Let \(\mathbf{x}_1, \mathbf{x}_2, \ldots\) be a sequence of the points of the product space \(\prod X_\alpha\). Show that this sequence converges to the point \(\mathbf{x}\) if and only if the sequence \(\pi_\alpha(\mathbf{x}_1), \pi_\alpha(\mathbf{x}_2), \ldots\) converges to \(\pi_\alpha(\mathbf{x})\) for each \(\alpha\). Is this fact true if one uses the box topology instead of the product topology?
Solution
1. In the product topology, if \(\mathbf x_n\to\mathbf x\), then \(\pi_\alpha(\mathbf x_n)\to\pi_\alpha(\mathbf x)\) for every \(\alpha\).
Each projection \(\pi_\alpha\) is continuous, and a continuous map carries a convergent sequence to a sequence converging to the image of the limit.
2. In the product topology, if \(\pi_\alpha(\mathbf x_n)\to\pi_\alpha(\mathbf x)\) for every \(\alpha\), then \(\mathbf x_n\to\mathbf x\).
A basic neighborhood of \(\mathbf x\) is \(\prod_\alpha U_\alpha\) with \(U_\alpha\) open and \(U_\alpha=X_\alpha\) except for \(\alpha\) in a finite set \(F\). For each \(\alpha\in F\) choose \(N_\alpha\) with \(\pi_\alpha(\mathbf x_n)\in U_\alpha\) for \(n\ge N_\alpha\). For \(n\ge\max_{\alpha\in F}N_\alpha\), every coordinate of \(\mathbf x_n\) lies in the corresponding \(U_\alpha\), so \(\mathbf x_n\in\prod_\alpha U_\alpha\).
3. In the box topology the equivalence fails: in \(\RR^\omega\) the sequence \(\mathbf x_n=(\frac1n,\frac1n,\frac1n,\ldots)\) converges to \(0\) in every coordinate but does not converge to \(\mathbf 0\).
The set \(B=\prod_{k\ge1}(-\frac1k,\frac1k)\) is a box neighborhood of \(\mathbf 0\). The \(n\)-th coordinate of \(\mathbf x_n\) is \(\frac1n\notin(-\frac1n,\frac1n)\), so \(\mathbf x_n\notin B\) for every \(n\). The implication of step 1 still holds, since projections are continuous in the box topology.