# Unit ball in a separable Banach Space

Suppose that $X$ is a separable Banach space. Then there exists a sequence $\{ x_n: n \in \mathbb{N} \}$ such that the sequence is dense in a unit ball $B_X$.

Question: Whenever we talk about separable Banach space, I notice that we always have a sequence dense in a unit ball. Why we talk about unit ball instead of the whole space $X$? Is it because all $x \in X$ can be normalised?

Two reasons: if $X$ is a metric space (as a Banach space is) and $X$ is separable (i.e. has a countable dense subset), then every subset of $X$ also has a countable dense subset. This holds because having a countable dense subset and having a countable base (for the topology) are equivalent in metric spaces. So $X$ separable implies $B_X$ separable by being metric alone.

But because of the Banach space structure, the reverse also holds: if $B_X$ has a countable dense subset $\{x_n: n \in \mathbb{N}\}$, then the set $\{m\cdot x_n: m,n \in \mathbb{N} \}$ is dense in $X$:

Let $x \in X$, $r>0$. Pick $m \in \mathbb{N}$ with $m > ||x||$. Then $\frac{1}{m} \cdot x \in B_X$, so we pick $x_n$ within distance $\frac{r}{m}$ of this point, by denseness. Then $||m \cdot x_n - x || = m||x_n - \frac{1}{m} \cdot x|| < m \frac{r}{m} = r$ as required.

And so $X$ also has a countable dense subset. And because all balls are homeomorphic, the unit ball isn't even special in that regard (if one open ball is separable, they all are).

(Note trivially that in a discrete metric space any unit ball is trivial, so separable, but only countable discrete metric spaces are separable, so in general we cannot go from separability of a ball to that of the whole space.)

• Why there always exist $m \in \mathbb{N}$ such that $m > \| x \|$? It means that all $x \in X$ has bounded norm? – Idonknow Oct 27 '15 at 7:25
• @Idonknow Yes, the norm of some fixed $x$ is always some fixed non-negative real number (not infinite), and the integers are unbounded in those. It's the Archimedean property of $\mathbb{R}$ (en.wikipedia.org/wiki/Archimedean_property) if you want to be very precise. – Henno Brandsma Oct 27 '15 at 7:52

Because if a set $A$ is dense in set $B$ and $X=\bigcup_n (nB)$ then the set $C=\bigcup_n (nA)$ is dense in $X.$