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.

In a metric space let $A$ be compact and $B$ - open, dense everywhere. Does it mean that $$ \overline{A} = \overline{A\cap B} $$ or there are counterexamples?

share|improve this question
1  
Note that compact sets in Hausdorff spaces are closed, so the left-hand side of your equation can be replaced by A. –  Dan Ramras Jun 30 '11 at 16:18
    
Metric spaces are Hausdorff, so compact sets are closed. Therefore, $\overline{A}=A$. I know what "dense" and what "nowhere dense" mean. What does "dense everywhere" mean? –  Arturo Magidin Jun 30 '11 at 16:19
    
I think you might need the interior of $A$ to be non-empty. –  Joel Cohen Jun 30 '11 at 16:24
    
Even with non-empty interior the statement is false - take $A = [0,1] \cup \{2\}$ and $B = \mathbb{R} - \{2\}$. You could ask that $A$ is equal to the closure of its interior, but then the statement is trivial. –  MartianInvader Jun 30 '11 at 16:47
    
Thanks for the post; it reminds me to not jump to the conclusion that $ \overline{A\cap B}$=$\overline{A}\cap\overline{B}$, and examples like A$=(0,1)$ and B=$(1,2)$ –  gary Jun 30 '11 at 17:48
add comment

1 Answer

up vote 13 down vote accepted

What about $A=\{0\}$ and $B=\mathbb R\setminus\{0\}$?

I assume you use everywhere dense as a synonym to dense - like here.

Perhaps this result is worth mentioning in this context: If $A$ is dense in $X$, then for every open $U\subset X$ we have $\overline U=\overline{U\cap A}$.

share|improve this answer
add comment

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.