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.
 A: Note that $(X,\tau)$ is homeomorphic to a subspace of $(X,\tau)^n$ by mapping $x \mapsto (x,x_0,x_0,\dots,x_0)$ for some fixed $x_0 \in X$ (ignoring the case $X = \emptyset$ which is even easier). Now use the fact that subspaces of Hausdorff spaces are Hausdorff spaces.
A: 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 $y\in V_{j^*}$.
A: I think this is true. For any two points $x\neq y$ of $(X,\tau)$, consider the two points
$$
(x,x_2,\dots,x_n)\neq(y,x_2,\dots,x_n)
$$
which have disjoint neighborhoods, say $U,V$. Judging from your notation, the product topology on $(X,\tau)^n$ is implied. So it is generated by basis of the form $W_1\times\cdots W_n$ where $W_i$ are open sets of $X$. By the definition of basis, exists two basis element $U_1\times\cdots\times U_n$ and $V_1\times\cdots\times V_n$ s.t.
$$
\begin{split}
(x,x_2,\dots,x_n)\in U_1\times\cdots\times U_n\subset U \\
(y,x_2,\dots,x_n)\in V_1\times\cdots\times V_n\subset U \\
\end{split}
$$
Since $U,V$ are disjoint:
$$
(U_1\times\cdots\times U_n)\cap(V_1\times\cdots\times V_n)=(U_1\cap V_1)\times\cdots\times (U_n\cap V_n)=\varnothing
$$
It is then obvious that $U_i\cap V_i\neq\varnothing$ for $i\geq 2$, and $U_1\cap V_1$ has to be empty. This is the required disjoint neighborhoods for $x,y$ in $(X,\tau)$.
