# The Unapologetic Mathematician

## Tychonoff’s Theorem

One of the biggest results in point-set topology is Tychonoff’s Theorem: the fact that the product of any family $\{X_i\}_{i\in\mathcal{I}}$ of compact spaces is again compact. Unsurprisingly, the really tough bit comes in when we look at an infinite product. Our approach will use the dual definition of compactness.

Let’s say that a collection $\mathcal{F}$ of closed sets has the finite intersection hypothesis if all finite intersections of members of the collection are nonempty, so compactness says that any collection satisfying the finite intersection hypothesis has nonempty intersection. We can then form the collection $\Omega=\{\mathcal{F}\}$ of all collections of sets satisfying the finite intersection hypothesis. This can be partially ordered by containment — $\mathcal{F}'\leq\mathcal{F}$ if every set in $\mathcal{F}'$ is also in $\mathcal{F}$.

Given any particular collection $\mathcal{F}$ we can find a maximal collection containing it by finding the longest increasing chain in $\Omega$ starting at $\mathcal{F}$. Then we simply take the union of all these collections to find the collection at its top. This is almost exactly the same thing as we did back when we showed that every vector space is a free module! And just like then, we need Zorn’s lemma to tell us that we can manage the trick in general, but if we look closely at how we’re going to use it we’ll see that we can get away without Zorn’s lemma for finite products.

Anyhow, this maximal collection $\mathcal{F}$ has two nice properties: it contains all of its own finite intersections, and it contains any set which intersects each set in $\mathcal{F}$. These are both true because if $\mathcal{F}$ didn’t contain one of these sets we could throw it in, make $\mathcal{F}$ strictly larger, and still satisfy the finite intersection hypothesis.

Now let’s assume that $\mathcal{F}$ is a collection of closed subsets of $\prod\limits_{i\in\mathcal{I}}X_i$ satisfying the finite intersection hypothesis. We can then get a maximal collection $\mathcal{G}$ containing $\mathcal{F}$. Then given an index $i\in\mathcal{I}$ we can consider the collection $\{\overline{\pi_i(G)}\}_{G\in\mathcal{G}}$ of closed subsets of $X_i$ and see that it, too, satisfies the finite intersection hypothesis. Thus by compactness of $X_i$ the intersection of this collection is nonempty. Letting $U_i$ be a closed set containing one of these intersection points $x_i$, we see that the preimage $\pi_i^{-1}(U_i)$ meets every $G\in\mathcal{G}$, and so must itself be in $\mathcal{G}$.

Okay, so let’s take the point $x_i$ for each index and consider the point $p$ in $\prod\limits_{i\in\mathcal{I}}X_i$ with $i$-th coordinate $x_i$. Then pick some set $V=\prod\limits_{i\in\mathcal{I}}V_i$ containing $p$ from the base for the product topology. For all but a finite number of the $i$, $V_i=X_i$. For those finite number where it’s smaller, the closure of $V_i$ contains the point $x_i\in X_i$, and so $\pi_i^{-1}(V_i)$ is in $\mathcal{G}$. So their finite intersection must be nonempty, and so is $V$ itself!

Now, since $V$ is in $\mathcal{G}$, it must intersect each of the closed sets in the original collection $\mathcal{F}$. Since the only constraint on $V$ is that it contain $p$, this point must be a limit point of each of the sets in $\mathcal{F}$. And because they’re closed, they must contain all of their limit points. Thus the intersection of all the sets in $\mathcal{F}$ is nonempty, and the product space is compact!

Advertisements

January 17, 2008