I am trying to prove that a countable product of finite topological spaces is compact without using Tychonoff’s theorem.
I tried to prove this through sequences, but I couldn't do it. Also, I don't know if it's possible to do it in this case.
How could this be achieved?
Let $X=\prod_{i\in\mathbb N}X_i$ where each $X_i$ is a nonempty finite topological space; let $\pi_i:X\to X_i$ be the projection map. Let $\mathcal U$ be an open cover of $X$. Let's say that a subset of $X$ is finitely covered if it's covered by a finite subfamily of $\mathcal U$.
Assume for a contradiction that $X$ is not finitely covered. Then, since $X_1$ is finite, we can choose $x_1\in X_1$ so that $\pi_1^{-1}(x_1)$ is not finitely covered. Next, since $X_2$ is finite, we can choose $x_2\in X_2$ so that $\pi_1^{-1}(x_1)\cap\pi_2^{-1}(x_2)$ is not finitely covered. Continuing in this way, we construct a point $x=(x_1,x_2,x_3,\dots)\in X$ such that, for each $n\in\mathbb N$, the set $\pi_1^{-1}(x_1)\cap\cdots\cap\pi_n^{-1}(x_n)$ is not finitely covered. It follows that no neighborhood of $x$ is finitely covered, but this contradicts the assumption that $\mathcal U$ is an open cover of $X$.
For each $n\in\Bbb N$ let $X_n$ be finite and non-empty. Let $T$ be the product topology on $X=\prod_{n\in\Bbb N}X_n.$
Now if each set $X_n$ is given the discrete topology then the product topology $T^*$ on $X$ will satisfy $T^*\supset T.$ So if $T^*$ is a compact topology, then so is $T.$
Let $d_n$ be "the" discrete metric on $X_n,$ i.e. if $p,q\in X_n$ with $p\ne q$ then $d_n(p,q)=1.$
A member of $X$ is a function $f$ on $\Bbb N$ with $f(n)\in X_n$ for each $n$. The metric $d(f,g)=\sum_{n\in\Bbb N}2^{-n}d_n(f(n),g(n))$ on $X$ generates the topology $T^*.$
Using $d,$ we can show that $T^*$ is sequentially compact. Let $(f_j)_{j\in\Bbb N}$ be a sequence of members of $X.$ Take $x_1\in X_1$ such that the set $J_1=\{j:f_j(1)=x_1\}$ is infinite. Take $M_1\in J_1.$ Recursively, if the set $J_n=\{j:\forall i\le n\,(f_j(i)=x_i)\}$ is infinite, take $x_{n+1}\in X_{n+1}$ such that the set $J_{n+1}=\{ j\in J_n:f_j(n+1)=x_{n+1}\}$ is infinite. This is possible because $X_{n+1}$ is finite (and not empty). Now take $M_{n+1}\in J_{n+1}$ with $M_{n+1}>M_n$.
Now let $f(n)=x_n$ for each $n.$ The subsequence $(f_{M_n})_{n\in\Bbb N}$ converges to $f,$ that is, $\lim_{n\to\infty}d(f_{M_n},f)=0$ because $d(f_{M_n},f)=\sum_{j>n}2^{-j}d_j(f_{M_n}(j),f(j))\le \sum_{j>n}2^{-j}=2^{-n}.$
