2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Jason Nov 6 '17 at 20:36
  • $\begingroup$ @Jason No, I think if we can find a $Y$ containing $X$ satisfying that $\endgroup$ – Xenidia Nov 6 '17 at 20:41
1
$\begingroup$

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”.

enter image description here

$\endgroup$
1
$\begingroup$

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...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.