Uncountable Product of Separable Spaces is not always Separable
Theorem
Let $I$ be an indexing set with uncountable cardinality.
Let $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$ be a family of topological spaces indexed by $I$.
Let $\ds \struct {S, \tau} = \prod_{\alpha \mathop \in I} \struct {S_\alpha, \tau_\alpha}$ be the product space of $\family {\struct {S_\alpha, \tau_\alpha} }_{\alpha \mathop \in I}$.
Let each of $\struct {S_\alpha, \tau_\alpha}$ be a separable space.
Then it is not necessarily the case that $\struct {S, \tau}$ is also a separable space.
Proof
Let $T = \struct {\Z_{\ge 0}, \tau}$ denote the topological space consisting of the set of positive integers $\Z_{\ge 0}$ under the discrete topology.
Let $I$ be an indexing set with uncountable cardinality.
Let $\ds T' = \struct {\prod_{\alpha \mathop \in I} \struct {\Z_{\ge 0}, \tau}_\alpha, \tau'}$ be the uncountable Cartesian product of $\struct {\Z_{\ge 0}, \tau}$ indexed by $I$ with the product topology $\tau'$.
From Countable Discrete Space is Separable, $T$ is a separable space.
But from Uncountable Cartesian Product of Discrete Topology on Positive Integers is not Separable, $T$ is not a separable space.
Hence the result.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $3$: Compactness: Invariance Properties