# A subset of $\bar{S}\backslash S$ contains an open ball in $\bar{S}$? (operator theory)

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!

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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

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.
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