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Let $(X_j)_{j\in J}$ be an indexed family of non-empty topological spaces.

If $X_j$ second countable for each j and $\coprod_{j\in J}X_j$ is second countable.

Show that $J$ is countable.

Note: $\coprod_{j\in J}X_j$ $=$ $\bigcup_{j\in J}X_j^*$ where $X_j^*$ is the image of $X_j$ under the canonical injection from $X_j$ to the disjoint union.

My attempt:

Observe that there exists a mapping $g: J \rightarrow \bigcup_{j\in J}X_j^*$ such that $g(j)\in X_j^*$ for each J. This mapping is injective since for $I\neq J$,$X_I^*\neq X_J^*$. Let $\mathbb{B}$ be a countable basis for the disjoint union. Define a map $f: \bigcup_{j\in J} X_j^* \rightarrow \mathbb{B}$ by assigning $x$ the basis $B_m \in \mathbb{B}$ which owns $x$ where $m$ is the least such natural number. This map is injective because for $\alpha \neq \beta$, $X_{\alpha}^* \cap X_{\beta}^*$ is empty. Thus $f \circ g$ is injective and since $\mathbb{B}$ is countable, it follows that so is $J$.

Is this proof correct? (please answer this)

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1 Answer 1

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It suffices to note that all $X^\ast_j, j \in J$ are non-empty, pairwise disjoint and open. So each of them contains a necessarily distinct base element $B_j \in \mathbb{B}$. So $J \to B_j$ is an injection from $J$ into a countable set.

Or if you know basic cardinal functions theory this argument can be represented as:

$$|J| \le c(\coprod_{j \in J} X_j) \le w(\coprod_{j \in J} X_j) = \aleph_0$$

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  • $\begingroup$ Thank you once again! $\endgroup$ Commented Jan 18, 2020 at 11:40

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