Prove that if $X$ is Hausdorff, $\Delta$ is closed in $X\times X$

Question

Prove that if $X$ is Hausdorff, $\Delta=\{(x, x)\mid x\in X\}$ is closed in $X\times X$ (with the product topology).

My attempt:

Let $x_1, x_2\in X$ s.t. $x_1\ne x_2$.

There exist neighborhoods $U_1$ and $U_2$ of $x_1$ and $x_2$ that are disjoint.

$U_1\times U_2$ is a basis element in the product topology on $X\times X$. So, $U_1\times U_2$ is open in $X\times X$.

Let $x\in X$.

$(x, x)\in U_1\times U_2\implies x\in U_1$ and $x\in U_2\implies x\in U_1\cap U_2$, which contradicts the fact that $U_1$ and $U_2$ are disjoint.

So, $(x, x)\notin U_1\times U_2$.

I feel that I'm on the right track but don't know how to proceed. Could someone please help me out?

Answer 1

Let $(x, y)\in X\times X-\Delta$

$\implies(x, y)\in X\times X$ and $(x, y)\notin\Delta$

$\implies x, y\in X\text{ and }x\ne y$

There exist neighborhoods $U_x$ and $U_y$ of $x$ and $y$ respectively that are disjoint.

$U_x\times U_y$ is a basis element in the product topology on $X\times X$.

$(x, y)\in U_x\times U_y$ --------------------------------------------- (1)

Let $(u, v)\in U_x\times U_y$. --------------------------------------------- (2)

$\implies u\in U_x$ and $v\in U_y$

Since $U_x\subset X$ and $U_y\subset X$,

$u, v\in X$

$\implies(u, v)\in X\times X$ --------------------------------------------- (3)

$u=v\implies u=v\in U_x\cap U_y$, a contradiction as $U_x$ and $U_y$ are disjoint.

So, $u\ne v$.

$\implies(u, v)\notin\Delta$ --------------------------------------------- (4)

From (3) and (4),

$(u, v)\in X\times X-\Delta$ --------------------------------------------- (5)

From (2) and (5),

$U_x\times U_y\subset X\times X-\Delta$ --------------------------------------------- (6)

Combining (1) and (6),

$(x, y)\in U_x\times U_y\subset X\times X-\Delta$

So, $X\times X-\Delta$ is open in $X\times X$.

So, $\Delta$ is closed in $X\times X$.

QED

Answer 2

Your work shows that $$ \Delta^c=\bigcup_{\substack{(x_1,x_2)\in X\times X\\x_1\neq x_2}} U_1(x_1)\times U_2(x_2), $$ where $U_1(x_1)$ and $U_2(x_2)$ are separating sets for $x_1,x_2$. This shows the complement of $\Delta$ is a union of open sets, so the complement of $\Delta$ is open, so $\Delta$ is closed.

Answer 3

You're basically there, you just need to interpret your result. You found that for any point $(x_1,x_2)\in X\times X-\Delta$, there exists a neighborhood of $(x_1,x_2)$ contained in $X\times X-\Delta$. That is, $X\times X-\Delta$ is open. Therefore $\Delta$ is...