Hausdorff spaces

1. Nov 18, 2005

TimNguyen

Hi all, here's the problem I'm working on.

Suppose (X,T) is a Hausdorff space and that f:(X,T) -> (Y,T') is a bijection such that f inverse is continuous.
1) Prove that (Y,T') is Hausdorff.
2) Suppose that (Y,T') is a Hausdorff space instead of (X,T). With the rest of the statement as given above, can we then prove that (X,T) is Hausdorff?

For 1), I could just show two distinct points in (X,T) such that there exists open sets U and V where x1 is in U, x2 is in V, where U intersect V = 0. From there, I map those distinct points onto Y, which will give unique f(x1) and f(x2) due to the bijection condition. Hence, I then map the U and V in a similar fashion which will lead to f(U) intersect f(V) = 0. It seems correct but I'm still not sure with this proof.

For 2), I don't think there's any difference in the statement if (Y,T') is Hausdorff instead of (X,T) but constructing a proof for the statement seems difficult...

2. Nov 18, 2005

matt grime

f inverse is continuous thus f sends open sets to open sets, but f is not continuous so the situation is definitely not symmetric in the arguments so there IS a difference between 1 and 2.

your proof of 1 is wrong. Indeed you suppose that the hausdorff condition is true in X to produce U and V. You can't do that.

3. Nov 18, 2005

Palindrom

I believe that 2) is wrong. Hint: Think of the discrete topology on X to be (Y, T') with f=id.

4. Nov 18, 2005

StatusX

To prove 1 you need to show that given any two points in X, there exist open sets U and V each containing one of these points and disjoint from each other. That is, pick two arbitrary points, and then construct the corresponding sets.

5. Nov 18, 2005

Palindrom

You mean in Y.

6. Nov 18, 2005

TimNguyen

Thanks for the help.

Yeah, I pretty much got the proof down but I was confused about the second part. I can't assume f is continuous judging from f inverse being continuous, so there's no way to prove that (X,T) is Hausdorff.

7. Nov 18, 2005

matt grime

You must give an example to demonstrate that it fails (it might n ot fail, there might be another way to prove it.. there isn't, as it is false, and a counter example exists, but just because one proof fails doesn't mean all proofs fail).

If by 'got a proof down' for the first you meant the one you gave then it is wrong, as has been explained.

8. Nov 18, 2005

StatusX

Yea, sorry about that. I'm used to the more reasonable way of doing this, by showing that if there's a continuous function from X to Y, then X is Hausdorff if Y is.

9. Nov 18, 2005

Palindrom

So am I, to be honest. For some reason this exercise seems to be 'backwards'.

Edit: That's as long as the function is 1:1, of course.

Last edited: Nov 18, 2005