TITLE: Hamel Dimension of Infinite Dimensional Separable Banach Space
QUESTION [6 upvotes]: I was trying to show this result. I have seen proofs that show that the dimension is at least $\mathfrak{c}$, however, I'm unable to prove that it is exactly $\mathfrak{c}$. The last line of this article is unclear. How does separability imply that dimension is $\mathfrak{c}$?

REPLY [6 votes]: If $X$ is separable, it has a countable dense subset $A$.  Every point of $X$ is a limit of some sequence in $A$, and there are only $|A|^{|\mathbb{N}|}=\mathfrak{c}$ different sequences in $A$.  So $|X|\leq\mathfrak{c}$.  A Hamel basis for $X$ is a subset of $X$, and so any Hamel basis has at most $\mathfrak{c}$ elements.