If $X$ is a separable Banach space $ \Rightarrow$ $ X=\overline{\cup_{n=1}^{\infty} X_n}$ where $\dim X_n=n$

Question

Let $X$ be a Banach space separable.

How can we prove that there is a sequence of subspaces:

$X_1\subset\ X_2 \subset \cdots \subset \ X_n \subset \cdots $ of $X$ such that $\displaystyle X=\overline{\bigcup_{n=1}^{\infty} X_n}$ where $\dim X_n=n$

Any hints would be appreciated.

Answer 1

Hints: How do you usually specify an $n$-dimensional space? Recall that the definition of separability gives you a sequence of points or vectors lying in $X$. How can you use these to build the $X_n$?

PS: Note that your result requires $X$ to be infinite dimensional because otherwise the sequence cannot exist. Consider the real line.