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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.

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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 $S_X$, $\{qx\in\Bbb Q\times D:0<q\le 1\}$ is dense in $B_X$, so $d(B_X)\le d(S_X)=d(X)$, and all three densities are equal.

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  • $\begingroup$ 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? $\endgroup$ – carciofo21 Jan 5 at 23:23
  • $\begingroup$ You are right about the weight, I was misremembering, I edited my comment, that observation was inappropriate. $\endgroup$ – carciofo21 Jan 5 at 23:29

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