Question

Does s/o want to check whether a proof I did is free of errors/omissions? Here's the proposition that should be proved: Every space that is both T1 and normal is also Hausdorff.

And here my tentative to prove this: Let {a} and {b} be two arbitrary singleton sets. In a T1-space these sets are closed. As the space is also normal there exist two disjoint open sets U,V with {a} a subset of U and {b} a subset of V. This also means that for arbitrary points a,b there exist disjoint open sets U,V with a in U and b in V which is precisely the Hausdorff condition.

Answer 1

as i recall "normal" is the same as T4 right?

Answer 2

normal + T1 is the same as T4

Answer 3

so disjoint closed sets A, B are contained in disjoint open sets

Answer 4

Right, that's the normal condition.

Answer 5

and you have to show that for every pair x, y that there are disjoint open sets U, V where x is one of them and y in the other

Answer 6

right, that would be the Hausdorff condition

Answer 7

well unless i miss my guess then you have done what is required

Answer 8

Question off topic: You'r in a topology group. I don't find it... Does it still exist?

Answer 9

as i recall what is harder is coming up with something that is T1 but not T2, T2 but not T3 etc
it has been a while but i am sure that T4 implies T3 etc

Answer 10

i am not sure what your second question is asking, but the last time i looked at tolpology i think reagan was president

Answer 11

Yes, I think you're right. I found a quite ingenious construction of a space that was normal but not T1 (and not Hausdorff), therefore not T4. I'll see whether I find it.

Answer 12

Haha, I was talking about a study group here on OS. I saw that you earned some medals there.

Answer 13

http://mathoverflow.net/questions/53300/locally-compact-hausdorff-space-that-is-not-normal

Answer 14

i forget everthing. i am fairly sure you are right though. good luck

Answer 15

Thanks!