Would appreciate some explicit solution to the following question as I already have the solution but I am not quite clear:
a. Show that a normed space X is separable if and only if there are finite dimensional subspaces $ X_1 \subset X_2 \subset ...$ of $X$ such that: $\bar\cup_{n\in N} X_n = X$.
b. I also need to find a concrete example of a separable normed space $X$ and finite dimensional spaces $X_1 \subset X_2 \subset ...$ of $X$ such that: $\bar\cup_{n\in N} X_n = X$ but $\cup_{n\in N} X_n \neq X$.
I know the following proposition: Let $X$ be a normed space, and $(x_n)_{n \in N}$ a sequence in X with X ={$\overline{span \{x_n: n \in N\}}$}. Then $X$ is separable.
Let $\overline{\{x_{n}: n=1,2,...\}}=X$ and $X_{n}=\left<x_{1},...,x_{n}\right>$, then $\overline{\displaystyle\bigcup_{n}X_{n}}=X$.
For an example, look up any separable Hilbert space $X$ and let $\mathcal{B}=\{x_{n}: n=1,2,...\}$ be its Hilbert basis, then with those $X_{n}$ defined, the union is not the whole $X$.