Product topology and $T_0,T_1,T_2$ topology spaces

Question

If I prove the following

Let $\{(X_\alpha,\tau_\alpha)\}$ be a family of topological spaces. $\displaystyle\Big (\prod_{\alpha\in I},\tau_{\pi}\Big)$ is $T_2\iff\forall\alpha\in I,(X_\alpha,\tau_\alpha)$ is $T_2.$

And as $T_2\implies T_1\implies T_0$, could I say that this

$\displaystyle\Big (\prod_{\alpha\in I},\tau_{\pi}\Big)$ is $T_0\iff\forall\alpha\in I,(X_\alpha,\tau_\alpha)$ is $T_0$ and this $\displaystyle\Big (\prod_{\alpha\in I},\tau_{\pi}\Big)$ is $T_1\iff\forall\alpha\in I,(X_\alpha,\tau_\alpha)$ is $T_1$

also holds (without the need to write the proof)? ?

Answer 1

No, the equivalence of for $T_2$ does not logically imply the same equivalence for $T_1$ or $T_0$.

All of these separation axioms are hereditary (if a space has it, so has every subspace in the subspace topology), and here I showed that for a product $\prod_{\alpha \in I} X_\alpha$ (with all factors $X_\alpha$ non-empty, of course), every $X_\alpha$ is homeomorphic to a subspace of $\prod_{\alpha \in I} X_\alpha$ and so if the product is $T_i$ ($i=0,1,2$) so is every $X_\alpha$, $\alpha \in I$.

So in each equivalence one direction is for free, anyway.

For $T_1$ e.g. if $x,y \in \prod_{\alpha \in I} X_\alpha$ with $x \neq y $, this means there is some fixed $\alpha_0 \in I$ such that $x_{\alpha_0} \neq y_{\alpha_0}$. Then we can find an open set $U_1 \subseteq X_\alpha$, with $x_{\alpha_0} \in U_1, y_{\alpha_0} \notin U_1$ and another open set $U_2 \subseteq X_{\alpha_0}$ with $x_{\alpha_0} \notin U_2, y_{\alpha_0} \in U_2$. Then for the (sub)basic open sets $V_i = \pi_{\alpha_0}^{-1}[U_i]$, $i=1,2$ we have that $x \in V_1, y \notin V_1$ and $x \notin V_2, y \in V_2$, so that $\prod_{\alpha \in I} X_\alpha$ is $T_1$.

The pattern is the same for all these axioms: find a coordinate where distinct points differ, apply the axiom in question in that coordinate, and use the corresponding product open set ($X_\alpha$ in all other coordinates) to fulfill the demand in the product. The above proof for $T_1$ can almost trivially be adapted to $T_2$ or $T_0$ as well. The implications between them are irrelevant: if we know the product of $T_1$ we cannot use any facts about $T_2$-ness. Just see that the structures of the proofs are identical.

Answer 2

Biconditional for $T_{1}$.

Only if part. For $x_{\alpha}\ne y_{\alpha}\in X_{\alpha}$, let $\eta_{\beta}\in X_{\beta}$ be fixed and $x_{\beta}=y_{\beta}=\eta_{\beta}$ for $\beta\ne\alpha$.

Then $(x_{i})\ne(y_{i})\in\displaystyle\prod_{i\in I}X_{i}$. Find open sets $G,H\in\tau_{\pi}$ be such that $(x_{i})\in G$, $(y_{i})\notin G$ and $(x_{i})\notin H$, $(y_{i})\in H$.

Since $(x_{i})$ and $(y_{i})$ differ only on one coordinate, it is not hard to see there are open sets $G_{\alpha},H_{\alpha}\in\tau_{\alpha}$ such that $x_{\alpha}\in G_{\alpha}$ and $y_{\alpha}\notin G_{\alpha}$ and similar in the symmetry way.

If part. For $(x_{i})\ne(y_{i})\in\displaystyle\prod X_{\alpha}$. Say, $x_{\alpha}\ne y_{\alpha}$, then find some $G_{\alpha},H_{\alpha}\in\tau_{\alpha}$ such that $x_{\alpha}\in G_{\alpha}$, $y_{\alpha}\notin G_{\alpha}$. Then the open sets \begin{align*} G&=G_{\alpha}\times\prod_{i\ne\alpha}X_{i},\\ H&=H_{\alpha}\times\prod_{i\ne\alpha}X_{i}, \end{align*} are such that $(x_{i})\in G$, $(y_{i})\notin G$ and $(x_{i})\notin H$, $(y_{i})\in H$.