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.

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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

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