Tychonov Theorem

(Remark: The proof presented in this post is a reorganization and interpretation of that given by James Munkres in his book "Topology".)

Theorem 37.3 (Tychonov Theorem) The product of arbitrary number of compact spaces is compact in the product topology, i.e.
\[
X = \prod_{i \in I} X_i
\]
is compact if $X_i$ is compact for all $i \in I$.

Basic thoughts about proving the Tychonov Theorem

Although compactness is originally defined via finite open covering, it has another equivalent formulation via finite intersection property (FIP) of a collection of closed sets. The compactness of the product space $X$ will be proved by following this FIP route.
A strict partial order based on the set inclusion relation will be assigned to the system or superset of collections of subsets in the product space which have the FIP. It can be envisioned that every such collection of subsets dwells in a chain which has an upper bound. The Zorn's Lemma can then be used to prove the existence of a maximal collection $\mathcal{D}$.
For any collection $\mathcal{A}$ of subsets in the product space $X$ having the FIP, a point $\vect{x}$ in $X$ will be constructed with each of its component belonging to the intersection of the closures of the corresponding component of each subset in $\mathcal{A}$.
The properties of the maximal collection $\mathcal{D}$ will be explored, which can be used to prove that any subbasis and hence basis element containing the constructed $\vect{x}$ belongs to $\mathcal{D}$. Based on these conclusions, this $\vect{x}$ can be proved to be in the intersection of all the elements in $\mathcal{A}$ and the Tychonov theorem is proved.

Compactness and finite intersection property

Lemma 26.9 A topological space $X$ is compact if and only if every collection $\mathcal{F}$ of closed sets in $X$ with the finite intersection property has a nonempty intersection.

Proof: This lemma will be proved by using reduction to absurdity.

Prove in the forward direction

Let $X$ be a compact space. For any of its opening covering $\{U_i\}_{i \in I}$, there exists an finite sub-covering $\{U_{i_k}\}_{k=1}^n$, from which we have
\[
\bigcap_{i \in I} U_i^c = \varPhi
\]

and

\[
\bigcap_{k=1}^n U_{i_k}^c = \varPhi,
\]

where each $U_i^c$ is a closed set in $X$. From this we learns that if a topological space $X$ is compact, for any collection of closed sets in $X$ having an empty intersection, it must have a finite sub-collection, which also has an empty intersection.

Then, assume that there exists a collection $\mathcal{F}$ of closed sets with their intersection being empty. $\mathcal{F}$ must have a finite sub-collection which also has an empty intersection. However, this contradicts the fact that $\mathcal{F}$ should have the FIP. Therefore, every collection $\mathcal{F}$ of closed sets must has a nonempty intersection.
Prove in the inverse direction

Assume $X$ is not compact, then there exists an open covering $\{U_i\}_{i \in I}$ of $X$, which has no finite sub-covering. From this, we have
\[
\bigcap_{i \in I} U_i^c = \varPhi
\]
and for any of its finite sub-collections $\{U_{i_k}\}_{k = 1}^n$,
\[
\bigcap_{k = 1}^n U_{i_k}^c \neq \varPhi.
\]
Therefore, the collection of closed sets $\{U_i^c\}_{i \in I}$ has the FIP. According to the given condition, the intersection of all $U_i^c$ should be nonempty, which contradicts the conclusion derived from the assumption.

Summarizing the above two steps, Lemma 26.9 is proved.

Construction of the maximal collection $\mathcal{D}$

Lemma 37.1 Let $X$ be a set and $\mathcal{A}$ be a collection of subsets of $X$ having the finite intersection property. Then there exists a collection $\mathcal{D}$ of subsets of $X$ such that $\mathcal{A} \subset \mathcal{D}$ and $\mathcal{D}$ has the finite intersection property. In addition, there is no larger collection, which also has the finite intersection property, containing $\mathcal{D}$.

Proof: Let $\mathbb{A}$ be a superset, each element of which has the FIP and contains the collection $\mathcal{A}$. Assign this superset $\mathbb{A}$ with a strict partial order based on the proper set inclusion relation $\subsetneq$. Then let $\mathbb{B}$ be a subset of $\mathbb{A}$, which itself is also a superset, such that $\mathbb{B}$ is simply ordered by $\subsetneq$.

Let $\mathcal{C}$ be the union of all the elements in $\mathbb{B}$, i.e.
\[
\mathcal{C} = \bigcup_{\mathcal{B} \in \mathbb{B}} \mathcal{B}.
\]
Next, we need to show that $\mathcal{C}$ belongs to $\mathbb{A}$ and is an upper bound of the chain $\mathbb{B}$.

