Is this topology problem correct? 
Let $X$ be a topological space and $n \in \mathbb{N}$. Prove that the set $\Delta = \{(a,a,\dots,a) \in X^n \;|\; a \in X\}$ is closed in $X^n$.

This is an exercise from my topology textbook. I think that the question, as stated, is wrong. That's because in the case of $n=2$ we know that $\Delta$ is closed iff $X$ is Hausdorff. Is this true or am I missing something? I know how to prove the result with the assumption that $X$ is Hausdorff. Assuming everything I said is correct does the converse also hold? That is, if $\Delta$ is closed in $X^n$ for some $n \geq 2$ is $X$ necessarily Haussdorf?
 A: 
Is this true or am I missing something?

It is true, and you're not missing something.

Assuming everything I said is correct does the converse also hold? That is, if $\Delta$ is closed in $X^n$ for some $n \ge 2$ is $X$ necessarily Hausdorff?

Suppose $\Delta$ is closed, and fix two points $x_1, x_2 \in X$. Then the point $x = (x_1, x_2, \ldots, x_2)$ lies in $X^n \setminus \Delta$, an open set. There must therefore exist a basic open neighbourhood of $\mathcal{U}_1 \times \mathcal{U}_2 \times \ldots \times \mathcal{U}_n$ of $x$, which gives us open neighbourhoods $\mathcal{U}_1$ and $\mathcal{U}_2 \cap \ldots \cap \mathcal{U}_n$ of $x_1$ and $x_2$ respectively. Note that there cannot be a point $y$ in the intersection $\mathcal{U}_1 \cap \mathcal{U}_2 \cap \ldots \cap \mathcal{U}_n$, because this will generate a point
$$(y, y, \ldots, y) \in (\mathcal{U}_1 \times \mathcal{U}_2 \times \ldots \times \mathcal{U}_n) \cap \Delta,$$
which contradicts our construction of $\mathcal{U}_1 \times \mathcal{U}_2 \times \ldots \times \mathcal{U}_n$. So, $X$ must be Hausdorff.
