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[SOLVED] locally compact hausdorff space. any open subset is locally compact.

caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
Suppose $X$ is locally compact and Hausdorff. Given $x \in X$ and an open set $U$ containing $x$. Find a compact set $C \subseteq U$ which contains an open set $V$ such that $x \in V$.

I tried to construct an open subset $K$ of $U$ such that $\overline{K} \subseteq U$ and $x \in K$. Since if this happens then $C=\overline{K}$ and $V=K$ do the job. But I have not been able to. I found a proof over the internet using "one point compactification" but this concept is not discussed in my book..so there must be another way. Please help.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Suppose $X$ is locally compact and Hausdorff. Given $x \in X$ and an open set $U$ containing $x$. Find a compact set $C \subseteq U$ which contains an open set $V$ such that $x \in V$.

I tried to construct an open subset $K$ of $U$ such that $\overline{K} \subseteq U$ and $x \in K$. Since if this happens then $C=\overline{K}$ and $V=K$ do the job. But I have not been able to. I found a proof over the internet using "one point compactification" but this concept is not discussed in my book..so there must be another way. Please help.
There are several inequivalent definitions of local compactness. I'll work with the weakest one, which is that a locally compact space is one where each point has a compact neighbourhood.

So suppose that $K$ is a compact neighbourhood of $x$. If $K\subseteq U$ then we are done. So suppose that is not the case. The idea is to trim off the parts of $K$ outside $U$.

For each point $y$ in $K\setminus U$ there exist disjoint open sets $V_y,\ W_y$ with $y\in V_y$ and $x\in W_y$. The sets $V_y$ together with $U$ form an open cover of $K$. By compactness there is a finite subcover $\{U, V_{y_1},\ldots,V_{y_n}\}.$ Then $C = K\setminus\bigcup_{j=1}^{\,n} V_{y_j}$ is a closed subset of $K$, hence compact. It is contained in $U$ (because the sets $V_{y_j}$ cover all of $K$ except the part in $U$). Finally, $C$ contains $K\cap\bigcap_{j=1}^n W_{y_j}$, which is a neighbourhood of $x$.
 
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caffeinemachine

Well-known member
MHB Math Scholar
Mar 10, 2012
834
There are several inequivalent definitions of local compactness. I'll work with the weakest one, which is that a locally compact space is one where each point has a compact neighbourhood.

So suppose that $K$ is a compact neighbourhood of $x$. If $K\subseteq U$ then we are done. So suppose that is not the case. The idea is to trim off the parts of $K$ outside $U$.

For each point $y$ in $K\setminus U$ there exist disjoint open sets $V_y,\ W_y$ with $y\in V_y$ and $x\in W_y$. The sets $V_y$ together with $U$ form an open cover of $K$. By compactness there is a finite subcover $\{U, V_{y_1},\ldots,V_{y_n}\}.$ Then $C = K\setminus\bigcup_{j=1}^{\,n} V_j$ is a closed subset of $K$, hence compact. It is contained in $U$ (because the sets $V_{y_j}$ cover all of $K$ except the part in $U$). Finally, $C$ contains $K\cap\bigcap_{j=1}^n W_j$, which is an open neighbourhood of $x$.
Thank you Opalg for your reply.
I take it by $\bigcup_{j=1}^{\,n}V_j$ you mean $\bigcup_{j=1}^{\,n}V_{y_j}$. Similarly for the same ting with $W_j$.
I have assumed "$K$ is a neighborhood of $x$" to mean "$K$ is a subset of $X$ containing an open set which contains $x$". (different books take different definitions of neignborhood).
I don't see how to show that $K\cap\bigcap_{j=1}^n W_j$ is an open set.
If here we use in place of $K$ the open set, say $O$, which $K$ contains ( $x \in O$), wouldn't that be correct too?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I take it by $\bigcup_{j=1}^{\,n}V_j$ you mean $\bigcup_{j=1}^{\,n}V_{y_j}$. Similarly for the same ting with $W_j$.
Correct. I have edited the post accordingly.

I have assumed "$K$ is a neighborhood of $x$" to mean "$K$ is a subset of $X$ containing an open set which contains $x$". (different books take different definitions of neignborhood).
Also correct. I find that concept of neighbourhood very useful in dealing with problems like this one.

I don't see how to show that $K\cap\bigcap_{j=1}^n W_j$ is an open set.
If here we use in place of $K$ the open set, say $O$, which $K$ contains ( $x \in O$), wouldn't that be correct too?
Correct yet again. I should have said that $K\cap\bigcap_{j=1}^n W_j$ is a neighbourhood of $x$ rather than an open neighbourhood (and I have edited the post to correct that too).