Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

If $X$ is a locally compact( every point has a compact nbd) $T_2$ space then $X^*$ is a compact $T_2$ space such that, if X is not compact, then $X$ is dense in $X^*$.

Earlier in the theorem is established that $X^*/ X$ is one point and $X^*$ is a compact $T_2$ space.

We will assume that X is not compact, thus there is a cover that it does not have a finite subcover, and will have to show that $X$ is dense in $X^*$. Let U be an open subset of $X^*$, I need to show that $U\cap X\neq \emptyset$. I need help starting the proof. Thank you for your patience. The text defines $X^*=X\cup \{ \infty \}$, my teacher said just another point out of $X$

share|improve this question
3  
What is your question? –  Chris Eagle Jan 29 '13 at 18:35
    
You should say what $X^*$ is. Presumably it is the one point compactification of $X$. –  Miha Habič Jan 29 '13 at 18:38
    
@Miha Yes it is the one point compactification of X –  Klara Jan 29 '13 at 18:39

3 Answers 3

up vote 3 down vote accepted

The only non-empty set that could miss $X$ would be $\{p\} = X^\ast \setminus X$. But this set is not open (the only open sets that contain $p$ must have a compact complement in $X$ and $X$ is not compact itself). So all non-empty open sets intersect $X$.

share|improve this answer

HINT: Let $p$ be the one point of $X^*$ that is not in $X$. The only subset of $X^*$ that does not meet $X$ is $\{p\}$. If $\{p\}$ were open in $X^*$, $X^*\setminus\{p\}=X$ would be closed in $X^*$. What do you know about a closed subset of a compact space?

share|improve this answer
    
X would be compact, as a closed subspace of a compact set, which is not true. –  Klara Jan 29 '13 at 18:52
    
@Klara: Exactly. And you’ve excluded that, so $\{p\}$ can’t be open. –  Brian M. Scott Jan 29 '13 at 18:53
    
@ Brian and then? –  Klara Jan 29 '13 at 18:55
    
@Klara: That means that if $U$ is a non-empty open subset of $X^*$, $U\ne\{p\}$, and therefore $U\cap X\ne\varnothing$. Remember, $X^*\setminus X=\{p\}$. –  Brian M. Scott Jan 29 '13 at 18:56
    
@ Brian, I now this should be easy to see, but where does this definition come from?.. –  Klara Jan 29 '13 at 18:58

Let $p$ be such that $X^*=X \cup \{p\}$ (so $p \notin X$ since $X$ is not compact). By definition of $X^*$, any neighborhood $U$ of $p$ contains $X \backslash K$ for some compact $K$ in $X$; therefore, $U \cap X \neq \emptyset$ since $X$ is not compact. You deduce that $X$ is dense in $X^*$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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