# Thread: Showing Hausdorff and Compact spaces are homeomorphic

1. ## Showing Hausdorff and Compact spaces are homeomorphic

Ok, I am totally stuck on this problem, I just am not really sure where to go with it.

Let X and Y be topological spaces and let $\displaystyle f:X \rightarrow Y$ be a one-to-one, onto function with a continuous inverse. Prove that if X is Hausdorff and Y is compact, then f is a homeomorphism.

The only thing I could really get is that having a continuous inverse takes the place of the function needing to be open, so all that is left is showing that the function is continuous. So I have to show for any set V in Y, $\displaystyle f^{-1}(V)$ is in X. I am not sure how compactness and Hausdorff can be used to prove this.

2. Originally Posted by okor
Ok, I am totally stuck on this problem, I just am not really sure where to go with it.

Let X and Y be topological spaces and let $\displaystyle f:X \rightarrow Y$ be a one-to-one, onto function with a continuous inverse. Prove that if X is Hausdorff and Y is compact, then f is a homeomorphism.

The only thing I could really get is that having a continuous inverse takes the place of the function needing to be open, so all that is left is showing that the function is continuous. So I have to show for any set V in Y, $\displaystyle f^{-1}(V)$ is in X. I am not sure how compactness and Hausdorff can be used to prove this.
Recall that $\displaystyle f^{-1}$ is continuous if and only if the preimage of closed sets is closed, said differently if and only if $\displaystyle \left(f^{-1}\right)^{-1}\left(C\right)=f\left(C\right)$ is closed for each closed $\displaystyle C\subseteq X$. Recall that a closed subspace of a compact space is compact and thus $\displaystyle C$ is compact. That said, compactness is invariant under continuous maps so that $\displaystyle f\left(C\right)\subseteq Y$ is compact, but since $\displaystyle Y$ is Hausdorff it follows that $\displaystyle f\left(C\right)$ is closed (since compact subspaces of Hausdorff spaces are closed). The conclusion follows by previous commment.

3. Originally Posted by Drexel28
Recall that $\displaystyle f^{-1}$ is continuous if and only if the preimage of closed sets is closed, said differently if and only if $\displaystyle \left(f^{-1}\right)^{-1}\left(C\right)=f\left(C\right)$ is closed for each closed $\displaystyle C\subseteq X$. Recall that a closed subspace of a compact space is compact and thus $\displaystyle C$ is compact. That said, compactness is invariant under continuous maps so that $\displaystyle f\left(C\right)\subseteq Y$ is compact, but since $\displaystyle Y$ is Hausdorff it follows that $\displaystyle f\left(C\right)$ is closed (since compact subspaces of Hausdorff spaces are closed). The conclusion follows by previous commment.
Thank you a ton for the help. I had a little trouble following at first, but I read through the theorems on compactness in my textbook and it makes perfect sense now.

Here is what I came up with:

Let A be a closed subset of Y. Since Y is compact, A is compact. Then $\displaystyle f^{-1}(A)$ is compact because $\displaystyle f^{-1}$ is continuous. Since X is Hausdorff, $\displaystyle f^{-1}(A)$ is closed, hence f is continuous and therefore a homeomorphism.

4. Originally Posted by okor
Thank you a ton for the help. I had a little trouble following at first, but I read through the theorems on compactness in my textbook and it makes perfect sense now.

Here is what I came up with:

Let A be a closed subset of Y. Since Y is compact, A is compact. Then $\displaystyle f^{-1}(A)$ is compact because $\displaystyle f^{-1}$ is continuous. Since X is Hausdorff, $\displaystyle f^{-1}(A)$ is closed, hence f is continuous and therefore a homeomorphism.
Hmm...I feel as though you've mixed up the functions. You know that since $\displaystyle f$ is continuous and $\displaystyle A$ closed that $\displaystyle f\left(A\right)=\left(f^{-1}\left(A\right)\right)$ is closed since compact subspaces of Hausdorff spaces are closed. Thus, the preimage of closed sets under $\displaystyle f^{-1}$ are closed...so $\displaystyle f^{-1}$ is continous and thus $\displaystyle f$ is a homeomorphism.

5. Originally Posted by Drexel28
Hmm...I feel as though you've mixed up the functions. You know that since $\displaystyle f$ is continuous and $\displaystyle A$ closed that $\displaystyle f\left(A\right)=\left(f^{-1}\left(A\right)\right)$ is closed since compact subspaces of Hausdorff spaces are closed. Thus, the preimage of closed sets under $\displaystyle f^{-1}$ are closed...so $\displaystyle f^{-1}$ is continous and thus $\displaystyle f$ is a homeomorphism.
Err, I am a little confused now.

I know $\displaystyle f^{-1}$ is continuous, I need to show f is continuous. I apologize if I have gotten things mixed up.

6. Originally Posted by okor
Err, I am a little confused now.

I know $\displaystyle f^{-1}$ is continuous, I need to show f is continuous. I apologize if I have gotten things mixed up.
Oh, my apologies. I did misread. The same concept works taking $\displaystyle g=f^{-1}$

Oh, my apologies. I did misread. The same concept works taking $\displaystyle g=f^{-1}$