The Claim:The product/sum of a sequence of separable metrizable spaces is separable.
Here the sum of a family $((X_i, d_i))_{i\in I}$ of metric spaces is defined (up to isometry) as follows: By copying the metric of each $X_i$ on a set of the same cardinality, we can assume that the sets $X_i$ are pairwise disjoint. Let $X=\cup_{i\in I}X_i$. We define a metric $d$ on $X$ by letting $d(x, y)=d_i(x, y)$ if $x, y\in X_i$, and $d(x, y)=1$ if otherwise. The topology of this metric space is the sum of the topologies of $((X_i, d_i))$.
Source: Classical Descriptive Set Theory by Alexander Kechris.
My Question: I am almost certain that the claim involves the lemma "countable union of countable sets is countable": Let $D_i\subseteq X_i$ be the countable dense set in each $X_i$, and let $D=\bigcup_{i\in\mathbb{Z}^+}D_i$. Then $D$ is a countable dense set of $X$. Thus, the Axiom of Choice is at play here. Is my suspicion justified, or is there a way to prove the claim without invoking AC? Any help would be greatly appreciated.
Best Answer
For the sum case, it is at least as strong as the claim that a countable union of countable sets is countable. If we have a choice of a countable dense set for each of the spaces, the problem is equivalent to the claim that the countable union of countable sets is countable.
For consider a countable family of countable sets $(X_i)_{i \in D}$, and suppose WLOG the $X_i$ are pairwise disjoint. Equip each $X_i$ with the discrete metric. Then $T = \bigcup\limits_{i \in D} X_i$ also has the discrete metric, so the only dense set in $T$ is $T$ itself. Thus, $T$ is separable if and only if it is countable.
Some amount of choice is needed to show the countable union of countable sets is countable. This is strictly weaker than countable choice.