# $X$ normed linear space separable $\Longleftrightarrow$ $\exists K \subset X$ compact s.t. $\overline{ \text{span}\{K\}}= X$

Let $X$ be a normed linear space. Show that $X$ is separable if and only if there is a compact subset $K$ of $X$ for which $\overline{ \text{span}\{K\}}= X$

I can't figure out how to solve this exercise:

My first problem is: how to control cardinality of $\overline{ \text{span}\{K\}}$? According to the definition of separable space it has to be countable. But if the field is $\mathbb{R}$ (or $\mathbb{C}$) |$\text{span}\{K\}$| is strictly greater than |$\mathbb{N}$| because the span contains finite linear combinations of elements of $K$ and $K$ has at least one element.

Maybe i'm not getting the point of this exercise (Real Analysis - Royden, $4$ ed pag. $262$ n. $28$) or missing some basic properties of compact sets in normed linear spaces.

Can someone help me? Thanks in advance

• A separable space need not be countable. It merely has a countable dense subset. For instance, $\mathbb{R}$ is separable, but not countable. – Prahlad Vaidyanathan Dec 3 '13 at 14:53
• Yeah but span (K) need to be countable in my case right? – Riccardo Dec 3 '13 at 14:57
• Not at all. $\mathbb{R} = \text{span}\{1\}$ – Prahlad Vaidyanathan Dec 3 '13 at 14:58
• Ah, I see. Take the linear combinations with rational coefficients to get a countable set. – Daniel Fischer Dec 3 '13 at 15:18
• Yeah but in this case the notation will be $\mathbb{R} = \overline{ \text{ span}_Q(1)}$ – Riccardo Dec 3 '13 at 15:32

In the one direction, to construct a compact set with dense span from a countable dense subset, note that if $(x_n)_{n\in\mathbb{N}}$ is a sequence converging to $x_\ast$, then the set $\{ x_n : n \in \mathbb{N}\} \cup \{x_\ast\}$ is compact. Construct a convergent sequence from the countable dense subset without changing the span.

For example, if $D = \{ y_n : n\in\mathbb{N}\}$ is a countable dense subset, set $x_n = \frac{1}{1+(n+1)\lVert y_n\rVert}y_n$ for $n\in\mathbb{N}$. We then have $\lVert x_n\rVert < \frac{1}{n+1}$, so $x_n \to 0$, whence $K = \{0\} \cup \{ x_n : n \in \mathbb{N}\}$ is compact, and since $y_n = (1+(n+1)\lVert y_n\rVert)y_n \in \operatorname{span} K$, it follows that $\overline{\operatorname{span} K} \supset \overline{D} = X$, so $K$ is a compact set with dense span.

In the other direction, note that a compact metric space is separable. Show that if $\operatorname{span} K$ is dense, and $D$ a countable dense subset of $K$, then $\operatorname{span}_{\mathbb{Q}[i]} D$ is also dense.

Namely, since $\mathbb{Q}[i]$ is dense in $\mathbb{C}$, it follows that $\operatorname{span} D \subset \overline{\operatorname{span}_{\mathbb{Q}[i]} D}$, and hence

$$\overline{\operatorname{span}_{\mathbb{Q}[i]} D} = \overline{\operatorname{span} D}$$

is a closed linear subspace containing $\overline{D} = K$, hence $\overline{\operatorname{span} K} = X$. But $\operatorname{span}_{\mathbb{Q}[i]} D$ is countable since $D$ and $\mathbb{Q}[i]$ are, so $X$ is separable.

• What have you got, and where are you stuck? – Daniel Fischer Dec 3 '13 at 21:43
• I proved working out the directions but i'm not going anywhere. First implication: can't build the compact set... Tried to put together the sequences $a_i \subset A$ (A si the dense countable set) which converge to an element of the hamel basis (in this way I obtain the dense span). But i can't manage to control the compactness of such union of sets. The other implication now is clear, just messed up indexes – Riccardo Dec 3 '13 at 21:53
• I somehow missed the notification, so a belated response. Don't care about a Hamel basis. Let $D = \{ d_n : n \in \mathbb{N}\}$ be the dense subset. Scale each $d_n$ to obtain a convergent sequence. – Daniel Fischer Dec 3 '13 at 23:36
• no problem! now it's late here (italy) I'll try tomorrow working out your hints! thanks in advance! – Riccardo Dec 3 '13 at 23:38
• If you get stuck, don't hesitate to ask for further guidance. – Daniel Fischer Dec 3 '13 at 23:40