Hi all, here's the problem I'm working on.(adsbygoogle = window.adsbygoogle || []).push({});

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...

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# Homework Help: Hausdorff spaces

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