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Let {($X_\alpha,\mathscr T_\alpha):\alpha\in J$} be a set of topological spaces. Let $\beta_1,\beta_2,...,\beta_n$ be a finite set of distinct members of $J$. For each i=1,2,...,n let $U_{\beta_{i}}\in\mathscr T_{\beta_{i}}$, and let $B=\cap _{i=1}^{n}\pi_{\beta_{i}}^{-1}(U_{\beta_{i}})$. Prove that $B=\prod_{\alpha\in J}B_\alpha$, where $B_{\beta_{i}}=U_{\beta_{i}}$ for i = 1,2,...,n and $B_\alpha = X_\alpha$ for all $\alpha$ different from $\beta_1,\beta_2,...,\beta_n$.

I want to do this with if and only if statements instead of showing containment both ways. Here is what I have.

Let $(x_\alpha)$ be a J-tuple such that $x_\alpha\in X_\alpha$ for all $\alpha\in J$. Let $(x_\alpha)\in\cap _{i=1}^{n}\pi_{\beta_{i}}^{-1}(U_{\beta_{i}})$ This holds if and only if $x_{\beta_{i}}\in U_{\beta_{i}}$ for i = 1,2,...,n.
How can i stretch this to get to the product.

Am I on the right track? Any and all help is appreciated.

1 Answers 1

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Note that a point in $\prod_{\alpha \in J} X_\alpha$ is by definition a function $f: J \rightarrow \cup_{\alpha \in J} X_\alpha$, such that for all $\alpha \in J$ : $f(\alpha) \in X_\alpha$ Also, $\pi_\alpha(f) = f(\alpha)$ (definition of the projections)

So $f \in \cap_{i=1}^n \pi^{-1}[U_{\beta_i}]$ iff for all $\alpha \in J$ $f(\alpha) \in X_\alpha$ (needed to be in the product) and for $i=1\ldots n$ $\pi_{\beta_i}(f) = f(\beta_i) \in U_{\beta_i}$

The last two conditions just state the definition for $f$ to be in $\prod_{\alpha \in J} B_{\alpha}$ where $B_\alpha = U_\alpha$ for $\alpha \in \{\beta_1, \ldots ,\beta_n\}$ and $B_\alpha = X_\alpha$ otherwise.

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    Ahh thank you. The second portion is where I was mixed up I think. I felt what i needed but couldn't say it. Also the function perspective cleared it up a bunch. – 2017-02-15
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    @Jerry glad I could clarify matters. These basic open sets are conditions on $f$ that certain restrictions to finite subsets have specific conditions. – 2017-02-15