Lemma 5.25.3.label Let $X$ be a LCH space and $(Y, \varphi)$ be a compactification of $X$, then $\varphi(X) \subset Y$ is open.
Proof. For each $x \in X$, let $U \in \cn_{X}(x)$ be a compact neighbourhood. Since $Y$ is a compact Hausdorff space, $\varphi(U)$ is closed by Proposition 5.16.3. As $\varphi \in C(X; Y)$ is an embedding, there exists $V \in \cn_{Y}(\varphi(x))$ such that $\varphi(U) = \varphi(X) \cap V$. Given that $\varphi(X)$ is dense in $Y$, $\varphi(U) = \ol{\varphi(X) \cap V}\supset V$. Therefore $\varphi(U) \in \cn_{Y}(\varphi(x))$, and $\varphi(X)$ is open in $Y$.$\square$
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