# Hausdorff space - proof (trying to improme my proof writing skills)

I'm working on the following problem from Munkres book on Topology:

"Show that X is Hausdorff if and only if the diagonal $$\Delta$$ = $$\{x \times x | x \in X \}$$ is closed in $$X \times X$$."

My question is: what does $$x_1$$ and $$x_2$$ mean? Are they equivalent to $$p_1$$ and $$p_2$$? I've written it in steps to make it more clear to my self.

1. step

Assume X is Hausdorff. Now prove that $$(X \times X) \setminus \Delta$$ is open.

2. step

Let $$p_1, p_2 \in (X \times X)\setminus \Delta$$ and therefore $$p_1 \neq p_2$$

3. step

Let $$U_1$$ and $$U_2$$ be disjoint open sets containing $$p_1$$ and $$p_2$$, respectively.

4. step

Then $$U_1 \times U_2$$ is an open set in $$X \times X$$ containing $$p_1$$ and $$p_2$$, such that $$U_1 \times U_2 \cap \Delta = \emptyset$$

5. step

Therefore, $$(p_1, p_2) \in (U_1 \times U_2) \subset (X \times X) \setminus \Delta$$. This implies that $$(X \times X)\setminus \Delta$$ is open.

6. step

Assume that $$\Delta$$ is closed (i.e. that $$(X \times X) \setminus \Delta$$ is open)

7. step

Since $$(X \times X) \setminus \Delta$$ is open, there is an open set $$U \times V \in X \times X$$ such that $$(x_1, x_2) \in U \times V \subset (X \times X)\setminus \Delta$$.

8. step

This implies $$x_1 \in U$$ and $$x_2 \in V$$, and $$U$$ and $$V$$ are disjoint, which shows that X is Hausdorff.

$$\square$$

• Be carefull, in step $2$ you take $p_1, p_2 \in X \times X$ that means that $p_1=(x_1,y_1) \in X \times X$ and $p_2=(x_2,y_2) \in X \times X$. In step $3$ you take two disjoint open sets but... you dont know if they exists, actually its what you want to prove. I think in step $3$ you want to take open sets in $X$, but $p_1$ and $p_2$ are in $X \times X$, so the open sets taken in $3$ never will containg $p_1$ and $p_2$. Feb 9, 2023 at 17:33
• With regards to step 2. isn't $p_1 = (x_1, x_1) \in X \times X$ and $p_2 = (x_2, x_2) \in X \times X$ equivalent to saying that $p_1 \neq p_2$? Feb 9, 2023 at 17:58
• Actually $p_1, p_2 \in (X \times X)\setminus \Delta$ means that if we write $p_1=(x_1,y_1)$ then $x_1 \neq y_1$, ande the same for $p_2$. On the other hand, $p_1 \in X \times X$ means $p_1 =(x_1,y_1)$ with $x_1,y_1 \in X$, but could be equals or not. Feb 9, 2023 at 18:00
• Ok, thank you. With regards to step 3: I should prove there is disjoint subsets in U and V in X $\times$ X before I can proceed? Feb 9, 2023 at 18:18
• Yes, using thath $X$ its Hausdorff. Feb 9, 2023 at 18:48