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Consider the following question:
Let $A$ be a normed space containing a closed subset $B\subseteq A$ and a dense subset $D\subseteq A$. Is $B \cap D$ necessarily a dense subset of $B$?

My conclusion is that it need not hold in general. To see this, take $A=\mathbb{R}$ (with the standard metric), $B=\{\pi\}$ and $D=\mathbb{Q}$. Then $B\cap D = \{\pi\} \cap \mathbb{Q} = \emptyset$, the closure of which is $\emptyset$. Therefore $B\cap D$ is not dense in $B$.

I am particularly interested in understanding what happens if we add the assumption that $A$ is a (unital) C*-algebra, $B$ is a sub-C*-algebra of $A$ (hence closed) and $D$ is a dense $*$-subalgebra of $A$. Is the answer affirmative then? Can anyone come up with a counterexample?

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Take $A = L^{\infty}[0,1]$ with pointwise multiplication and let $B = C[0,1]$ be the closed $C^{\ast}$-subalgebra of continuous functions. The $\ast$-subalgebra $D \subset A$ consisting of the simple functions (finite linear combinations of characteristic functions of measurable sets) is dense in $L^{\infty}[0,1]$ but $D \cap B = \{\text{constant functions}\}$ is as far from dense in $B$ as it gets. Note that all $*$-algebras are commutative and unital here.

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  • $\begingroup$ That's a nice example! $\endgroup$ – Nate Eldredge Oct 17 '11 at 2:23

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