## Abstraction The goal of this blog post is to provide a novel proof of the following result:

Proposition: Let $$A$$ be a finite set and let $$B$$ be a subset of $$A$$. Then $$B$$ is finite.

Proof: Suppose otherwise, that $$B$$ is an infinite subset of $$A$$. This means precisely that the complement of $$B$$ is not an element of the Fréchet filter $$\mathcal F$$ in $$A$$, hence $$\mathcal F$$ is a proper filter on $$A$$. Since it’s a proper filter, we can apply the ultrafilter lemma to show that it is contained in some ultrafilter $$\mathcal U$$. A standard result states that any ultrafilter containing Fréchet filter is nonprincipal. So $$U$$ is a nonprincipal ultrafilter on $$A$$. This implies that the Stone-Čech compactification $$\beta A$$ is a proper superset of $$A$$ (under the standard identification of elements of $$A$$ as principal ultrafilters in $$A$$). In particular, by pigeonhole principle, $$\beta A$$ and $$A$$ are not bijective.

Now note that $$\beta A$$, as a topological space, is the Stone-Čech compactification of $$A$$ considered as a topological space with discrete topology. Since $$A,\beta A$$ are not bijective, they are surely not homeomorphic as topological spaces. We will reach a contradiction as soon as we show that $$A$$ is homeomorphic to its own Stone-Čech compactification.

To show that, we will use the characterization of Stone-Čech compactification as the unique, up to homeomorphism, compact Hausdorff topological space $$X$$ containing $$A$$ as a subspace and satisfying the universal property: any continuous function $$f:A\rightarrow Y$$, where $$Y$$ is some compact Hausdorff space, can be uniquely extended to a continuous function $$\widetilde f:X\rightarrow Y$$ such that, for all $$a\in A,f(a)=\widetilde f(a)$$. We need to verify all these properties.

Compactness: Let $$\displaystyle A\subseteq \bigcup_{i\in I}U_i$$, where $$U_i$$ are all open. Then for any $$a\in A$$ there is an $$i_a\in I$$ such that $$a\in U_{i_a}$$. Since $$A$$ is finite, this gives us a finite set of indices $$i_a,a\in A$$ such that $$\displaystyle A\subseteq\bigcup_{a\in A} U_{i_a}$$. This gives us a finite subcover of any open cover of $$A$$. This means that $$A$$ is compact.

Hausdorffness: Let $$a,b\in A$$ be two distinct elements. Then $$\{a\},\{b\}$$ are open, because we consider $$A$$ with the discrete topology. Moreover, these two sets are disjoint, because every element in their intersection would be both $$a$$ and $$b$$, and these are different. This means precisely that $$A$$ is Hausdorff.

Universal property: Let $$f:A\rightarrow Y$$ be any continuous function, where $$Y$$ is compact and Hausdorff. We define $$\widetilde f:A\rightarrow Y$$ by $$\widetilde f(a)=f(a)$$ for every $$a\in A$$. Then for every $$a\in A$$ we have $$f(a)=\widetilde f(a)$$. Moreover, for any open set $$U\subseteq Y$$ we have $$\widetilde f^{-1}(U)=f^{-1}(U)$$, which is open, since $$f$$ is continuous by assumption, so $$\widetilde f$$ is continuous. We have shown existence, so we only need to confirm uniqueness. Suppose $$\widetilde{\widetilde f}$$ is another such function. In particular, for all $$a\in A, \widetilde{\widetilde f}(a)=f(a)=\widetilde f(a)$$. This means that $$\widetilde{\widetilde f}=\widetilde f$$, as we wanted.

So we have a contradiction. Hence $$B$$ is finite. $$\square$$

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1. #### David V Feldman

Re the first line of your proof:
As the Frechet filter contains *co-finite* sets, not lying in it equates with having an infinite *complement.*
(Perhaps some symbol isn’t rendering, but I tried two browsers.)

• #### Wojowu

Very much right! Thank you for spotting the mistake.