I have been tasked with proving the following:
Let $X$ be a compact, Hausdorff space, and let $U$ be a proper opens subset of $X$. Prove that $$ U^{\infty} \cong X / \left( X - U \right)$$
Note that $U^{\infty} = U \cup \left\lbrace \infty \right\rbrace$ and $X / \left(X - U \right)$ denotes $X/\sim$ where $\sim$ is the equivalence relation on $X$ given by $$x \sim x' \text{ if and only if } x= x' \text{ or } x,x' \in X - U.$$
My attempt so far:
Since $U$ is open in $X$, $X - U$ is a closed subset of $X$, and thus $X / \left(X - U\right)$ is a compact, Hausdorff space. Also, by a previous theorem, $U^{\infty}$ is compact. Now, consider the mapping $h : U^{\infty} \to X/(X-U)$ given by $h(u) = p(u)$ for each $u \in U$ and $h(\infty)= p(X- U)$ where $p: X \to X/(X-U)$ is the natural projection. If $u, v \in U$ and $h(u) = h(v)$, then $u = v$, so $f$ one-to-one. Next, $h(X) = p(U) \cup p(X-U) = X/(X- U)$ since $p$ is onto, so $h$ is onto, and hence $h$ is a homeomorphism.
I am pretty sure my logic is sound, although I do not think my proof is neither thorough nor concise enough. So, I welcome all suggestions.
You still have to argue that $h$ is continuous. You just showed it is a bijection.
It's clearly continuous at all $u \in U$ as $U$ is open and $p$ is continuous. So show continuity at $\infty$.