# 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 $$0$$.

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 $$\mathbb{C}$$. $$\Box$$