Which of the following spaces is Hausdorff?

- anonymous

Which of the following spaces is Hausdorff?

- chestercat

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- anonymous

- anonymous

The discrete space

- anonymous

The indiscrete space

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## More answers

- anonymous

R
with the finite complement topology

- anonymous

X = [a,b] endowed with the topology \[\tau \] =
(\[ϕ \],X,[a])

- 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

discrete topology

- anonymous

- anonymous

A set is nowhere dense if the set
\[barA \]
has empty

- anonymous

i think interior ????????????????

- anonymous

- zzr0ck3r

Sorry, I was helping someone before you texted me and I did not think it would take this long.

- zzr0ck3r

wait, are you asking another question now?

- anonymous

yes

- anonymous

i think the answer to the first question is the discrete space

- zzr0ck3r

it is not

- zzr0ck3r

why do you think that?

- zzr0ck3r

dont just move on....answer me :)

- anonymous

because discrete topology is one that has every set open

- anonymous

so sorry if i am wrong

- zzr0ck3r

oh sorry, I thought you asked for the non Hausdorff one...you are correct. But let us not move on...

- zzr0ck3r

why does every set being open mean that it is Hausdorff?

- anonymous

ok thank you sir . a big hug

- anonymous

because there intersection will give the empty set

- zzr0ck3r

what intersection?

- zzr0ck3r

who is they?

- anonymous

a space is Hausdorff if for any two elements, there is an open nbhd around each that does not intersect each other

- anonymous

that was what you said

- 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

ok. thank you sir

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