Answer to Question #133628 in Differential Geometry | Topology for PRATHIBHA ROSE C S

Question

Prove that a space X is homeomorphic to an open subspace of a compact Hausdorff space if and only if X is locally compact.

Answer 1

Solution: First we assume that $X$ is a homeomorphic to an open subset of a compact Hausdorff space $C$ . Identifying $X$ with that subset, we may assume that $X \subset C$ and an open subset. Since $C$ is Hausdorff, $X$ is also Hausdorff. Let $x \isin X$. Since $X$ is an open neighbourhood $V$ of $x$ in $C$ such that $x \isin V \subseteq \overline{V} \subset X$. Since $\overline{V}$ is compact and $V$ is open in $X$ (as it is open in $C$ ), we have verified that $X$ is locally compact.

Now, conversely, suppose that $X$ is locally compact Hausdorff. Then $X$ imbeds in the 1-point compactification $X^+$, which is compact Hausdorff. By definition of topology on $X^+$, $X$ is an open subset of $X^+$

Answer: A space X is homeomorphic to an open subspace of a compact Hausdorff space if and only if X is locally compact.