Proof. Let $\infty$ be a point not in $X$, $X^{*} = X \sqcup \bracs{\infty}$, and $\mathcal{T}^{*} \subset 2^{X^*}$ such that for each $U \in \mathcal{T}^{*}$, one of the following holds:

1. (a)

   $U \in \mathcal{T}$.

2. (b)

   $\infty \in U$ and $U^{c} \subset X$ is compact.

Let $\seqi{U}\subset \mathcal{T}^{*}$ be an open cover of $X$, then there exists $i \in I$ such that $\infty \in U$. In which case, $U_{i}$ must satisfy (b), so there exists $J \subset I$ finite such that $\bigcup_{j \in J}U_{j} \supset U_{i}^{c}$, and $\bracsn{U_j|j \in J \cup \bracs{i}}$ is a finite subcover. Now, let $x \in X$, then since $X$ is locally compact, there exists a precompact neighbourhood $U \in \cn_{X}^{o}(x)$. In which case, $\ol{U}^{c} \in \cn_{X^*}(\infty)$ with $U \cap \ol{U}^{c} = \emptyset$. Therefore $X^{*}$ is a compact Hausdorff space.

Let $\iota: X \to X^{*}$ be the inclusion map. For each $U \in \mathcal{T}^{*}$ satisfying (b), $\iota^{-1}(U) = U \cap X$. Since $U^{c} \subset X$ is compact, $U \cap X$ is open by Proposition 5.16.3, so $\iota \in C(X; X^{*})$. By (a), $\iota$ is an embedding.

Finally, let

Let $U \subset X^{*}$ with $\infty \not\in U$, then $(\varphi^{*})^{-1}(U) = \varphi(U)$ is open in $\varphi(X)$ because $\varphi$ is an embedding, and open in $Y$ by Lemma 5.25.3. On the other hand, for each $V \in \cn_{X^*}^{o}(\infty)$,

Since $\varphi \in C(X; Y)$ is an embedding, $V$ is relatively open in $\varphi(X)$, so $V \cup (Y \setminus \varphi(X))$ is open in $Y$.$\square$