If $(X,\tau)^n$ is Hausdorff, is $(X,\tau)$ also Hausdorff? 
If $(X,\tau)^n$ is Hausdorff, is $(X,\tau)$ also Hausdorff?

I know that product of Hausdorff space is Hausdorff, but I want to know if this weaker converse of it is true. Thanks.

Consider $x,y\in X$ such that $x\neq y$. Let $\mathbf x=(x,x,\ldots,x)$ and $\mathbf y=(y,y,\ldots,y)$. Since $\mathbf x\neq \mathbf y$ and $X^n$ is Hausdorff, there exist disjoint subsets $A$ and $B$ of $X^n$ that are open in the product topology, and $\mathbf x\in A$ and $\mathbf y\in B$. By the definition of the product topology, there exist open subsets $\{U_j,V_j\}_{j=1}^n$ of $X$ such that


*

*$\mathbf x\in\prod_{j=1}^n U_j\subseteq A,$

*$\mathbf y\in\prod_{j=1}^n V_j\subseteq B,$

*$\left(\prod_{j=1}^n U_j\right)\bigcap\left(\prod_{j=1}^n V_j\right)=\prod_{j=1}^n(U_j\cap V_j)=\varnothing$ (given that $A\cap B=\varnothing$).


The third line implies that $U_{j^*}\cap V_{j^*}=\varnothing$ for at least one $j^*\in\{1,\ldots,n\}$ (based on the finite version of the axiom of choice). The first two lines and the definitions of $\mathbf x$ and $\mathbf y$ imply that $x\in U_{j^*}$ and $y\in V_{j^*}$.
