If $E_i$ is a normed space, does $\overline{\prod_iA_i}=\prod_i\overline{A_i}$ hold in $\{x∈\prod_iE_i:\sup_i\left\|x_i\right\|_{E_i}<∞\}$?

Question

Let $I$ be a countable set, $E_i$ be a topological space for $i\in I$ and $$E:=\prod_{i\in I}E_i.$$ Now let $A_i\subseteq E_i$ for $i\in I$. We know that $\overline{\prod_{i\in I}A_i}=\prod_{i\in I}\overline{A_i}$ when $E$ is considered as being equipped with the product topology.

However, suppose the the topology on each $E_i$ is induced by a norm and let $$\tilde E:=\left\{x\in E:\sup_{i\in I}\left\|x_i\right\|_{E_i}<\infty\right\}$$ be equpped with $$\left\|x\right\|_{\tilde E}:=\sup_{i\in I}\left\|x_i\right\|_{E_i}<\infty\;\;\;\text{for }x\in X.$$ Does t he desired identity remain to hold in $\tilde E$? To be precise: $\overline{\tilde E\cap\prod_{i\in I}A_i}=\tilde E\cap\prod_{i\in I}\overline{A_i}$?

Answer 1

The desired result does hold. Here is one way to see that.

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Claim: The subspace topology on $\tilde{E}$ is coarser than the norm topology.

Proof: Balls of the form $B_j(x,n^{-1})$ for $x \in E_j$ and $n \geq 1$ form a basis for the topology of $E_j$. Hence, sets of the form $$U = \bigg(\prod_{i=1}^n B_i(x_i,n_i^{-1}) \bigg) \times \bigg(\prod_{j>n} E_j \bigg)$$ form a basis for the product topology on $E$. Then just note that for $y \in U \cap \tilde{E}$, for $i \leq n$ there is $\varepsilon_i$ such that $B_i(y_i, \varepsilon_i) \subseteq B_i(x_i,n_i^{-1})$ and so
$$B_{\tilde{E}}(y, \min_{i \leq n} \varepsilon_i) \subseteq U \cap \tilde{E}.$$ It follows that $U \cap \tilde{E}$ is open in the norm topology on $E$ from which the claim follows.

Corollary: $\operatorname{clos}_{\tilde{E}} (\tilde{E} \cap \prod_{i \in I} A_i) \subseteq \prod_{i \in I} \overline{A_i} \cap \tilde{E}$

Proof: By the last claim $\operatorname{clos}_{\tilde{E}} (\tilde{E} \cap \prod_{i \in I} A_i)$ is contained in the closure of the same set in the subspace topology. It is a standard result that this coincides with $\overline{\prod_{i \in I} A_i} \cap \tilde{E} = \prod_{i \in I} \overline{A_i} \cap \tilde{E}$.

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For the other inclusion, we rely on the norm structure. Suppose that $y = (y^{(1)}, y^{(2)}, y^{(3)}, \dots) \in \prod_{i \in I} \overline{A_i} \cap \tilde{E}$. We want to show that we can approximate $y$ by elements of $\prod_{i \in I} A_i \cap \tilde{E}$ in the norm topology on $\tilde{E}$.

For each $i \in I$, pick a sequence $(y_n^{(i)})_{n \geq 1}$ in $A_i$ converging to $y^{(i)}$ for the norm on $E_i$. It is not automatic that $(y_n^{(1)}, y_n^{(2)}, \dots)$ converges to $y$ in $\tilde{E}$ since the convergence may not be uniform in the coordinates but by passing to subsequences we can arrange this.

Let $t_{0,i} = 1$ and let $$t_{n,i} = \inf\{ k > t_{n-1,i}: \|y^{(i)} - y_k^{(i)}\|_{E_i} \leq 2^{-n}\}.$$ Then by construction, $(y_{t_{n,1}}^{(1)}, y_{t_{n,2}}^{(2)}, \dots)$ is in $\prod_{i \in I} A_i \cap \tilde{E}$ and converges to $y$ in the norm on $\tilde{E}$ as $n \to \infty$. Hence $y \in \operatorname{clos}_{\tilde{E}} (\tilde{E} \cap \prod_{i \in I} A_i)$