Suppose that the sequence converges in the product topology. Let $U$ be an open neighborhood of $\pi_{\alpha}(\mathbf{x})$ in $X_{\alpha}$ . Then the product of $U$ and all other spaces is an open neighborhood of $\mathbf{x}$ in the product topology and all members of the sequence starting from some $N$ lie in the neighborhood. Hence, $\pi_{\alpha}(\mathbf{x}_{n})\in U$ for $n\ge N$ . This works for the box topology as well.

The other direction. Suppose, that the projections of the sequence converge. Take any neighborhood of $\mathbf{x}$ , it contains a basis set $U$ containing $\mathbf{x}$ . $U$ is the product of open sets $U_{\alpha}$ in each coordinate space, and the space is in the product topology, for all but a finite number of $\alpha$ ’s: $U_{\alpha}=X_{\alpha}$ . For every $U_{\alpha}$ there is a number $N_{\alpha}$ such that for $n\ge N_{\alpha}$ , $\pi_{\alpha}(\mathbf{x}_{n})\in U_{\alpha}$ . If $U_{\alpha}=X_{\alpha}$ , we choose $N_{\alpha}=1$ . Then, for the product topology, there is only a finite number of $N_{\alpha}$ ’s that are greater than $1$ , and we can take the maximum of all $N_{\alpha}$ ’s, $N=\max_{\alpha}N_{\alpha}$ . All elements of the sequence starting from $N$ are contained in $U$ . For the box topology this direction may fail: there can be infinite number of $N_{\alpha}>1$ and no greatest $N_{\alpha}$ . Some examples are in Exercise 4(b) of §20 (note that in all examples every projection converges to 0).