One point compactification and closed subspace

Question

Im having some trouble on where to start with the following problem:
Let X be a locally compact Hausdorff space where $X^+=X \cup \{\infty\}$ is its one-point compcatification and $Y \subseteq X$. Show that:

1. Y closed in $X^+$ if and only if Y is compact.
2. If Y is closed in X, then $Y \cup \{\infty\}$ is closed in $X^+$

I know that X being locally compact means that at every $x \in X$ there is a compact subspace of X that contains a neighborhood of x. And X being Hausdorff ($T_2$) means that for every two distinct points $x,y \in X$ there exist a neighborhood $U \subseteq X$ of x og $V \subseteq X$ of y s.t $U \cap V = Ø$.

I Also know the definiton of compactification is as follows:
If Y is compact Hausdorff and $X \subseteq Y$ whole closure equals Y, then Y is Said to be a compactification of C. If Y\X equals a single point, then Y is called one-point compactification of X (denoted by $X^+$.
So I know all the definitons, but if someone could give me a hint on how to start with the statements, it would be greatly appreciated!

Answer 1

1. You can use the following theorem: The subset of a (Hausdorff) compact space is compact iff it's closed.
2. Use the definition of the topology of the subspace. The set $Y$ is closed in $X$ (which is a subspace of $X^+$) iff $Y=\hat Y\cap X$ where $\hat Y$ is closed in $X^+$. There can be only two possibilities:

• $\hat Y=Y$. Then $Y$ is closed in $X^+$ and then $Y\cup\{\infty\}$ also is.
• $\hat Y=Y\cup\{\infty\}$ and we are done.

Answer 2

$X$ is locally compact Hausdorff space $\iff$ $\exists X^+$ such that $X$ is a subspace of $X^+$, $X^+ -X=\infty$, $X^+$ is compact Hausdorff space. For proof see this.

(1) ($\Rightarrow$)$X^+$ is compact. Since closed subset of compact space is compact, $Y$ is compact. ($\Leftarrow$) $X^+$ is Hausdorff. Since compact subspace of Hausdorff space is closed, $Y$ is closed.

(2) By elementary set theory, $X^+ -(Y\cup \{\infty\})=X-Y$. Since $Y$ is closed in $X$, $X-Y \in \mathcal{T}_X \subseteq \mathcal{T}_{X^+}= \mathcal{T}_X \cup \{X^+ -C| C\subseteq X$ and $C$ is compact$\}$. So $X^+ -(Y\cup \{\infty\})\in \mathcal{T}_{X^+}$. Thus $Y\cup \{\infty\}$ is closed in $X^+$.