Let $f: S^1 \to T$ be an embedding into the torus (from the circle). We define $X_f=T \setminus f(S^1)$.
Show that $X_f$ is not compact, but it is locally compact.
Context
In question a) we proved that $S^1$ and $T$ are not homeomorphic, since removing two points from the circle leaves it disconnected into two parts, but after removing two points from the torus we see that it is still connected.
In question b) we proved that $f$ is never surjective. This is because an embedding is always a homeomorphism to $f(S^1)$. If $f$ is surjective we get $f(S^1)=T$ and we see that this would imply that $S^1$ and $T$ are homeomorphic. This is a contradiction to a).
So we know that $f: S^1 \to T$ is a non-surjective embedding, which means $X_f$ is non-empty, how does this help us with compactness? any pointers?
Best Answer
First of all observe the $T$ is Hausdorff and compact (and so locally compact).
Now if $X_{f}$ were to be compact, it would be closed because compact sets of Hausdorff spaces are closed. So we would have that $T\backslash X_{f} = f(S^{1})$ is open as it is the complement of a closed set. Now $f(S^{1})$ is compact because $S^{1}$ is and $f$ is continuous, so $f(S^{1})$ must be closed too. We just proved that $f(S^{1})$ is both open and closed, but this is impossible in a connected space such as $T$. So $X_{f}$ is not compact. Now to prove locally compactness: we already know (by the previous argument) that $f(S^{1})$ is closed, so $X_{f}\subset T$ is open, and an open subspace of a locally compact Hausdorff space ($T$ is one of those) is againg locally compact.
If anything of what I said is unclear, feel free to ask.