# Math Help - first-countable...

1. ## first-countable...

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

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

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