Prove $\dim X = \mathfrak{c}$ for every infinite dimentional Banach space

Question

Let $X$ be an infinite-dimensional Separable Banach Space. Prove that $\dim X=\mathfrak{c}$.

On the direction of $\dim X\ge \mathfrak{c}$, I thought taking the subset of all elements $x\in X$ which can be represented as an convergent expansion $$X\ni x = \sum_{n=1}^\infty c_ne_n$$ where $$e_n^i=\cases{0\quad i\neq n \\1\quad i=n,}$$ which will be mapped to $\ell^\infty$, a space of cardinality $\mathfrak c$ by the isomorphism $x\mapsto(c_1,c_2,\dots)$.

About the other direction I don't have any clue. We know that a Banach space with a basis must be separable but how can we use that fact to find out that $\dim X\le\mathfrak{c}$?

Answer 1

Let $Q$ be a countable dense subset of your Banach space $X$. In particular, every element of $X$ may b written as the limit of a sequence of points in $Q$, so there is an onto map $Q^{\mathbb{N}} \to X$, hence $$|X| \leq |Q^{\mathbb{N}}| = \mathfrak{c}.$$ On the other hand, $$|X| = |\mathbb{K}| \cdot \dim (X)$$ where $\mathbb{K}= \mathbb{R}$ or $\mathbb{C}$ is your field. Therefore, $$\mathfrak{c} \geq |X|= \max( \mathfrak{c}, \dim(X))$$ hence $\dim(X) \leq \mathfrak{c}$.