The upper bound property based on the proper set inclusion relation has already been ensured by the construction of $\mathcal{C}$ via the union operation. We only need to prove $\mathcal{C} \in \mathbb{A}$, which requires the following two conditions to be fulfilled:

$\mathcal{C}$ contains $\mathcal{A}$.

This is obvious, because for every $\mathcal{B}$ in $\mathbb{B}$, $\mathcal{A}$ is contained in $\mathcal{B}$ and $\mathcal{B}$ is contained in $\mathcal{C}$.
$\mathcal{C}$ has the FIP.

Let $\{C_1, \cdots, C_n\}$ be a finite collection selected from $\mathcal{C}$. Because $\mathcal{C}$ is the union of all the collections in the superset $\mathbb{B}$, there exists a $\mathcal{B}_i$ in $\mathbb{B}$ such that $C_i \in \mathcal{B}_i$ for all $i = 1, \cdots, n$. Then $\{\mathcal{B}_1, \cdots, \mathcal{B}_n\}$ is a finite chain which must have a largest element and we here let it be $\mathcal{B}_k$. Hence we have $C_i \in \mathcal{B}_k$ for all $i = 1, \cdots, n$.

Because each element of $\mathbb{B}$ including $\mathcal{B}_k$ is also an element of $\mathbb{A}$ which has the FIP and the finite collection $\{C_1, \cdots, C_n\}$ is contained in $\mathcal{B}_k$, $\mathcal{C}$ has the FIP.

Summarizing the above, we know that $\mathcal{C}$ is really an upper bound in $\mathbb{A}$ of the chain $\mathbb{B}$. According to the Zorn's Lemma, there is a maximal element, let it be $\mathcal{D}$, such that $\mathcal{D}$ has the FIP and contains $\mathcal{A}$.

Properties of the maximal collection $\mathcal{D}$

Lemma 37.2 Let $X$ be a set and $\mathcal{D}$ be a collection of subsets of $X$ which is maximal with respect to the FIP and strict partial order based on the proper set inclusion relation. Then:

Any finite intersection of elements of $\mathcal{D}$ is also an element of $\mathcal{D}$.
If a subset of $X$ intersects every element of $\mathcal{D}$, this subset is also an element of $\mathcal{D}$.

Proof:

Let $B$ be the finite intersection of elements in $\mathcal{D}$. Let's see if we can construct a collection which is larger than $\mathcal{D}$ by appending the element $B$ to the collection $\mathcal{D}$. If this is not achievable, we know that the appended element $B$ must belong to $\mathcal{D}$.

Let $\mathcal{E} = \mathcal{D} \cup \{B\}$. Because $\mathcal{D} \subset \mathcal{E}$, $\mathcal{D}$ and $\mathcal{E}$ belong to the same chain. Then we check if $\mathcal{E}$ has the FIP.
- When the finite number of elements extracted from $\mathcal{E}$ are all selected from $\mathcal{D}$, their intersection is not empty due to the FIP of $\mathcal{D}$.
- When the selected finite number of elements include $B$, their intersection has the following formulation:
  \[
  D_1 \cap \cdots \cap D_m \cap B,
  \]
  where $D_1, \cdots, D_m \in \mathcal{D}$.
Because $B$ is also a finite intersection of elements in $\mathcal{D}$, $D_1 \cap \cdots \cap D_m \cap B$ is a finite intersection of elements in $\mathcal{D}$ as well, which is of course not empty.

Summarizing the above, we know that $\mathcal{E}$ has the FIP and $\mathcal{D} \subset \mathcal{E}$. Because $\mathcal{D}$ is a maximal element, we have $\mathcal{D} = \mathcal{E}$. Therefore, $B \in \mathcal{D}$.
Let $A$ be the subset of $X$ which intersects every element of $\mathcal{D}$. Similarly as above, let $\mathcal{E} = \mathcal{D} \cup \{A\}$ and we have the following two facts:
- When the finite number of elements are all selected from $\mathcal{D}$, their intersection is nonempty due to the FIP of $\mathcal{D}$.
- When the selected finite number of elements include $A$, their intersection has the following formulation:
  \[
  D_1 \cap \cdots \cap D_m \cap A.
  \]
  According to the already proved conclusion 1 in this lemma, $D = D_1 \cap \cdots \cap D_m$ is also an element of $\mathcal{D}$. Because $A \cap D \neq \varPhi$ from the given condition, the intersection $D_1 \cap \cdots \cap D_m \cap A$ is not empty.
Summarizing the above, we know that $\mathcal{E}$ has the FIP and is larger than $\mathcal{D}$. Because $\mathcal{D}$ is a maximal element, we have $\mathcal{D} = \mathcal{E}$ and $A \in \mathcal{D}$.

Tychonov Theorem

Finally, we come to the proof of the Tychonov Theorem.

