If \(X_n\) is metrizable with metric \(d_n\), then
\[ D(\mathbf{x}, \mathbf{y}) = \sup\ts{\bar{d}_i(x_i, y_i)/i} \]
is a metric for the product space \(X = \prod X_n\). Show that \(X\) is totally bounded under \(D\) if each \(X_n\) is totally bounded under \(d_n\). Conclude without using the Tychonoff theorem that a countable product of compact metrizable spaces is compact.
Solution
Fix \(\varepsilon>0\). Choose \(N\) so large that \(1/(N+1)<\varepsilon\). For each \(1\le i\le N\), total boundedness of \((X_i,d_i)\) (equivalently of \((X_i,\bar d_i)\)) gives a finite \(\varepsilon i\)-net \(F_i\subset X_i\) for \(\bar d_i\) (if \(\varepsilon i\ge1\), one point suffices). Choose arbitrary basepoints \(a_i\in X_i\) for \(i>N\). The finite set \[ F=F_1\times\cdots\times F_N\times\prod_{i>N}\{a_i\} \] is an \(\varepsilon\)-net for \(X\): given \(x\), choose \(y_i\in F_i\) with \(\bar d_i(x_i,y_i)<\varepsilon i\) for \(i\le N\) and put \(y_i=a_i\) afterwards. Then for \(i>N\), \[ \bar d_i(x_i,y_i)/i\le1/i<\varepsilon, \] so \(D(x,y)<\varepsilon\). Thus \((X,D)\) is totally bounded.
If every \(X_i\) is compact, each is complete. A \(D\)-Cauchy sequence is Cauchy in every coordinate because \(\bar d_i(x_i,y_i)\le iD(x,y)\); hence it converges coordinatewise to some \(x\in X\). Given \(\varepsilon>0\), choose \(N\) with \(1/(N+1)<\varepsilon\); the terms with \(i>N\) in \(D(x^{(k)},x)\) are at most \(1/(N+1)\), and the finitely many terms with \(i\le N\) are less than \(\varepsilon\) for large \(k\). So \(x^{(k)}\to x\) in \(D\), and \(X\) is complete. A complete totally bounded metric space is compact. Hence the countable product of compact metrizable spaces is compact, without invoking Tychonoff.