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  1. #1
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    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 Im wrong...
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  2. #2
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    Quote Originally Posted by Inti View Post
    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 Im 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.
    Last edited by aliceinwonderland; Jun 14th 2009 at 02:13 AM. Reason: Correction
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