Proof of Tychonoff's Theorem Using Ultrafilters

11 Aug 2025

This post was machine translated and has not yet been proofread. It may contain minor errors or unnatural expressions. Proofreading will be done in the near future.

Tychonoff’s Theorem. The product of compact spaces is compact.

This is an extremely simple yet powerful theorem, but its proof is rather complex in undergraduate topology texts such as Munkres. Many books omit the proof entirely. However, using ultrafilters, one can prove this theorem very simply.

1. Preliminary Concepts

For detailed explanations and proofs of the concepts introduced here, refer to Chapter 4 of Stephen Willard, General Topology.

Filters and Ultrafilters

In a previous post, we introduced filters in Boolean algebra structures. Since topological spaces form Boolean algebra structures when $\cup$ and $\cap$ correspond to disjunction and conjunction respectively, we can define filters on topological spaces.

Definition. For a topological space $X$, a filter $\mathcal{F}$ on $X$ is a collection of subsets of $X$ satisfying the following conditions:

1. $X \in \mathcal{F}, \varnothing \notin \mathcal{F}$
2. $A, B \in \mathcal{F} \implies A \cap B \in \mathcal{F}$
3. $A \in \mathcal{F}, A \subseteq B \implies B \in \mathcal{F}$

Additionally, if the following is satisfied, it is an ultrafilter:

4. $A \notin \mathcal{F} \implies X \setminus A \in \mathcal{F}$

Neighbourhood Filters

One important filter in topology is the neighbourhood filter. First, we define neighbourhood as follows:

Definition. For a topological space $X$, we say that $N \subseteq X$ is a neighbourhood of $x \in X$ if there exists some open set $U$ such that $x \in U \subseteq N$.

Note that neighbourhoods need not be open sets. When a neighbourhood is an open set, it is called an open neighbourhood. This contrasts with the convention in some textbooks that use neighbourhood and open neighbourhood synonymously, so care must be taken.

Notation. The collection of neighbourhoods of $x$ is denoted $\mathcal{N}(x)$ or $\mathcal{N}_x$.

One can easily verify that $\mathcal{N}(x)$ is a filter. Accordingly, $\mathcal{N}(x)$ is called the neighbourhood filter.

Generation of Filters

Definition. A collection $\mathcal{C}$ of subsets of a topological space $X$ is called a prefilter or filter base when it satisfies the following:

1. $\varnothing \notin \mathcal{C}$
2. $A, B \in \mathcal{C} \implies \exists C \in \mathcal{C} : C \subseteq A \cap B$

That is, whilst filters are closed under intersection, prefilters are “weakly” closed under intersection. When $\mathcal{C}$ is a prefilter, one can generate a filter $\mathcal{F}$ from $\mathcal{C}$ as follows:

\[\mathcal{F} = \{ F \subseteq X : C \subseteq F \text{ for some } C \in \mathcal{C} \}\]

Moreover, given a mapping $f: X \to Y$ and a filter $\mathcal{F}$ on $X$, the following collection is a prefilter:

\[\mathcal{C} = \{ f(F) : F \in \mathcal{F} \}\]

Notation. The filter generated by $\mathcal{C}$ is denoted $f(\mathcal{F})$.

When $f$ is onto, we have $\mathcal{C} = f(\mathcal{F})$, which is convenient to handle. Also, when $f$ is onto, the following holds (trivially):

Theorem. If $f: X \to Y$ is onto and $\mathcal{F}$ is an ultrafilter on $X$, then $f(\mathcal{F})$ is an ultrafilter on $Y$.

Extension to Ultrafilters

Every filter can be extended to an ultrafilter. For the proof, refer to the previous post (the axiom of choice is used).

Theorem. If $\mathcal{F}$ is a filter on $X$, then there exists an ultrafilter $\mathcal{U}$ on $X$ such that $\mathcal{U} \supseteq \mathcal{F}$.

Convergence of Filters

Definition. When a filter $\mathcal{F}$ contains $\mathcal{N}(x)$, we say that $\mathcal{F}$ converges to $x$, and write $\mathcal{F} \to x$.

