existence of $W \subset X^*$ which separates the points in separable normed space $X$.

Question

Let $X$ be a separable normed space. I have to see that there exists a numerable subset $W \subset X^*$ such that $W$ separates the points in $X$. I could start the proof but I have a problem: what does "separates the points in $X$" mean? Thanks

Answer 1

Let $A, B$ be sets, and $\mathscr{F}$ a family of functions from $A$ to $B$. Then $\mathscr{F}$ separates points in $A$, if for all $a_1, a_2 \in A$ with $a_1 \neq a_2$ there is an $f \in \mathscr{F}$ such that $$f(a_1) \neq f(a_2)\,.$$ Thus a point of $A$ is determined by the indexed family of its images under elements of $\mathscr{F}$, in other words the map $\Phi \colon A \to B^{\mathscr{F}}$ given by $$\Phi(a) = \langle f(a)\rangle_{f \in \mathscr{F}}$$ is injective.

If $A$ and $B$ are vector spaces over the field $k$, and $\mathscr{F} \subset \mathscr{L}(A,B)$ is a family of $k$-linear maps, then $\mathscr{F}$ separates points if and only if for every $a \in A \setminus \{0\}$ there is an $f \in \mathscr{F}$ such that $f(a) \neq 0$.

Vector spaces and linear maps are more than we need for this characterisation by the way, it holds (with a slight modification of notation, replace "$0$" with the identity elements "$e$") for arbitrary groups and a family of group homomorphisms.

The proof that this characterises families separating points is easy. On the one hand, $a \in A \setminus \{e\}$ (I'm using the group setting, $e = 0$ for vector spaces) means $a \neq e$, so if $\mathscr{F}$ separates points there is an $f \in \mathscr{F}$ such that $f(a) \neq f(e) = e$. On the other hand, if $a_1 neq a_2$, then $a_1^{-1}a_2 \neq e$, so by assumption there is $f \in \mathscr{F}$ with $$e \neq f(a_1^{-1}a_2) = f(a_1)^{-1}f(a_2)\,,$$ which is equivalent to $f(a_1) \neq f(a_2)$, and thus $\mathscr{F}$ separates points.

For your specific problem, pick a countable dense subset $D = \{x_n : n \in \mathbb{N}\}$ of $X$ (it may be convenient to assume $0 \notin D$) and for every $n$ choose an $f_n \in X^{\ast}$ (it can be convenient to choose them such that $\lVert f_n\rVert = 1$ for all $n$) based on $x_n$ alone (hint: Hahn-Banach). If you made the right choices, $W = \{ f_n : n \in \mathbb{N}\}$ will separate points in $X$.