# Prove that $\prod X_\alpha$ is first (second)-countable if and only if $X_\alpha$ is first (second)-countable, $\forall \alpha\in I.$

Let $$I\neq\emptyset$$ numerable and $$(X_\alpha,\tau_\alpha)$$ a family of topological spaces. Prove the following.

1. $$\displaystyle\prod X_\alpha$$ is first-countable if and only if $$X_\alpha$$ is first-countable, $$\forall \alpha\in I.$$

2. $$\displaystyle\prod X_\alpha$$ is second-countable if and only if $$X_\alpha$$ is second-countable, $$\forall \alpha\in I.$$

All I have are the definitions and I do not know how to proceed to do the proofs.

Could you give me the idea of the proof for 1. and 2. please ? ? ?

Note that I am not asking for the entire proof since I know this site does not work like that.

Definition

$$(X,\tau)$$ is first-countable if $$\forall x\in X,$$ has a countable local basis of neighborhoods.

$$(X,\tau)$$ is second-countable if $$\tau$$ has a countable basis, i.e. if $$\exists\beta\subset\tau:\beta$$ is countable and basis for $$\tau.$$

• Hint on 2: A space is second-countable if it has a countable subbasis. This because the collection of finite intersections of its elements is a basis and is countable. Jan 21 '18 at 7:50

2) If part. Let $\beta_{\alpha}$ be a countable basis for $\tau_{\alpha}$, then try to argue that the following set \begin{align*} \beta=\left\{G_{\alpha_{1}}\times\cdots\times G_{\alpha_{n}}\times\prod_{\alpha\ne\alpha_{1},...,\alpha_{n}}X_{\alpha}: G_{\alpha_{k}}\in\beta_{\alpha_{k}},~\alpha_{k}\in I,~k=1,...,n,~n\in{\bf{N}}\right\} \end{align*} is a countable basis for $\displaystyle\prod X_{\alpha}$.

Only if part. Let $\beta$ be a countable basis for $\displaystyle\prod X_{\alpha}$, then try to argue that the following set \begin{align*} \beta_{\alpha}=\{\pi_{\alpha}(G): G\in\beta\} \end{align*} is a countable basis for $\tau_{\alpha}$, where \begin{align*} \tau_{\alpha}&:\prod X_{\alpha}\rightarrow X_{\alpha},~~~~(x_{\alpha})\rightarrow x_{\alpha}. \end{align*}

Every subspace of a second countable space is second countable. Likewise for first countable. As each factor is homeomomorphic to a subsapce of the product we get all left to right implications for free. We could also have used that these properties are preserved by open, continuosu surjective maps instead (the projections are such maps).

If each $X_\alpha$ is second countable, pick a countable base $\mathcal{B}_\alpha$ for $X_\alpha$ for each $\alpha \in I$.

Then define $$\mathcal{B} = \{\prod_{\alpha \in I} O_\alpha: \exists F \subseteq I \text{ finite }: \forall \alpha \in F: O_\alpha \in \mathcal{B}_\alpha \text{ and } \forall \alpha \notin F: O_\alpha = X_\alpha\}$$

All members of $\mathcal{B}$ are basic open sets (by the definition of the product topology: they are product open sets that essentially depend on a finite subset of the index set). $\mathcal{B}$ is countable by set theory: there are countably many finite subsets of $I$ as $I$ is countable, and for each finite subset we have finitely many choices from a countable set $\mathcal{B}_\alpha$.

And it's pretty easy to see that $\mathcal{B}$ is a base as all $\mathcal{B}_\alpha$ are, using the definition of a base as : for each $x \in O$ , where $O$ is open, there is some basic open set containing $x$ inside $O$.

For first countable we esentially use the same argument at each point, using local bases instead of global ones. It's basically the same idea.