$X_{\alpha}$ is normal when $\prod{X_{\alpha}}$ is normal

Question

If $\prod{X_{\alpha}}$ is normal in product topology then $X_{\alpha}$ is normal for all $\alpha$
assuming $X_{\alpha}\neq\emptyset$

Answer 1

let $X=\prod{X_{\alpha}}$ be a normal space in product topology$\implies$ X is regular $\implies X_{\alpha}$ is regular $\implies$ one point sets in $X_{\alpha}$ are closed set
consider space $X_{\beta}$ for some index $\beta$. let $A_{\beta}$ and $B_{\beta}$ be two disjoint closed set in $X_{\beta}$
for each $\alpha\neq\beta$ choose $x_{\alpha}\in X_{\alpha}$ set $$A=\prod{A_{\alpha}}$$ and $$B=\prod{B_{\alpha}}$$ where $A_{\alpha}=B_{\alpha}=\{x_{\alpha}\}$ when $ \alpha\neq\beta$ as for all $\alpha$ $A_{\alpha}$ and $B_{\alpha}$ are closed set in $X_{\alpha}$ $\implies$ A and B are closed in in X, also as $A_{\beta}\cap B_{\beta}=\emptyset$ $\implies A\cap B= \emptyset$
then by normality of X choose sets U and V open in X such that $$U\cap V=\emptyset$$ and $$A\subset U ; B\subset V$$ now set $$C=\prod{C_{\alpha}}$$ where $C_{\alpha}=\{x_{\alpha}\}$ when $\alpha\neq\beta$ and $C_{\beta}=X_{\beta}$
let $\pi_{\beta}$ be the $\beta$ coordinate map and let $f$ be restriction of $\pi_{\beta}$ on $C$ it is easy to check that $f:C\to X_{\beta}$ is a homeomorphism and as $U\cap C$ and $V\cap C$ are open and disjoint in subspace topology of C $\implies$ $U_{0}=f(U\cap C)$ and $V_{0}=f(V\cap C)$ are open and disjoint in $X_{\beta}$ also observe that $$A_{\beta}=f(A)\subset f(U\cap C)=U_{0}$$ and $$B_{\beta}=f(B)\subset f(V\cap C)= V_{0}$$ and the normality of $X_{\beta}$ follows