# 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.

## Answer

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.

- answered
- 817 views
- $15.00

### Related Questions

- Rouche’s Theorem applied to the complex valued function $f(z) = z^6 + \cos z$
- Prove that $S \subseteq X$ is nowhere dense iff $X-\overline{S}$ is dense.
- Show that ${(x,\sin(1/x)) : x∈(0,1]} ∪ {(0,y) : y ∈ [-1,1]}$ is closed in $\mathbb{R^2}$ using sequences
- [ Banach Fixt Point Theorem ] $\frac{dy} {dx} = xy, \text{with} \ \ y(0) = 3,$
- Find $\lim\limits _{n\rightarrow \infty} n^2 \prod\limits_{k=1}^{n} (\frac{1}{k^2}+\frac{1}{n^2})^{\frac{1}{n}}$
- A Real Analysis question on convergence of functions
- Banach's fixed point theorem application
- Prove that every compact Hausdorff space is normal