A proof of Tychonoff theorem from Alexander subbase theorem

Question

I'm trying to prove Tychonoff theorem from Alexander subbase theorem. I posted my proof as an answer below. Could you have a check on my attempt?

Let $(E_i, \tau_i)_{i\in I}$ be an arbitrary collection of compact topological spaces. We endow $E:= \prod_{i\in I} E_i$ with the product topology. Then $E$ is compact.

PS: I posted my proof separately so that I can accept my own answer to remove my question from unanswered list. Surely, if other people post answers, I will happily accept theirs.

Answer 1

A more direct alternative: let $\mathcal{U}$ be a subbasic cover of $E$, so that each $U \in \mathcal{U}$ is of the form $\pi_{i(U)}^{-1}[O_U]$ for some $i(U) \in I$ and some $O_U \in \mathcal{T}_{E_i}$.

If for some $i \in I$ we have that $\{O_U\mid i(U)=i\}$ is an open cover of $E_i$, by compactness of $E_i$ we can find a finite subcover of $\mathcal{U}$.

So we can assume WLOG that this is not the case and using AC we can pick $p_i \in E_i\setminus \bigcup \{O_U\mid i(U)=i\}$. This $p$ is then covered by no member of $\mathcal{U}$. Contradiction, so we must actually be in the previous case and we're done.

Answer 2

Let $\pi_i:E \to E_i$ be the canonical projection for each $i\in I$. Then $\mathcal B := \left \{ \pi_i^{-1}[A_i] \,\middle\vert\, A_i \in \tau_i \right \}$ is a subbase of $E$. Let $(B_j)_{j\in J}$ with $B_j = \pi_{i_j}^{-1}[A_j] \in \mathcal B$ be an open cover of $E$. Here $i_j \in I$ and $A_j \in \tau_{i_j}$ for all $j\in J$. WLOG, we assume $A_j \neq E_{i_j}$ and thus $B_j \neq E$ for all $j\in J$. Assume the contrary that $(B_j)_{j\in J}$ has no finite subcover. WLOG, we assume $J:=\mathbb N$. We have $$\bigcup_{k \le n} \pi_{i_k}^{-1}[A_k] \subsetneq E, \quad \forall n \in \mathbb N.$$

Let $\mathcal I := \{i_n \mid n \in \mathbb N \}$. For each $i \in \mathcal I$, let $N_i := \{n \in \mathbb N \mid i_n = i\}$.

We claim that there exists at least one $i^\star \in \mathcal I$ such that $(A_{n})_{n \in N_{i^\star}}$ is an open cover of $E_{i^\star}$. If not, $\bigcup_{n \in N_{i}} A_{n} \subsetneq E_{i}$ for all $i \in \mathcal I$. By axiom of choice, we can pick $e_i \in \bigcap_{n \in N_{i}} A^c_{n}$ for each $i\in \mathcal I$. Let $e \in E$ such that the $\pi_i(e) = e_i$ for all $i\in \mathcal I$. Because $B_n \neq E$ for all $n \in \mathbb N$, we get $e$ is not covered by $(B_n)_{n \in \mathbb N}$. This is a contradiction.

As such, there exists $i^\star \in \mathcal I$ such that $(A_{n})_{n \in N_{i^\star}}$ is an open cover of $E_{i^\star}$. Because $E_{i^\star}$ is compact, $N_{i^\star}$ has a finite subset $N'_{i^\star}$ such that $(A_{n})_{n \in N'_{i^\star}}$ is an open cover of $E_{i^\star}$. It follows that $(B_n)_{n \in N'_{i^\star}}$ is a finite subcover. This is again a contradiction. It follows that $E$ is compact by Alexander subbase theorem.

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Alexander subbase theorem: Let $E$ be a topological space and $\mathcal B$ its subbase. If every open cover by elements from $\mathcal B$ has a finite subcover, then $E$ is compact.

Proof: Assume the contrary that $E$ is not compact. Let $\mathcal C$ be the collection of all open covers without finite subcover. Then $\mathcal C$ is non-empty and partially ordered by inclusion. By Zorn's lemma, we extract the maximal element $O:= (O_i)_{i\in I}$ of $\mathcal C$.

We have $O' :=O \cap \mathcal B$ can not cover $E$. If not, $O' \subseteq \mathcal B$ and thus $O$ would have a finite subcover. It follows that there is $x \in (\bigcup O')^c$. There is $i \in I$ such that $x \in O_i$. There are $B_1, \ldots, B_n\in \mathcal B$ such that $x \in B:= \bigcap_{j=1}^n B_j \subseteq O_i$. We claim that $B_j \notin O$ for $j = 1, \ldots, n$. If not, $B_j \in O'$ and thus $x \in \bigcup O'$, which is a contradiction.

Then each $O \cup \{B_j\}$ has a finite subcover $\{B_j,O_{j,1}, \ldots, O_{j,n_j}\}$ with $O_{j,1}, \ldots, O_{j,n_j} \in O$. If $y \notin \bigcup_{t=1}^{n_j} O_{j, t}$, then $y \in B_j$. If $y \notin \bigcup_{j=1}^n \bigcup_{t=1}^{n_j} O_{j, t}$, then $x \in \bigcap_{j=1}^n B_j = B$. It follows that $\{B\} \cup\{ O_{j, t} \}{_{j=1}^n}{_{t = 1}^{n_j}}$ is a finite cover of $E$. We have $B \subseteq O_i$, so $\{O_i\} \cup\{ O_{j, t} \}{_{j=1}^n}{_{t = 1}^{n_j}}$ is also a finite cover of $E$. This contradicts the fact that $O$ has no finite subcover. Hence $E$ is compact.