Question

Which of the following spaces is Hausdorff?

Answer 1

@zzr0ck3r

Answer 2

The discrete space

Answer 3

The indiscrete space

Answer 4

R
with the finite complement topology

Answer 5

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

Answer 6

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?

Answer 7

discrete topology

Answer 8

@zzr0ck3r

Answer 9

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

Answer 10

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

Answer 11

@zzr0ck3r

Answer 12

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

Answer 13

wait, are you asking another question now?

Answer 14

yes

Answer 15

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

Answer 16

it is not

Answer 17

why do you think that?

Answer 18

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

Answer 19

because discrete topology is one that has every set open

Answer 20

so sorry if i am wrong

Answer 21

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

Answer 22

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

Answer 23

ok thank you sir . a big hug

Answer 24

because there intersection will give the empty set

Answer 25

what intersection?

Answer 26

who is they?

Answer 27

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

Answer 28

that was what you said

Answer 29

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 \(\{x_0\}\cap\{x_1\}=\emptyset\).

Answer 30

ok. thank you sir