# Prove that every simply ordered set is a Hausdorff space in the order topology. The product of two Hausdorff space is again a Hausdorff space

In the book of General Topology by Munkres, at page 100, it is asked to prove that

Every simply ordered set is a Hausdorff space in the order topology.The product of two Hausdorff space is again a Hausdorff space.

Proof of the first statement:

Let X be a topological space with the order topology and $a,b \in X$. Without loss of generality, assume $a < b$, then since $(-\infty, b)$ is a $a$ neighbourhood and $(a,\infty)$ is a $b$ neighbourhood, and the intersection of these intervals are empty, we have the desired result.

For the second statement:

Let $X ,Y$ be Hausdorff spaces, then consider $(a,b), (c,d) \in X\times Y$, then by our hypothesis, $\exists U_a, V_b, U_b, V_d$ such that $U_a \cap U_c = \emptyset$ and $V_b \cap V_d = \emptyset$. Then $U_a \times V_b \cap U_c \times V_d = \emptyset$, hence $X\times Y$ is Hausdorff.

So, is there any problem, flow in the proof ? or any point that you advise me to clarify?

I mean even though the proofs are not complex, I have started Hausdorff spaces just today, and I want to make sure that that I'm not missing anything.

• even you can take $V_b=V_d=X$. Even You don't need order topology and two or finite product. Look at here math.stackexchange.com/questions/487626/… – S.S.Danyal Nov 21 '17 at 14:13
• @S.S.Danyal You are right, thanks for pointing out. – onurcanbektas Nov 21 '17 at 14:25
• For the first one look at here math.stackexchange.com/questions/526586/…. Please before asking a little bit search your question. – S.S.Danyal Nov 21 '17 at 14:47
• 1 is wrong. 0 < 1, 0 in (-oo,1), 1 in (0,oo) and those two intervals are not disjoint. You yet to consider the case when they are not disjoint. – William Elliot Nov 21 '17 at 22:58
• 2 is wrong. What you wrote is correct but there are no two points that you are separating by open sets. – William Elliot Nov 21 '17 at 23:03