I have been scratching my head trying to figure out my mistake for this problem.
The problem statement is the following: "Let $U$ be an open subset of a compact Hausdorff space $X$ and $U^*$ its one-point compactification. If $\phi: X \rightarrow U^*$ is defined by $\phi(x) = x$ if $x \in U$ and $\phi(x) = \infty$ if $x \in U^c$, prove that $\phi$ is continuous.
So considering all the possible open sets in $U^*$, let $V$ be open in $U^*$ and does not contain $\infty$, then $V$ is an open set in $U$m then $f^{-1}(V) = V$ and is open in $U$ and since $U$ is open in $X$, $V$ is open in $X$.
If $V$ contains $\infty$, then $V^c$ is a compact subset of $U$, and so compact in $X$, since $X$ is hausdorff $V^c$ is closed in $X$. now since $f^{-1}(V) = (f^{-1}(V^c))^c$ = $X \setminus V^c$, which is an open. Thus $\phi$ is continuous.
I am strongly doubting my solution, I think I make some mistakes in the argument for the second case but I can't figure out what. But no where in the proof did I use the fact that $X$ is COMPACT. If someone can help point out my mistake that would be great.
thank you!!
You need not use compactness of $X$ in your proof. However, it is an important requirement. The spaces having a one-point compactification are precisely the locally compact Hausdorff spaces. So we need to know that $U$ is locally compact. This is true because it is an open subset of a compact space.
Therefore, your doubts are unnecessary: Your proof is correct.
In fact, you have proved a more general theorem: If $U$ is an open locally compact subspace of a Hausdorff space $X$, then $\phi$ is continuous.