anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
as i recall "normal" is the same as T4 right?
anonymous
  • anonymous
normal + T1 is the same as T4
anonymous
  • anonymous
so disjoint closed sets A, B are contained in disjoint open sets

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anonymous
  • anonymous
Right, that's the normal condition.
anonymous
  • anonymous
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
anonymous
  • anonymous
right, that would be the Hausdorff condition
anonymous
  • anonymous
well unless i miss my guess then you have done what is required
anonymous
  • anonymous
Question off topic: You'r in a topology group. I don't find it... Does it still exist?
anonymous
  • anonymous
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
anonymous
  • anonymous
i am not sure what your second question is asking, but the last time i looked at tolpology i think reagan was president
anonymous
  • anonymous
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.
anonymous
  • anonymous
Haha, I was talking about a study group here on OS. I saw that you earned some medals there.
anonymous
  • anonymous
http://mathoverflow.net/questions/53300/locally-compact-hausdorff-space-that-is-not-normal
anonymous
  • anonymous
i forget everthing. i am fairly sure you are right though. good luck
anonymous
  • anonymous
Thanks!

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