Question about pointwise convergence of sequences in the box and product topologies.

Question

Can someone please verify my proof or offer suggestions for improvement? I'm aware that there may be answers floating elsewhere, but I need help with my proof in particular.

$\textbf{Note:}$ This is not homework.

Let $x_1, x_2, \ldots$ be a sequence of points in the product space $\prod X_\alpha$. Show that this sequence converges to the point $x$ if and only if the sequence $\pi_\alpha(x_1), \pi_\alpha(x_2), \ldots$ converges to $\pi_\alpha(x)$ for each $\alpha$. Is this fact true if one uses the box topology instead of the product topology?

$(\Rightarrow)$ Let $\beta$ be an index. Let $\pi_\gamma$ denote the projection onto the $\gamma^{th}$ factor. Let $U_\beta$ be a neighborhood of $\pi_\beta(x)$. Then, $\pi_\beta^{-1}(U_\beta)$ is a neighborhood of $x$. So, there exists a natural number $N$ such that for all $n > N$, $x_n \in \pi_\beta^{-1}(U_\beta)$. But then, for all $n > N$, $\pi_\beta(x_n) \in U_\beta$. Therefore, for all $\alpha$, $\pi_\alpha(x_n)$ converges to $\pi_\alpha(x)$.

$(\Leftarrow)$ Let $\prod U_\alpha$ be a basis element of the product topology containing $x$. Let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be the indices for which $U_\alpha \neq X_\alpha$. Then, for each $1 \leq k \leq n$, there exists a natural number $N_i$ such that for all $n > N_i$, $\pi_{\alpha_k(x)} \in U_{\alpha_k}$ . Set $N = \max \{N_k: 1 \leq k \leq n\}$. Then, for all $n > N$, $x \in \prod U_\alpha$.

This is not true if we use the box topology instead of the product topology. Consider the box topology on $\mathbb{R}^\omega$, the countably infinite cartesian product of $\mathbb{R}$ with itself. Let $$x_n = \left( \frac{1}{n}, \frac{2}{n}, \frac{3}{n}, \ldots \right)$$

Clearly, for any $k \in \mathbb{N}$, $\pi_k(x_n) \longrightarrow 0$. However, it is not the case that $x \longrightarrow \textbf{0}$. To see this, consider the basis element $U = \displaystyle{\prod_{k=1}^\infty (-1, 1)}$ containing $\textbf{0}$. Then, for all $x_n$, $x_n \notin U$, since we can always find a $k$ such that $\pi_k(x_n) > 1$.

Answer 1

The proof seems correct to me. There is just a typo in the backwards implication paragraph. There should be $x_n ∈ ∏U_α$ instead of $x ∈ ∏U_α$.

For the first implication, note that you actually prove that any continuous function preserves convergence of sequences from which the implication immediately follows. Also note that the same proof works even for transfinite sequences or any nets.

Just a side note about notation. I think that one should not omit the index in notation like product (sum, union, …). I.e. write $∏_α U_α$ instead of $∏ U_α$. In theory if one had $U_α = ⟨X_i: i ∈ I⟩$ then $∏ U_α = ∏⟨X_i: i ∈ I⟩ = ∏_{i ∈ I} X_i$ instead of $∏_α U_α$.