Elements in the basis of Product topology determined by sub-basis other than sub-basis elements.

Question

I could prove the result for $|\Lambda|$ finite. Here $|\Lambda|$ is arbitrary.

My attempt:-

Let $\langle x_{\alpha}\rangle_{\alpha\in \Lambda}\in B \implies \langle x_{\alpha}\rangle_{\alpha\in \Lambda}\in \pi_{\beta_i}^{-1}(U_{\beta_i}), \forall i=1,2,3,...,n. $ So, $\pi_{\beta_i}(\langle x_{\alpha}\rangle_{\alpha\in \Lambda})\in U_{\beta_i},i=1,2,3..,n \implies \pi_{\beta_i}(\langle x_{\alpha}\rangle_{\alpha\in \Lambda}) \in \prod_{\alpha\in \Lambda} B_{\alpha}, \begin{cases} B_{\alpha}=X_{\alpha},\alpha \neq \beta_i,\forall i=1,2,...,n\\ B_{\alpha}=U_{\beta_{i}}, \alpha=\beta_i, \forall i=1,2,...,n \end{cases}$. hence, $B\subset \prod_{\alpha\in \Lambda} B_{\alpha}, \begin{cases} B_{\alpha}=X_{\alpha},\alpha \neq \beta_i,\forall i=1,2,...,n\\ B_{\alpha}=U_{\beta_{i}}, \alpha=\beta_i, \forall i=1,2,...,n \end{cases}$.

Let $\langle x_{\alpha}\rangle_{\alpha\in \Lambda}\in \prod_{\alpha\in \Lambda} B_{\alpha},\begin{cases} B_{\alpha}=X_{\alpha},\alpha \neq \beta_i,\forall i=1,2,...,n\\ B_{\alpha}=U_{\beta_{i}}, \alpha=\beta_i, \forall i=1,2,...,n \end{cases}$. It is obvious that $\langle x_{\alpha}\rangle_{\alpha\in \Lambda}\in B.[\because,\langle x_{\alpha}\rangle_{\alpha\in \Lambda}$ is of the form $x_\alpha \in U_{\beta_{i}}, \forall i=1,2,3,...,n,$ So, obviously $\pi_{\beta_i}(\langle x_{\alpha}\rangle_{\alpha\in \Lambda})\in U_{\beta_i}, \forall i=1,2,3,...,n,$]. Can you rectify any error creep in to my proof?

Answer 1

There really is nothing to prove; it's true by definitions:

$x \in \bigcap_{i=}^n \pi_{\beta_i}^{-1}[U_{\beta_i}]$ iff

$\forall \beta \in \{\beta_1,\ldots, \beta_n\}: \pi_\beta(x) \in U_{\beta_i}$

For any set in the product we already have (by the definition of a Cartesian product) that $\forall \alpha \in \Lambda: \pi_\alpha(x) \in X_\alpha$. So the definition of $B$ is just stating the only extra condition that a point of the product has to obey to be in $\bigcap_{i=}^n \pi_{\beta_i}^{-1}[U_{\beta_i}]$.

(note that $x \in \prod_\alpha B_\alpha$ iff for all $\alpha$: $\pi_\alpha(x) \in B_\alpha$ which is a void condition except for those $\alpha \in \{\beta_1, \ldots, \beta_n\}$). )