How to show that if $\prod X_\alpha$ is Hausdorff or normal then so is $X_\alpha $? How to show that if $\prod X_\alpha$ is Hausdorff  or normal then so is $X_\alpha $?
I will include where I am getting stuck.


*

*When $\prod X_\alpha$ is Hausdorff :: To show $X_\alpha $ is Hausdorff let $(x_\alpha),(y_\alpha )$ be two points in  $X_\alpha $. How should I take the corresponding points in $\prod X_\alpha$ in order to use it is Hausdorff?


(Note:I don't want to use that Hausdorff is a hereditary property and $X_\alpha $is a subspace of $\prod X_\alpha$).


*When $\prod X_\alpha$ is normal :: let $U_\alpha,V_\alpha$ be two disjoint closed sets in $X_\alpha$. Since the projection map is continuous so $p_\alpha^{-1}(U_\alpha)$ and $p_\alpha^{-1}(V_\alpha)$ are closed in $\prod X_\alpha$. Then we will get two disjoint closed sets $U,V\in \prod X_\alpha$ such that $p_\alpha^{-1}(U_\alpha)\subset U$ and $p_\alpha^{-1}(V_\alpha)\subset V$ ;$U\cap V=\emptyset\ .$


Can I now use that $(U_\alpha)\subset p_\alpha (U)$ and  $(V_\alpha)\subset p_\alpha (V)$ and the fact that $p_\alpha $ is open?
Please say the problems involved here and suggest proper changes.
 A: Part 1: Assume $\Pi_{\beta \in J} X_{\beta}$ is Hausdorff. Fix $\alpha \in J$, and choose two points $x,y \in X_{\alpha}$.
Now choose $f,g \in \Pi_{\beta} X_{\beta}$ such that $f(\alpha)=x$ and $g(\alpha)=y$, and such that $f(\beta) = g(\beta)$ for all $\beta \neq \alpha$. Since $f \neq g$ and $\Pi_{\beta} X_{\beta}$ is Hausdorff, we can find open neighborhoods $U \ni f$ and $V \ni g$ such that $U \cap V = \emptyset$.
By definition of the product topology, there exist collections $\{ U_{\beta}\}$ and $\{V_{\beta}\}$ of open sets such that $f \in \Pi_{\beta} U_{\beta} \subset U$, and $g \in \Pi_{\beta} V_{\beta} \subset V$, and $U_{\beta} = X_{\beta} = V_{\beta}$ for all but finitely many $\beta$.
Then $U_{\alpha}$ and $V_{\alpha}$ must be disjoint, since we assumed that $f(\beta) = g(\beta)$ for all $\beta \neq \alpha$. Since $x \in U_{\alpha}$ and $y \in V_{\alpha}$, it follows that $X_{\alpha}$ is Hausdorff, and we are done.
Part 2: Assume $\Pi_{\beta \in J} X_{\beta}$ is normal. Fix some $\alpha \in J$, and choose disjoint closed sets $A, B \subset X_{\alpha}$.
By continuity of the projection map $\pi_{\alpha}: \Pi_{\beta \in J} X_{\beta} \to X_{\alpha}$, the sets $\pi_{\alpha}^{-1}(A), \pi_{\alpha}^{-1}(B)$ are disjoint closed subsets of $\Pi_{\beta \in J} X_{\beta}$, so there exist some open disjoint sets $U \supset \pi_{\alpha}^{-1}(A)$ and $V \supset \pi_{\alpha}^{-1}(B)$.
Now fix some $f_0 \in \Pi_{\beta \in J} X_{\beta}$. Now define a map $g: X_{\alpha} \to \Pi_{\beta \in J} X_{\beta}$ by $$(g(x))_{\beta} := \begin{cases} 
      x & \text{ if } \beta = \alpha \\
      f_0(\beta) & \text{ if } \beta \neq \alpha
   \end{cases}$$ Then define $P = g(X_{\alpha})$. Then $P$ (with the subspace topology from $\Pi_{\beta} X_{\beta}$), is homeomorphic to $X_{\alpha}$. Indeed, the map $\pi_{\alpha}:P \to X_{\alpha}$ is a homeomorphism with inverse $g: X_{\alpha} \to P$.
Then, easily enough, it is checked that $\pi_{\alpha}(U \cap P) = g^{-1}(U)$ and $\pi_{\alpha}(V \cap P) = g^{-1}(V)$ are disjoint open subsets of $X_{\alpha}$ containing $A$ and $B$, respectively. Hence $X_{\alpha}$ is normal.
