Let $f:X\to Y$ be a continuous bijection.
If $Y$ is Hausdorff, then can we conclude that $X$ is also Hausdorff?
What I learned so far is that:
- If $X$ is compact, then $f(X)$ is also compact;
- If, $X$ is compact and $Y$ is Hausdorff, then $f^{-1}$ is continuous.
I've no idea how to approach the problem. Is there any counterexample for the statement above?