TITLE: Is it true that $\dim(X) \leq \dim(X^{\ast})$ for every infinite dimentional Banach space $X$?
QUESTION [9 upvotes]: Given an arbitrary infinite dimensional Banach space $X$, can we deduce that it's dimension $\dim(X)$ (the cardinality of one of its Hamel bases) is less or equal of the dimension  $\dim(X^{\ast})$ of its dual space (the space of all continuous linear functionals $f:X\to\mathbb{R}$)?

REPLY [2 votes]: That is a great answer to a very interesting problem. Given your current solution, I would like to suggest a possible generalization to the case of normed spaces.
Since in the finite dimensional case everything is working well, let $X$ be an infinite dimensional normed space and consider the canonical embedding of $X$ to its double dual $X^{**}$ given by:
$T:X\rightarrow X^{**},\hspace{10pt}  T(x)=T_x \hspace{5pt}$ where for $x\in X,$ we define $T_x:X^*\rightarrow\mathbb{R}$ such that $ \hspace{7pt}  {T_x}(x^*)=x^*(x)$ 
Of course, T is an isometric embedding.
Consider the spaces:
$T(X)\hspace{5pt}$ (which is isometric to $X$ and obviously dense in $\overline{T(X)}$) and
$\overline{T(X)}\hspace{5pt}$ (which is a Banach space, as it is a closed subspace of the Banach space $X^{**}$)
(This process is of course standard when considering the completion of a normed space $X$).
Since $\overline{T(X)}$ is a Banach space, by your argument we must have that $\dim(\overline{T(X)})\leq \dim(({\overline{T(X)}})^*)$
But, since $T(X)$ is dense in $\overline{T(X)},$ we must have that the spaces ${(T(X))}^*$ and $({\overline{T(X)}})^*$ are isometric.
(Indeed, consider this result when stated in the following more general fashion:
Let $X$ be a normed space, $Z$ a dense subspace of $X$ and $Y$ a Banach space. Then, the spaces $\mathcal{B}(X,Y)$ and $\mathcal{B}(Z,Y)$ are isometric.
I will be glad to give hints to the proof of this fact, if anyone is interested).
Now combine all the previous arguments together to get:
$\dim(X)= \dim(T(X))\leq \dim({\overline{T(X)}})\leq \dim(({\overline{T(X)}})^*) = \dim(({T(X)})^*)= \dim(X^*)$
(Notice that we have also implicitly used the fact that if two spaces are isometric, then their duals must be isometric as well).