Answer to Question #217682 in Differential Geometry | Topology for Prathibha Rose

Question

Let (X,(tua)) be a space and ((X) infinity, (tau) infinity) its one point compactification .prove that ((X)infinity, (tau)infinity) is compact

Answer 1

Let $\{U_i \sub X_\infty\}_{i\in I}$ be an open cover. We need to show that this has a finite subcover.

Since we have an open cover, $\exist i_\infty \in I$ such that $U_{i_\infty}$ is an open neighbourhood of $\infty$. But the only open subsets containing $\infty$ are of the form $(X \setminus C) \cup \{\infty\}$, so $\exist$ a compact closed subset $C$ of $X$ such that $X \setminus C \sub U_{i_\infty}$.

Also, $\{U_i \sub X\}_{i\in I}$ is an open cover of $C$. Since $C$ is compact, $\exist$ a finite subcover $\{U_i \sub X\}_{i\in J}$ where $J \sub I$.

Hence we have that $\{U_i⊂X\}_{i∈J}∪U_{i_∞}$ is a finite subcover of the original cover. Thus our one-point compactification is compact.