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Let $\{X_i\}_{i\in\mathbf N}$ be separable spaces. Show that $X = \prod_{i\in \mathbf N}X_i $ is also separable.

My work:

Let $D_i$ be dense in $X_i$ for all $i$. We need to produce a countable dense subset in $X = \prod_{i\in \mathbf N}X_i $

Regards

tomasz
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Morata
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2 Answers2

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(Note: In this answer, I assume that the ${\bf N}$ is the set of natural numbers. The same idea works if it is any other countable set, and the conclusion is also true if it has cardinality of the continuum (or less), although the argument is a little bit more tricky in that case. For any ther ${\bf N}$, the conclusion is false (for nontrivial $X_i$).)

For each $i\in \mathbf N$, pick any one point $z_i\in D_i$. Then check that $$ D:=\{(x_i)_i\mid \textrm{each }x_i\in D_i\textrm{ and for all but finitely many }i, x_i=z_i \} $$ is a countable dense subset of the product.

tomasz
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  • This works if $N$ is countable. The OP could not provide any info about $N$. – drhab Dec 26 '17 at 08:33
  • @drhab: It's almost certainly supposed to be the set of natural numbers. For arbitrary "$N$" bigger than the continuum, this is not even true. – tomasz Dec 26 '17 at 11:23
  • I know. Maybe it is an idea to start your answer with something like "preassuming that $N$ denote the natural numbers..". That will make me upvote your answer. I asked the OP for information about $N$ and he/she replied negative on that. – drhab Dec 26 '17 at 11:27
  • @drhab: Is that what you had in mind? – tomasz Dec 27 '17 at 03:21
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The product $X$ is separable if and only if $N$ is countable. This question provides one counterexample for uncountable $N$.

Below we proceed with $N = \mathbb{N}$, as the finite case is similar, only easier.

As @tomasz suggests, let $(y_1, y_2, \ldots) \in \prod_{n\in\mathbb{N}} D_n$ be arbitrary and define$$D = \left\{(x_n)_{n=1}^\infty \in \prod_{n\in\mathbb{N}} D_n : x_i = y_i \text{ for all but finitely many } i \in \mathbb{N}\right\}$$ which is a countable set. Let's check that $D$ is dense in $X$.

If suffices to verify that $D$ intersects all nonempty basis elements for the product topology on $X$.

Take an arbitrary $$U_1 \times U_2 \times \cdots \times U_n \times X_{n+1} \times X_{n+1} \times \cdots \ne \emptyset$$ where $U_i \subseteq X_i$ are open, for $i = 1, \ldots, n$.

Since $D_i$ is dense in $X_i$ and $U_i$ is a nonempty open subset of $X_i$, let $x_i \in D_i \cap U_i \ne \emptyset$, for $i = 1, \ldots, n$.

Now notice that $$(x_1, x_2, \ldots, x_n, y_{n+1}, y_{n+2}, \ldots ) \in D \cap \big(U_1 \times U_2 \times \cdots \times U_n \times X_{n+1} \times X_{n+1} \times \cdots\big)$$

Hence $$D \cap \big(U_1 \times U_2 \times \cdots \times U_n \times X_{n+1} \times X_{n+1} \times \cdots\big) \ne \emptyset$$

We conclude that $D$ is dense in $X$. Therefore, $X$ is separable.

mechanodroid
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  • Why is $D$ countable? – kimchi lover Dec 26 '17 at 01:31
  • @kimchilover: It isn't. – tomasz Dec 26 '17 at 01:43
  • @kimchilover Oh, i thought that $N = \mathbb{N}$. Obviously $D$ is countable if and only if $N$ at most countable, in which case we may assume $D \subseteq \mathbb{N}$, so the same proof works. – mechanodroid Dec 26 '17 at 09:41
  • $N$ is countable, but the product of $N$ copies of (say) a two element set is uncountable. $X$ is the set of all sequences, whose $i$-th terms are in $X_i$. You should know that the set of infinite binary sequences is not countable. – kimchi lover Dec 26 '17 at 12:09
  • @kimchilover Yeah, I noticed that. I have fixed the error by introducing something like a direct sum of the sets $D_n$. – mechanodroid Dec 26 '17 at 12:11