Does the unit sphere $S$ consist of its boundary only? I want to prove that a normed space is separable iff its unit sphere is separable. I am trying to build up the proof by analyzing unit sphere and its density.

I would like to know if unit sphere $S$ consist of its boundary only?
Because the definition of separable is given as follows:

A normed space $X$ is separable if it contains dense, countable set $A$.

Now density of set is defined as $\bar A = \partial A \cup A$.

I am not sure if the sphere contains only its boundary $ S =\partial S$?

If so, then $\partial S \cup S = S = \bar S.$
Yes, it holds $\partial S = S$.

We have $$\partial S = \overline{S} \cap \overline{X\setminus S} = S \cap \overline{X\setminus S} \subseteq S$$

since $\overline{S} = S$, because $S$ is closed.

For the converse, take $x \in S$ and $r &gt; 0$. Note that $x \in B(x,r) \cap S$, and $$\left(1 + \frac{r}2\right)x \in B(x,r) \cap S^c$$

Therefore, every ball around $x$ intersects both $S$ and $S^c$, hence $x \in \partial S$.

We conclude $\partial S = S$.
Yes, it holds $\partial S = S$.

We have $$\partial S = \overline{S} \cap \overline{X\setminus S} = S \cap \overline{X\setminus S} \subseteq S$$

since $\overline{S} = S$, because $S$ is closed.

For the converse, take $x \in S$ and $r &gt; 0$. Note that $x \in B(x,r) \cap S$, and $$\left(1 + \frac{r}2\right)x \in B(x,r) \cap S^c$$

Therefore, every ball around $x$ intersects both $S$ and $S^c$, hence $x \in \partial S$.

We conclude $\partial S = S$.