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.


  • $\begingroup$ 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 $p_2$. $\endgroup$ Feb 9, 2023 at 17:33
  • $\begingroup$ 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 $? $\endgroup$
    – bestmate21
    Feb 9, 2023 at 17:58
  • $\begingroup$ 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. $\endgroup$ Feb 9, 2023 at 18:00
  • $\begingroup$ 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? $\endgroup$
    – bestmate21
    Feb 9, 2023 at 18:18
  • $\begingroup$ Yes, using thath $X$ its Hausdorff. $\endgroup$ Feb 9, 2023 at 18:48


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