Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 5 down vote accepted

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.

share|cite|improve this answer
That's a nice example! – Nate Eldredge Oct 17 '11 at 2:23

Your Answer


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.