$A_1, A_2 \subset X$ with disjoint closures in $X$, $cl_X(A_1)$ compact $\Rightarrow A_1, A_2$ have disjoint closures in $X_{\infty}$?

Question

Let $X$ be a locally compact Hausdorff space and $A_1,A_2 \subset X$. Suppose that $\overline{A_1}^{X} \cap \overline{A_2}^{X}=\emptyset$ and $\overline{A_1}^{X}$ is compact. Let $X_\infty$ be the one-point compactification of $X$. Does this imply that $\overline{A_1}^{X_\infty} \cap \overline{A_2}^{X_\infty}=\emptyset$?

The only thing I could think was that $\overline{A_1}^{X}$ is closed in $X_\infty$. I was trying to work with the closures in $X$ and $X_\infty$ to see if I could get a relationship but unfortunately I couldn't find anything.

Answer 1

The open sets in $X_\infty = X \cup \{ \infty\}$ are

• the open sets in $X$ and
• sets of the form $X_\infty \setminus A$ where $A \subseteq X$ is closed and compact in $X$.

By taking the complements, one sees that the closed sets in $X_\infty$ are

• sets of the form $A \cup \{\infty\}$ where $A \subseteq X$ is closed in $X$ and
• the closed and compact sets from $X$.

As you noted, $\overline{A_1}^X$ is closed in $X_\infty$. Therefore, $\overline{A_1}^{X_\infty} \subseteq \overline{A_1}^X$.

On the other hand, $\overline{A_2}^X \cup \{\infty\}$ is a closed set in $X_\infty$. This follows (by taking the complements) from the fact that $X \setminus \overline{A_2}^X$ is open in $X$ and thus open in $X_\infty$ as well. Hence, $\overline{A_2}^{X_\infty} \subseteq \overline{A_2}^X \cup \{ \infty\}$.

Putting things together we obtain $$ \overline{A_1}^{X_\infty} \cap \overline{A_2}^{X_\infty} \subseteq \overline{A_1}^X \cap (\overline{A_2}^X \cup \{ \infty\}) = \emptyset. $$