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anonymous
 one year ago
Which of the following spaces is Hausdorff?
anonymous
 one year ago
Which of the following spaces is Hausdorff?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0The indiscrete space

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0R with the finite complement topology

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0X = [a,b] endowed with the topology \[\tau \] = (\[ϕ \],X,[a])

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0ok, a space is Hausdorff if for any two elements, there is an open nbhd around each that does not intersect each other. If every set is open, what do we call this space?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A set is nowhere dense if the set \[barA \] has empty

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think interior ????????????????

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, I was helping someone before you texted me and I did not think it would take this long.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0wait, are you asking another question now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i think the answer to the first question is the discrete space

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0why do you think that?

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0dont just move on....answer me :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because discrete topology is one that has every set open

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so sorry if i am wrong

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0oh sorry, I thought you asked for the non Hausdorff one...you are correct. But let us not move on...

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0why does every set being open mean that it is Hausdorff?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok thank you sir . a big hug

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0because there intersection will give the empty set

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0a space is Hausdorff if for any two elements, there is an open nbhd around each that does not intersect each other

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that was what you said

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0but you did not show me the sets. Here is the proof, and this is what I am looking for. Suppose \(x_0, x_1\in X\). Then \(\{x_0\}\) and \(\{x_1\}\) are both open sets, and \(\{x_0\}\cap\{x_1\}=\emptyset\).
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