Proof: Let $\{X_i\}_{i \in I}$ be a collection of compact spaces. Their product is
\[
X = \prod_{i \in I} X_i.
\]
Let $\mathcal{A}$ be any collection of subsets of $X$ having the FIP. Then the closures of the elements in $\mathcal{A}$ also have the FIP. As long as we can prove
\[
\bigcap_{A \in \mathcal{A}} \bar{A} \neq \varPhi,
\]
$X$ is compact according to Theorem 26.9.

According to Lemma 37.1, there exists a maximal element $\mathcal{D}$ in the sense of FIP and strict partial order based on the proper set inclusion relation. Because $\mathcal{D}$ contains $\mathcal{A}$, if
\[
\bigcap_{D \in \mathcal{D}} \bar{D} \neq \varPhi,
\]
there is also
\[
\bigcap_{A \in \mathcal{A}} \bar{A} \neq \varPhi.
\]
Let $\pi_i: X \rightarrow X_i$ for all $i \in I$ be the projection map. Because $\mathcal{D}$ has the FIP, for a specific index $i$, the collection of the component sets $\{\pi_i(D) \vert D \in \mathcal{D} \}$ also has the FIP.

Because the component space $X_i$ is compact, according to Theorem 26.9, there exists an $x_i$ such that
\[
\begin{equation}
x_i \in \bigcap_{D \in \mathcal{D}} \overline{\pi_i(D)} \quad (\forall i \in I).
\label{eq:x_i_range}
\end{equation}
\]
Let all these $\{x_i\}_{i \in I}$ form an element $\vect{x}$ in the product space $X$. We will prove that this $\vect{x}$ really belongs to $\cap_{D \in \mathcal{D}} \bar{D}$ in the following, during which the concept of subbasis is adopted, whose finite intersection forms the topological basis of the product space.

Let $U_i$ be any open set in $X_i$ containing $x_i$ and $\pi_i^{-1}(U_i)$ be the corresponding subbasis element. For all $D \in \mathcal{D}$, we have
\[
D \cap \pi_i^{-1}(U_i) = \prod_{j \in I} \pi_j(D) \cap \pi_j(\pi_i^{-1}(U_i)).
\]
There are the following two cases:

When $j = i$, the corresponding component in the above product is $\pi_i(D) \cap U_i$. Because of equation \eqref{eq:x_i_range} and $U_i$ being any open set in $X_i$ containing $x_i$, there exists a $y_i \in \pi_i(D) \cap U_i$, i.e. $\pi_i(D) \cap U_i \neq \varPhi$.

For if $\pi_i(D) \cap U_i = \varPhi$, because $U_i$ being any open set in $X_i$ containing $x_i$, then $x_i \in \left( \overline{\pi_i(D)} \right)^c$, which contradicts the fact that $x_i \in \overline{\pi_i(D)}$.
When $j \neq i$,
\[
\pi_j(D) \cap \pi_j(\pi_i^{-1}(U_i)) = \pi_j(D) \cap X_j = \pi_j(D).
\]

Summarizing the above, we know that as long as we select a $\vect{y}$ from $D$ with its $i$th component being $y_i \in \pi_i(D) \cap U_i$, we have $\vect{y} \in D \cap \pi_i^{-1}(U_i)$ for all $D \in \mathcal{D}$. According to Lemma 37.2 (2), the subbasis element $\pi_i^{-1}(U_i)$ for all $i \in I$ belongs to $\mathcal{D}$. We also see that $\vect{x} \in \pi_i^{-1}(U_i)$.

In addition, the finite intersection of these subbasis elements, which is a basis element of the product space, is not empty due to the FIP of $\mathcal{D}$. Then according to Lemma 37.2 (1), the basis elements of $X$ which contain $\vect{x}$ also belong to $\mathcal{D}$.

Because $\mathcal{D}$ has the FIP, for all basis element $B$ containing $\vect{x}$ and for all element $D$ of $\mathcal{D}$, their intersection is not empty. Assume there exists a $D_0 \in \mathcal{D}$ such that $\vect{x} \notin \bar{D}_0$, or rather $\vect{x} \in X - \bar{D}_0$, which is open in $X$. Then there exits a basis element $B_0$ containing $\vect{x}$ such that $B_0 \cap \bar{D}_0 = \varPhi$ and hence $B_0 \cap D_0 = \varPhi$. This contradicts the fact that for all such basis element $B$, $B \cap D_0 \neq \varPhi$.

Now we arrive at the conclusion that for all $D \in \mathcal{D}$, $\vect{x} \in \bar{D}$, and for any collection $\mathcal{A}$ of subsets in $X$ having the FIP with $\mathcal{D}$ being the maximal element on the same chain, $\cap_{A \in \mathcal{A}} \bar{A} \neq \varPhi$. According to Lemma 26.9, the product space $X$ is compact.