Could you tell me how to prove that

a normed space $X$ is separable $\iff$ there exists an at most countable set of linearly independent vectors $X_0 \subset X$ such that $ \ \text{lin} X_0$ is dense in$ X $?

I know how to prove that if a normed space is separable, then there exists such a set.

But I don't know what to do about the opposite direction. Could you help?

Thank you


2 Answers 2


Given a countable set with dense span: $$\#A_0\leq\mathfrak{n}:\quad\overline{\langle A_0\rangle}=X$$

Consider the rational span: $$S_0:=\langle A_0\rangle_{\mathbb{Q}+\imath\mathbb{Q}}:\quad\#S_0\leq\mathfrak{n}$$

Its closure captures the linear span as: $$a,a'\in A_0:\quad\|\lambda a+\lambda'a'-\lambda_na-\lambda'_na'\|\leq|\lambda-\lambda_n|\cdot\|a\|+|\lambda'-\lambda'_n|\cdot\|a'\|\to0$$

So it was already dense: $$\langle A_0\rangle\subseteq\overline{\langle S_0\rangle}\implies X=\overline{\langle A_0\rangle}=\overline{\langle S_0\rangle}$$ (The trick here is that the closure is closed; cf. Kuratowski.)


In the earlier version of this proof there is flaw in the argument. So I have modified the proof. Take a look into this.

For the opposite direction, suppose that $X_0=\{e_n:n\in\mathbb N\}\subset X$ is a countable linearly independent subset such that $\overline{\text{Span}( X_0)}=X$. Let $S=\text{Span}(X_0)$.

Now consider $M=\bigcup\limits_{k\in\mathbb N}\{\sum\limits_{j=1}^k{b_j e_{n_j}:b_1,\dotsc,b_{n_k}\in \mathbb Q}\}$ (i.e. $M$ is set of all rational-linear combinations of elements of $X_0$). If the field is $\mathbb C$, then take $b_j$ to be a complex number with both real and imaginary parts as rational. Clearly $M$ is countable. Now we show that $M$ is dense in $X$. For this, take $x\in X$. Choose $\varepsilon >0$.

Case-1: If $x\in S$ then say $x=\sum\limits_{j=0}^{k}a_j e_{n_j}$. Choose $b_j\in \Bbb Q$ (or in $\mathbb Q+i\mathbb Q$) such that $|a_j-b_j|<\frac{\varepsilon}{kc}$ for each $j=0,\dotsc,k$, where $c=\max\limits_{j=0}^k {\|e_{n_j}\|}$. Let $x^\prime=\sum\limits_{j=0}^{k}b_j e_{n_j}\in M$. Then clearly $\|x-x^\prime\|<\varepsilon$.

Case-2: There is a sequence $x_n\in S$ such that $\|x-x_n\|\to 0$. Consequently $\exists ~n_0\in \mathbb N$ such that $\|x-x_n\|<\frac{\varepsilon}{2}$, for all $n\ge n_0$. Since $x_{n_0}\in S$, say $x_{n_0}=\sum\limits_{j=1}^{p}a_j e_{n_j}$. For $1\le j\le p,$ choose $b_j\in \mathbb Q$ (or in $\mathbb Q+i\mathbb Q$) such that $|a_j-b_j|<\frac{\varepsilon}{2pc}$, where $c=\max\limits_{j=1}^p {\|e_{n_j}\|}$. Take $x^\prime=\sum\limits_{j=1}^p {b_j e_{n_j}}\in M$. Then $\|x-x^\prime\|\le \|x-x_{n_0}\|+\|x_{n_0}-x^\prime\| <\frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon$ and hence $M$ is dense in $X$. Since $M$ is a countable dense subset, $X$ is separable.

  • $\begingroup$ A sequence of scalars might be not sufficient. As an example consider polynomials within the Banach space of continuous functions: $f(x):=|x|:\|f-\sum_{k=1}^Ka_ke_k\|_{[-1,1]}\not\to0(e_k(x):=x^k)$ $\endgroup$ Feb 15, 2015 at 22:04
  • $\begingroup$ @ Freeze_S: I presumed that $X_0$ is a Schauder basis, which is not and that is why I thought about a unique sequence of scalars. Anyway thanks for pointing out the mistake. I have modified the proof. I think this version is ok. $\endgroup$
    – user149418
    Feb 18, 2015 at 16:54

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