## Product of sequence of separable spaces is separable (p3 (A))

The product of a sequence of separable topological spaces is separable.

Proof.
Let $(X_n, \tau_n)$ be a sequence of separable spaces.  For each $X_n$, let $A_n$ denote a countable dense subset. Let $X := \prod_{n=0}^{\infty} X_n$ have the product topology, $\tau$.  Let $x \in X$, $x = (x_n)_{n=0}^{\infty}$.  Let $A := \bigcup_{n = 0}^{\infty} \prod_{i = 0}^n A_n \times \prod_{i = n+1}^{\infty} \{ x_i \}$, which is clearly countable, since it is a countable union of countable sets.

Now, I will show that $A$ is dense in $(X, \tau)$.  Let $y \in X$, $U$ be a basis element of $\tau$, such that $y \in U$.  Then, $U = \prod_{i = 0}^{\infty} U_i$, where $U_i$ is basis element of $\tau_i$, or $U_i = X_i$.  Furthermore, there exists $N$, such that $i > N$ implies $U_i = X_i$.  Now, for each $i \le N$, $\exists a_i \in A_i$ with $a_i \in U_i$, since $A_i$ is dense in $\tau_i$.  Then, $(a_0, a_1, \ldots, a_N, x_{N + 1}, x_{N + 2}, \ldots) \in A \cap U$.  Thus, $y \in \overline{A}$ and so $\overline{A} = X$.

This entry was posted in Descriptive Set Theory, Kechris. Bookmark the permalink.

### One Response to Product of sequence of separable spaces is separable (p3 (A))

1. Cenzer says:

This solution is good. You might say fix some element x \in X