Prove that $p_B :\prod_{\alpha \in A} X_\alpha \to \prod_{\alpha \in B} X_\alpha$ is a continuous map

Suppose $A$ is any set and $B \subseteq A$. Suppose for each $\alpha \in A$ we have a topological space $X_\alpha$. Prove that the projection \[ p_B :\prod_{\alpha \in A} X_\alpha \to \prod_{\alpha \in B} X_\alpha \] is continuous when we give both these spaces their product topology.


Let $U$ be an open set in $\prod\limits_{\alpha \in B} X_\alpha$. First assume

\[  U=\prod_{\alpha \in B}U_\alpha, \ \ \text{where } U_\alpha \text{ is open in } X_\alpha.   (1) \] Then \[ P_B^{-1}(U)=\prod_{\alpha \in A}V_\alpha, \] where $V_\alpha=U_\alpha$ if $\alpha \in B$ and $V_\alpha=X_\alpha$ if $\alpha \in A\setminus B$. Thus $P_B^{-1}(U)$ is open. Now suppose \[ O=\bigcup\limits_{i \in I}U_i     (2)\] such that $U_i$ is of the form (1). We have \[ P_B^{-1}(O)=P_B^{-1}(\bigcup\limits_{i \in I}U_i)=\bigcup\limits_{i \in I}P_B^{-1}(U_i). \] Thus $P_B^{-1}(O)$ is also open. Finally note that if $U$ is a finite intersection of open sets of type (2), then by a similar argument, $P_B^{-1}(U)$ is also open and therefore $P_B$ is continuous.

The answer is accepted.