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