of work. I won't go into rigorous detail, but the point will be more to get a visual understanding of the proof.

Theorem 2. A degree $n$-polynomial with complex coefficients $ p \in \mathbb{C}[x] $ has $ n $ roots in $ \mathbb{C}$.

Proof. If we show that one root of $p$ exists, say $\alpha$, then we can factor out $p(x)/(x-\alpha)$ to get another polynomial in $\mathbb{C}$. So by some inductive argument, it suffices to show that a non-constant polynomial has a root in $\mathbb{C}$.

Suppose for the sake of contradiction that $p$ has no root in $\mathbb{C}$. So $p$ is a function from $ \mathbb{C} $ to $\mathbb{C} \setminus \{0\}$. We'll visualize the input and output spaces as follows:

Consider the circle around the origin $$ C_r = \{re^{i\theta}\:|\:\theta \in [0, 2\pi)\}. $$ First, if the circle has radius 0, then $C_0 = 0$. So $ p(C_0) $ is some point in the output space:

Next, recall that when $x$ gets large, $p(x)$ starts to look like its highest order term. So if $p(x) = a_nx^n+\dots +a_1x +a_0$, then at large values of $x$, $p(x)$ acts like $a_n x^n$ (in an informal, handwavy manner of speaking). So for $r >> 0$, we get that $p(C_r)$ looks like $ a_n(C_r)^n $, i.e. a big circle. So we should be able to find a value $R$ large enough so that $ p(C_R) $ encircles $p(C_0)$ and the point $0$.

take C_R to a circular thing in the output

Finally, observe that varying $r$ from $0$ to $R$ defines a homotopy from $p(C_R)$ to $ p(C_0) $, because polynomials are continuous. But this means that the (loop homeomorphic to the) circle $ p(C_R) $ contracts to the point $ p(C_0) $. The one-point-complement of $\mathbb{C}$ is homotopy equivalent to $S^1$, which has nontrivial fundamental group, and in particular $p(C_R)$ describes a nontrivial element of this fundamental group, so this is a contradiction. In other words, we want to shrink the red curve $p(C_R)$ to the blue point $p(C_0)$, but analysis of $S^1$ says that we can't do so without 'getting stuck' on the point 0. So $p$ must have a root in $\mathbb{C}$. $\Box$