(a) A subspace of a Hausdorff space is Hausdorff; a product of Hausdorff spaces is Hausdorff.
(b) A subspace of a regular space is regular; a product of regular spaces is regular.
My attempt:
(a) It is generalizaction of Exercise 11, Section 17 of Munkres’ Topology. Precisely, theorem 19.4 of Munkres’ Topology.
(b) Let $Y\subseteq X$. Let $\{y\}$ be a singleton set in $Y$. Since $X$ is $T_1$, $\{y\}$ is closed in $X$. By theorem 17.2, $\{y\}= Y\cap \{y\}$ is closed in $Y$. Hence $Y$ is $T_1$. Let $x\in Y$ and $B$ is closed in $Y$ such that $x\notin B$ (i.e. $\{x\} \cap B =\emptyset$). By theorem 17.2, $B=C\cap Y$; $C$ is closed in $X$. If $x\in C$, then $x\in Y\cap C=B$. So $x\notin C$. Since $X$ is $T_3$, $\exists U,V\in \mathcal{T}_X$ such that $x\in U$, $C\subseteq V$ and $U\cap V=\emptyset$. So $\exists U\cap Y, V\cap Y \in \mathcal{T}_Y$ such that $x\in U\cap Y$, $B=C\cap Y\subseteq V\cap Y$ and $(U\cap Y) \cap (V\cap Y)=\emptyset$. Hence $Y$ is $T_3$. Is this proof better version of Munkres proof? Munkres used theorem 17.4, $(\overline{B})_Y =\overline{B}\cap Y$. Which is special case of theorem 17.2.
Let $X=\prod_{\alpha \in J} X_\alpha$ and $X_\alpha$ is regular, $\forall \alpha \in J$. Let $\{p\}$ be a singleton set in $\prod X_\alpha$. Then $p_\alpha \in X_\alpha$, $\forall \alpha \in J$. Since $X_\alpha$ is $T_1$, we have $\overline{\{p_\alpha\}}=\{p_\alpha \}$, $\forall \alpha \in J$. By theorem 19.5, $\overline{\{p\}}= \overline{\prod_{\alpha \in J} \{p_\alpha\}}= \prod_{\alpha \in J} \overline{\{p_\alpha\}}= \prod_{\alpha \in J}\{p_\alpha \}=\{p\}$. Hence $\prod X_\alpha$ is $T_1$. Let $x=(x_\alpha)\in \prod X_\alpha$ and $B$ is closed in $\prod X_\alpha$ such that $x\notin B$. So $x\in (\prod X_\alpha) -B \in \mathcal{T}_X$. $\exists \prod U_\alpha \in \mathcal{B}_X$ such that $x\in \prod U_\alpha \subseteq (\prod X_\alpha)-B$. So $B\subseteq (\prod X_\alpha) - (\prod U_\alpha) =\bigcup_{\alpha \in J} \pi_{\alpha}^{-1}(X_\alpha \setminus U_\alpha)$, by Theorem 19.5 of Munkres’ Topology. $x=(x_\alpha) \in \prod U_\alpha$. Then $x_\alpha \in U_\alpha \in \mathcal{T}_{X_\alpha}$, $\forall \alpha \in J$. So $x_\alpha \notin X_\alpha -U_\alpha$ and $X_\alpha -U_\alpha$ is closed in $X_\alpha$, $\forall \alpha \in J$. Since $X_\alpha$ is $T_3$, $\exists V_\alpha , W_\alpha \in \mathcal{T}_{X_\alpha}$ such that $x_\alpha \in V_\alpha$, $X_\alpha -U_\alpha \subseteq W_\alpha$ and $V_\alpha \cap W_\alpha =\emptyset$. Since $X_\alpha -U_\alpha \subseteq W_\alpha$, we have $\pi_{\alpha}^{-1}(X_\alpha -U_\alpha) \subseteq \pi_{\alpha}^{-1}(W_\alpha)$, $\forall \alpha \in J$. So $B \subseteq \bigcup_{\alpha \in J} \pi_{\alpha}^{-1}(X_\alpha \setminus U_\alpha)\subseteq \bigcup_{\alpha \in J}\pi_{\alpha}^{-1}(W_\alpha)$. Let $V=\prod_{\alpha \in J} V_\alpha$ and $W= \bigcup_{\alpha \in J} \pi_{\alpha}^{-1}(W_\alpha)$. It’s easy to check $V,W \in \mathcal{T}_X$. We show $V\cap W = \emptyset$. Assume towards contradiction, $\exists z\in V\cap W$. $z\in \pi_{\beta}^{-1} (W_\beta)$, for some $\beta \in J$. So $\pi_{\beta} (z)=z_\beta \in W_\beta$. $z_\beta \in V_\beta \cap W_\beta \neq \emptyset$. Thus we reach contradiction. Hence $V\cap W=\emptyset$. Thus $\exists V,W\in\mathcal{T}_X$ such that $x\in V$, $B\subseteq W$ and $V\cap W=\emptyset$. Hence $\prod X_\alpha$ is $T_3$. Is this proof correct? Munkres used lemma 31.1 of Munkres’ Topology.
