Hausdorffness -quotient space of a hausdorff space

[Tomer]

Homework Statement

Let (X,$\tau$) be a hausdorff space and let ~ be an equivalence relation, so that all equivalence classes are finite. Prove: X/~ is also a hausdorff space.

Homework Equations

Here's the definition of the standard topology of X/~:
Let q:X-> X/~ be the function that sends x to [x] (its equivalence class). U is called "open" in X/~ if q^-1(U) is open in X.

The Attempt at a Solution

Sounds like a simple enough exercise , but unfortunately I can't solve it.
I started out of course with taking two distinct points, [x], [y] in X/~.
We know that [x], [y] are finite.
Now, at this point I'll assume [x] = {x} and [y] = {y}. It's not a necessary simplification, it's just going to save some messy writing... otherwise I'd have to deal with lots of sets and indexes. If I can solve it for this example I should be able to solve the general case (that [x] and [y] contain n and m elements each).
My goal is to find two open distinct sets around [x] and [y] in X/~.
We know that since [x] ≠ [y], x ≠ y. Therefore, we have two distinct open sets U, V in X around x and y respectively.
That's the part where I'm not sure anymore what to do.
I tried taking the obvious parallel sets in X/~ according to this definition:
For a set U in X define: = {[x] : $x \in U$}.
So I would expect that and [V] would be the two desired sets in X/~. Unfortunately, I cannot prove that they are distinct, nor do I think it's necessarily correct.

Does anyone have an idea here?

Thanks a lot!

Tomer.

 

[mighty2000]

Dear Tomer,

Thank you for posting this question on the forum. I am a scientist who specializes in topology and I would be happy to help you with this problem.

First, let me clarify the definition of the standard topology on X/~. It is defined as follows:
A subset U of X/~ is open if and only if q^-1(U) is open in X.

Now, for your proof, you are on the right track. Let [x] and [y] be two distinct points in X/~. We know that [x] and [y] are finite, so let [x] = {x1, x2, ..., xn} and [y] = {y1, y2, ..., ym}. Since X is a Hausdorff space, we can find two open sets U and V in X such that x1 ∈ U and y1 ∈ V, and U and V are disjoint. Now, let's look at the inverse images of U and V under q.

q^-1(U) = {x1, x2, ..., xn} and q^-1(V) = {y1, y2, ..., ym}

Since U and V are disjoint, their inverse images are also disjoint. So we can choose two distinct open sets and [V] in X/~ such that [x] ∈ and [y] ∈ [V], and and [V] are disjoint.

Therefore, we have found two distinct open sets and [V] in X/~ such that [x] ∈ and [y] ∈ [V], and and [V] are disjoint. This proves that X/~ is a Hausdorff space.

I hope this helps! Let me know if you have any further questions.