How to show that $\infty$ is isolated?

Question

A point $p$ in a topological space $Y$ is isolated if there exists an open set $O$ such that $p \in O$, but $(Y \setminus \{p\}) \cap O=\emptyset$

Suppose that $X$ is compact. Show that $\infty$ is isolated in the one-point extension of $X_\infty$ of $X$. Is $X$ dense in $X_\infty$? Prove that $X_\infty$ is disconnected.

Answer 1

By definition of the Alexandroff extension, every subset $(X \setminus C) \cup \lbrace \infty \rbrace$, where $C$ is compact in $X$, is an open set of $X_{\infty}$. So here, with $C = X$, you get that $\lbrace \infty \rbrace$ is open, so $\infty$ is isolated in $X_{\infty}$.

$X$ is not dense in $X_{\infty}$ because the open set $\lbrace \infty \rbrace$ does not intersect $X$.

$X_{\infty}$ is disconnected because it is the disjoint union of the two closed sets $X$ and $\lbrace \infty \rbrace$.

Answer 2

The definition of the topology on $X_\infty$ tells us that all sets of the form $(X\setminus C) \cup \{\infty\}$ are open in $X_\infty$, where $C$ is compact and closed in $X$, (plus all open subsets of $X$ as well).

If $X$ is compact, we can choose it as $C$ in the previous paragraph, and we get that $\emptyset \cup \{\infty\} = \{\infty\}$ is open in $X_\infty$. This $O=\{\infty\}$ we can take as the $O$ in your definition of isolated point.

$X$ is not dense in $X_\infty$ because $\{\infty\}$ is a non-empty open set of $X_\infty$ that does not intersect $X$ (or we can say too that $\infty \notin \overline{X}$ if you prefer).

As $X$ is open in $X_\infty$ too and also closed (as its complement $\{\infty\}$ is open), the sets $X$ and $\{\infty\}$ form a decomposition of $X_\infty$ and so the latter space is disconnected.