# Compact Hausdorff Spaces and their local compactness

Let $Y$ be a compact Hausdorff space. Then for a $X \subset Y$ compact if say $\exists x_0 \in Y -X$ Then I can find disjoint open neighborhoods containing $X$ and $x_0$.

Q1:

Then can I say the following :

Say I'm given a open neighborhood $U \subset Y$ where $Y$ is compact Hausdorff. If $Y- U \neq \emptyset$ then for a given point $x \in U$ I can find a compact neighborhood contained in $U$.

I believe we can not. Since theorem states that $U$ is locally compact Hausdorff if and only if $Y - U$ contains only one element. However check the theorem and the corresponding proof below:

Prove that a space $X$ is homeomorphic to an open subset of a compact Hausdorff space if and only if X is locally compact Hausdorff.

Suppose first that $X$ is homeomorphic to an open subset of a compact Hausdorff space $C$. Identifying $X$ with that subset, we may assume that $X \subset C$ and is an open subset. Since $C$ is Hausdorff, $X$ is also Hausdorff. Let $x \in X$. Since $X$ is an open neighborhood of $x$ in $C$, and $C$ is Hausdorff, there exists an open neighborhood $V$ of $x$ in $C$ such that $x \in V \subset \bar{V} \subset X$ . Since $V$ iscompact,and $V$ is open in $X$ (since It is open in $C$), we have verified that $X$ is locally compact.

I couldn't get how the bold part was derived. Bold simply implies that $X$ is locally compact Hausdorff, and this is if and only if $C - X$ contains only one element. But we are not given that condition. If $C$ was locally compact Hausdorff then I can say that every open subspace of $C$ is locally compact Hausdorff, but hypothesis states only that $C$ is compact Hausdorff. Not necessarily locally compact. Then:

Q2:

Are compact spaces locally compact? Can I take the improper subset as the compact neighborhood?

The thing is I'm following munkres' book on topology and the sign $\subset$ is used both for improper and proper subsets, which leads me to confusion.

• Wait, why do you think a subspace of a compact Hausdorff space is locally compact if and only if its complement has only one point? – Jason Nov 6 '17 at 20:36
• @Jason No, I think if we can find a $Y$ containing $X$ satisfying that – Xenidia Nov 6 '17 at 20:41

Can I take the improper subset as the compact neighborhood?

Yes.

Are compact spaces locally compact?

Yes.

Say I'm given a open neighborhood $U \subset Y$ where $Y$ is compact Hausdorff. If $Y- U \neq \emptyset$ then for a given point $x \in U$ I can find a compact neighborhood contained in $U$.

Yes, even when $Y- U=\emptyset$, by Theorem 3.1.6 from Engelking’s “General topology”.

You are confused by the theorem you are quoting. The theorem states the following:

Let $X$ be a Hausdorff space. Then $X$ is locally compact if and only if there exists a compact Hausdorff space $Y$ with an open subspace $\tilde X$ such that $X$ is homeomorphic to $\tilde X$ and $Y\setminus\tilde X$ consists of a single point.

You'll note that the theorem doesn't claim that any compact space $Z$ containing $X$ as a subspace must be of the form $Z=X\cup\{a\}$ for some $a\notin X$. So in the proof of the theore, there is no reason why $C\setminus X$ should contain only one point. Example: $(0,1)$ is locally compact and contained in $[0,1]$, but $[0,1]\setminus(0,1)=\{0,1\}$ which contains two points.

By the way, the theorem you have been quoting is actually a stronger version of what you are trying to prove...