$X$ in its one point compactification

Question

Suppose $X$ is a non-compact space, denote its one point compactification by $X^*$.

Since $X$ is open in $X$, thus $X$ is open in $X^*$, can I say $X$ is the interior of $X^*$? Moreover since $X$ is closed in $X$, is $X$ closed in $X^*$?

Answer 1

$X$ is non-compact iff $X$ is dense in $X^\ast$ (so the closure of $X$ is $X^\ast$) and so $X$ is not closed in $X^\ast$ in your case; if $X$ were closed in the compact space $X^\ast$ it would be compact itself.

$X$ is open in $X^\ast$, always.

Of course the interior of $X^\ast$ is $X^\ast$ (this is the largest open subset of $X^\ast$, trivial in any space).

Both statements follow directly from the definition of $X^\ast = X \cup \{\infty\}$, where $\infty \notin X$ and the topology on $X^\ast$ being given by $$\{A \subseteq X: A \text{ open in } X\} \cup \{ (X\setminus K) \cup \{\infty\}: K \subseteq X \text{ closed and compact}\}$$

Answer 2

No. Consider the one-point compactification of $\Bbb R^2$, the Riemann sphere.

And in general, if $X$ were clopen in $X^*$, then $X^*$ wouldn't be connected. This certainly isn't always true (I will have to think about whether it's ever true. Apparently you can compactify some discrete spaces, for instance.)

$X$ wouldn't be closed in $X^*$, because the "point at infinity" is a limit point of $X$ (not in $X$).