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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.

Martin Sleziak
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Our
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    even you can take $V_b=V_d=X$. Even You don't need order topology and two or finite product. Look at here https://math.stackexchange.com/questions/487626/the-product-of-hausdorff-spaces-is-hausdorff – S.S.Danyal Nov 21 '17 at 14:13
  • @S.S.Danyal You are right, thanks for pointing out. – Our Nov 21 '17 at 14:25
  • For the first one look at here https://math.stackexchange.com/questions/526586/order-topology-is-hausdorff. 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
  • @WilliamElliot In the first case, originally, I consider the cases where between $a$ and $b$ there is at least one element and there is not element separately, but then I did not see the need for such a distinction. You are right, I need to consider those cases seperately, thanks for pointing out. – Our Nov 22 '17 at 06:08
  • @WilliamElliot In the second statement, I did not understand what is wrong exactly. I mean I'm taking two arbitrary elements in $X \times Y$, and showing that there is at least one neighbourhoods for each element such that these neighborhoods are disjoint. – Our Nov 22 '17 at 06:10
  • @onurcanbektas. You are correct as written. – William Elliot Nov 22 '17 at 10:24

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