anonymous
  • anonymous
Which of the following spaces is Hausdorff?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@zzr0ck3r
anonymous
  • anonymous
The discrete space
anonymous
  • anonymous
The indiscrete space

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More answers

anonymous
  • anonymous
R with the finite complement topology
anonymous
  • anonymous
X = [a,b] endowed with the topology \[\tau \] = (\[ϕ \],X,[a])
zzr0ck3r
  • zzr0ck3r
ok, 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
  • anonymous
discrete topology
anonymous
  • anonymous
@zzr0ck3r
anonymous
  • anonymous
A set is nowhere dense if the set \[barA \] has empty
anonymous
  • anonymous
i think interior ????????????????
anonymous
  • anonymous
@zzr0ck3r
zzr0ck3r
  • zzr0ck3r
Sorry, I was helping someone before you texted me and I did not think it would take this long.
zzr0ck3r
  • zzr0ck3r
wait, are you asking another question now?
anonymous
  • anonymous
yes
anonymous
  • anonymous
i think the answer to the first question is the discrete space
zzr0ck3r
  • zzr0ck3r
it is not
zzr0ck3r
  • zzr0ck3r
why do you think that?
zzr0ck3r
  • zzr0ck3r
dont just move on....answer me :)
anonymous
  • anonymous
because discrete topology is one that has every set open
anonymous
  • anonymous
so sorry if i am wrong
zzr0ck3r
  • zzr0ck3r
oh sorry, I thought you asked for the non Hausdorff one...you are correct. But let us not move on...
zzr0ck3r
  • zzr0ck3r
why does every set being open mean that it is Hausdorff?
anonymous
  • anonymous
ok thank you sir . a big hug
anonymous
  • anonymous
because there intersection will give the empty set
zzr0ck3r
  • zzr0ck3r
what intersection?
zzr0ck3r
  • zzr0ck3r
who is they?
anonymous
  • anonymous
a space is Hausdorff if for any two elements, there is an open nbhd around each that does not intersect each other
anonymous
  • anonymous
that was what you said
zzr0ck3r
  • zzr0ck3r
but 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 \(\{x_0\}\cap\{x_1\}=\emptyset\).
anonymous
  • anonymous
ok. thank you sir

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