$X$ is Hausdorff iff $X×X$ is Hausdorff, where $X$ is a topological space?

Question

One side is true i.e. if $X$ is Hausdorff then $X\times X$ is Hausdorff. But is it true that if $X\times X$ is Hausdorff then $X$ is also Hausdorff?

My proof: since $X\times X$ is Hausdorff, take two points $(x_1,y)$ and $(x_2,y)$ hence there are to basis elements say $U_1\times V_1$ and $U_2\times V_2$ containing them respectively and are disjoint. Hence $U_1$ and $U_2$ are required open sets in $X$ which are disjoint and contain $x_1,x_2$ respectively.

Since $x_1$, $x_2$ are arbitrary hence $X$ is Hausdorff.

Answer 1

Any subspace of a Hausdorff space is Hausdorff, and $X$ can obviously be embedded in $X\times X$, so if $X\times X$ is Hausdorff, so is $X$. Even more general, if there exists a non-empty topological space $Y$ such that $X\times Y$ is Hausdorff, then $X$ is Hausdorff. So we even find for two non-empty topological spaces $X$ and $Y$ that $X\times Y$ is Hausdorff if, and only if, both $X$ and $Y$ are Hausdorff.

Answer 2

Assuming that $X\times X$ is Hausdorff, take $x,x^\ast\in X$. You want to prove that there are open sets $A,A^\ast\subset X$ such that $x\in A$, $x^\ast\in A^\ast$, and $A\cap A^\ast=\emptyset$. Consider $(x,x),(x^\ast,x)\in X\times X$. There are open sets $B,B^\ast\subset X\times X$ such that $(x,x)\in B$, $(x,x^\ast)\in B^\ast$, and $B\cap B^\ast=\emptyset$. You can write $B$ as an union $\bigcup_{\lambda\in\Lambda}B_\lambda\times B_\lambda'$, where each $B_\lambda$ and each $B_\lambda'$ is an open subset of $X$. So, fix $\lambda\in\Lambda$ such that $(x,x)\in B_\lambda\times B_\lambda^\ast\subset B$. By the same argument, $(x,x^\ast)$ belongs to some product $C\times C^\ast$, where $C$ and $C^\ast$ are open subsets of $X$ and $C\times C^\ast\subset B^\ast$. Since $B\cap B^\ast=\emptyset$, $(B_\lambda\times B_\lambda^\ast)\cap(C\times C^\ast)=\emptyset$. But then$$\bigl(B_\lambda\times(B_\lambda^\ast\cap C^\ast)\bigr)\cap\bigl(C\times(B_\lambda^\ast\cap C^\ast)\bigr)=\emptyset$$and, since $B_\lambda^\ast\cap C^\ast\neq\emptyset$, you can deduce from this that $B_\lambda\cap C=\emptyset$.

Answer 3

If $X \times X$ is Hausdorff, and $x \neq x'$ in $X$ we can separate $(x,x)$ and $(x',x)$ by disjoint open sets in $X \times X$, so there are basic open sets $U \times V$ containing $(x,x)$ and $U' \times V'$ containing $(x',x)$ such that $(U \times V) \cap (U' \cap V')= \emptyset$. But then $U$ is an open neighbourhood of $x$ and $U'$ is an open neighbourhood of $x'$ such that $U \cap U' = \emptyset$ (if $p$ would be in the intersection, $(p,x)$ would be in $(U \times V) \cap (U' \cap V')$, contradiction.). So $X$ is Hausdorff.

Alternatively, apply a theorem that $X$ is an open image of $X \times X$ and as such also Hausdorff. Or use that fact that $X$ embeds into $X \times X$ (the diagonal e.g.) and subspaces are Hausdorff.