1. ## first-countable...

Let $\displaystyle \{X_i\}_{i \in I}$ a family of topological spaces. Prove that the product space $\displaystyle \prod X_i$ is first-countable if and only if each $\displaystyle X_i$ are first-countable, and every space $\displaystyle X_i$, except for a countable quantity, are indiscrete.

I have no idea... Maybe something for the converse, but probably I´m wrong...

2. Originally Posted by Inti
Let $\displaystyle \{X_i\}_{i \in I}$ a family of topological spaces. Prove that the product space $\displaystyle \prod X_i$ is first-countable if and only if each $\displaystyle X_i$ are first-countable, and every space $\displaystyle X_i$, except for a countable quantity, are indiscrete.

I have no idea... Maybe something for the converse, but probably I´m wrong...
Definition. A space X is first countable provided that there is a countable local basis at each point of X.

$\displaystyle (\Rightarrow)$
Assume $\displaystyle \prod_{i \in I}{X_i}$ is first countable. Then, for each $\displaystyle x = {(x_i)}_{i \in I}$ lies in $\displaystyle \prod_{i \in I}{X_i}$, there is a countable local basis at $\displaystyle x = {(x_i)}_{i \in I}$. Let a countable union of countable product set $\displaystyle \prod{(U_{x_i})}_i$ be such a countable local basis at $\displaystyle x = {(x_i)}_{i \in I}$, where $\displaystyle {(U_{x_i})}_i$ is a local basis at each $\displaystyle x_i \in X_i$ for finitely many values of $\displaystyle i$ and $\displaystyle {(U_{x_i})}_i = X_i$ for all other values of i. It follows that $\displaystyle {(U_{x_i})}_i$ should be all countable since $\displaystyle \prod_{i \in I}{X_i}$ has a countable local basis for each x lies in $\displaystyle \prod_{i \in I}{X_i}$ by hypothesis. Thus, each $\displaystyle {(U_{x_i})}_i$ can be the required countable local basis at each $\displaystyle x_i \in X_i$ and we conclude that $\displaystyle X_i$ is first-countable. Since a countable union of $\displaystyle \prod{(U_{x_i})}_i$ is countable, $\displaystyle {(U_{x_i})}_i = X_i$ for all but countable values of i. Thus, $\displaystyle X_i$ should be indiscrete for all but countable values of i.

$\displaystyle (\Leftarrow)$
Assume each $\displaystyle X_i$ is first-countable, and every space $\displaystyle X_i$, except for a countable quantity, are indiscrete.
Let $\displaystyle x = {(x_i)}_{i \in I}$ lies in $\displaystyle \prod_{i \in I}{X_i}$. If $\displaystyle (\beta_{x_i})_i$ is a countable local basis at $\displaystyle x_i$ for the space $\displaystyle X_i$, then the countable union of products $\displaystyle \prod{(U_{x_i})}_{i}$, where $\displaystyle {(U_{x_i})}_i = {(\beta_{x_i})}_i$ for finitely many values of $\displaystyle i$ and $\displaystyle {(U_{x_i})}_i = X_i$ for all the other values of $\displaystyle i$ is the desired countable local basis for $\displaystyle x = {(x_i)}_{i \in I}$ lies in $\displaystyle \prod_{i \in I}{X_i}$. Thus, $\displaystyle \prod_{i \in I}{X_i}$ is first countable.