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.

E and S are subsets of a metric space. $E$ is a subset of $\bar{S}\backslash S$. Then $\overline{E}\subset(\overline{S}\backslash S^{o})$, but I wonder whether there is some condition that guarantees or forbids $\overline{E}$ to contain an open ball in $\overline{S}$.

For people interested in the background, I am looking at something in operator algebras. $S\subset\mathcal{L}(X)$ is a subspace of operators on a Banach space and I want to use the properties of operators in $S$ to get some result about $\overline{S}$.

However, the argument clashes if the exceptional set $E$ is not nowhere dense. So I wonder whether there is some condition on $S$ or $X$ or whatever under which we can eliminate this possibility.

The question is actually quite open. I think any condition on either the underlying space $X$, or the operators on $S$ would be of great help. Actually even a condition that would imply that $\overline{E}$ contains an open ball would lead to something interesting in the other direction.

Thanks!

share|improve this question
    
Since everything is living in a nice algebra, do you know anything about the algebraic properties of $E$? –  Nate Eldredge Jun 6 '12 at 4:25
add comment

1 Answer 1

up vote 2 down vote accepted
+50

I don't think you can expect any condition like that. For any closed set $F$ with nonempty interior, most often you can write $F=E\cup S$ for disjoint dense sets $E,S$ (for example $F=[0,1]$, $E$ the rationals in the interval, $S$ the irrationals; or $F=C[0,1]$, $S$ the polynomials, $E=F\setminus S$). Then the sets satisfy your conditions but $\overline E=F$ will always have balls.

Note that in the second example mentioned above, $S$ is a subspace as in your problem.

share|improve this answer
    
Well, the kind of $\textit{condition}$ I am thinking about is some restrictions on the space in the operator theoretic sense. Like a subspace generated by commutative elements? Normal operators? Something like that. –  Hui Yu Jun 6 '12 at 1:30
    
My second example can be seen as one generated by a single selfadjoint operator, and it is then an abelian algebra. What you need is for $S$ to be "big" so that $E$ is "small", but this kind of reasoning leads one to impose rather than deduce your condition. –  Martin Argerami Jun 6 '12 at 5:32
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.