For example, the filter generated by the prefilter $\lbrace (0, \epsilon) : \epsilon > 0 \rbrace $ converges to $0$ (this example shows that $\mathcal{F} \to x$ does not require all elements of $\mathcal{F}$ to contain $x$ as an element).

The main properties of filter convergence are as follows:

Theorem.

1. For $E \subseteq X$, $x \in \overline{E}$ if and only if there exists some filter $\mathcal{F}$ such that $E \in \mathcal{F}$ and $\mathcal{F} \to x$.
2. $f: X \to Y$ is continuous if and only if for any filter $\mathcal{F}$, if $\mathcal{F} \to x$ then $f(\mathcal{F}) \to f(x)$.

This generalises the characterisation of closure and continuous functions by sequences in first countable spaces.

Convergence of filters in product spaces is equivalent to convergence in each component space. That is, the following holds:

Theorem. A filter $\mathcal{F}$ on the product space $\prod_{i \in I}X_i$ converges to $x$ if and only if $\pi_i(\mathcal{F}) \to \pi_i(x)$ for each $i \in I$.

2. Proof of Tychonoff’s Theorem

We now prove Tychonoff’s theorem. The key is the following lemma:

Lemma. $X$ is compact if and only if every ultrafilter on $X$ converges.

Proof

$(\Rightarrow)$ Let $\mathcal{F}$ be an ultrafilter on $X$ that does not converge. That $\mathcal{F}$ does not converge to any point means that for any $x \in X$, there exists some open set $U_x$ such that $U_x \notin \mathcal{F}$. Since $\mathcal{F}$ is an ultrafilter, $X \setminus U_x \in \mathcal{F}$. Now consider the collection $\mathcal{C} = \lbrace U_x : x \in X \rbrace $. $\mathcal{C}$ covers $X$, and since $X$ is compact, it has a finite subcover $\lbrace U_{x_1}, \dots, U_{x_n} \rbrace $. Therefore,

\[\begin{align} &X \setminus (U_{x_1} \cup \dots \cup U_{x_n}) \\ &= (X \setminus U_{x_1}) \cap \dots \cap (X \setminus U_{x_n}) \\ &= \varnothing \end{align}\]

But since $X \setminus U_{x_i} \in \mathcal{F}$ for each $1 \leq i \leq n$, by closure of filters under intersection, $\varnothing \in \mathcal{F}$, which is a contradiction. □

$(\Leftarrow)$ Suppose $X$ is not compact. Then there exists some open cover $\mathcal{C}$ of $X$ that has no finite subcover. Therefore, when we define $\mathcal{P}$ as follows:

\[\mathcal{P} = \{ X \setminus (U_1 \cup \dots \cup U_n) : U_i \in \mathcal{C} \}\]

$\mathcal{P}$ is closed under finite intersection and does not contain the empty set, so it is a prefilter. Accordingly, $\mathcal{P}$ can be extended to a filter, and this filter can be extended to an ultrafilter. Let such an ultrafilter be $\mathcal{F}$. By assumption, $\mathcal{F}$ converges to some point $x$. That is, $\mathcal{N}(x) \subseteq \mathcal{F}$. Therefore, the following holds:

\[x \in \bigcap_{F \in \mathcal{F}} \overline{F}\]

But this contradicts $\bigcap_{P \in \mathcal{P}} \overline{P} = \bigcap_{P \in \mathcal{P}} P = \varnothing$ (since $\mathcal{C}$ is a cover of $X$). ■

Tychonoff’s theorem now follows almost trivially.

Theorem. The product of compact spaces is compact.

Proof

Let each $X_i$ be a compact space for $i \in I$. $X$ is compact if and only if any ultrafilter $\mathcal{F}$ on $\prod_{i \in I}$ converges, and $\mathcal{F}$ converges if and only if $\pi_i(\mathcal{F})$ converges in $X_i$. Since $\pi_i$ is onto, $\pi_i(\mathcal{F})$ is an ultrafilter, and since $X_i$ is compact, $\pi_i(\mathcal{F})$ converges. Accordingly, $X$ is compact. ■