# Density of a normed space equals density of unit ball (or of the unit sphere)

Let X be a normed space, define $$\mathrm{dens} \,X$$, the density character of $$X$$, to be the smallest cardinality of a dense subset of $$X$$. (E.g.: X separable iff $$\mathrm{dens}\, X= \lvert \mathbf{N} \rvert$$). How can I prove that $$\mathrm{dens}\, X=\mathrm{dens}\, B_X=\mathrm{dens}\, S_X,$$ where $$B_X$$ denote the (closed, but I think this is not relevant) unit ball of $$X$$ and $$S_X$$ the unit sphere of $$X$$? I know how to do this in the case in which one of this spaces is known to be separable but I can't extend the proof to this general case.

## 1 Answer

For the record, the usual notation, at least in topology, is $$d(X)$$.

Suppose that $$D$$ is dense in $$S_X$$; Then $$\Bbb Q^+D=\{qx\in\Bbb Q\times D:q>0\}$$ is dense in $$X$$, and since $$D$$ is infinite, $$|\Bbb Q^+D|=\aleph_0\cdot|D|=|D|$$. Thus, $$d(X)\le d(S_X)$$.

If $$D$$ is dense in $$X$$, $$\left\{\frac1{\|x\|}x:x\in D\setminus\{0\}\right\}$$ is dense in $$S_X$$, so $$d(S_X)\le D(X)$$, and hence $$d(S_X)=d(X)$$.

If $$D$$ is dense in $$X$$, $$D\cap B_X$$ is dense in $$B_X$$, so $$d(B_X)\le d(X)$$. Finally, if $$D$$ is dense in $$S_X$$, $$\{qx\in\Bbb Q\times D:0 is dense in $$B_X$$, so $$d(B_X)\le d(S_X)=d(X)$$, and all three densities are equal.

• Maybe I'm saying something stupid, but I only can see that this proves $d(X)\le d(S_X)$, what about the other inequalities? – carciofo21 Jan 5 at 23:23
• You are right about the weight, I was misremembering, I edited my comment, that observation was inappropriate. – carciofo21 Jan 5 at 23:29