Is there always an injective map from a space in its dual space? Today our teacher said that dual spaces are "big" and told us that this is a consequence by Hahn-Banach's theorem. So I was wondering whether the dual space of a space is always "bigger" or equal compared with the space itself? I thought a good way to check the adjective "big" by mathematical methods is to ask whether there is an injective map in the dual space. 
I am very interested in any comment on this.
 A: If you're talking about normed spaces and their continuous duals, this is not necessarily true. 
Consider the sequence space $\ell_p$, where $1 < p < \infty$ and $p \neq 2$. Then the continuous dual $\ell_p^*$ is $\ell_q$, where $1/p + 1/q = 1$. It is possible to prove that $\ell_p$ does not embed into $\ell_q$ as a Banach space if $p \neq q$. In fact, even more is true. It is a theorem that $\ell_p$ and $\ell_q$ are totally incomparable: if $X$ is an infinite-dimensional subspace of $\ell_p$, then $X$ does not embed into $\ell_q$ as a normed space (similarly for infinite-dimensional subspaces of $\ell_q$). This is corollary 2.1.6 in Topics in Banach Space Theory by Albiac and Kalton.
For vector spaces in general, see this MO question. If $V$ is an infinite-dimensional vector space, then $V$ embeds into $V'$ as a vector space and the dimension of $V$ is strictly less than the dimension of $V'$ (as cardinalities). In the finite-dimensional case $V$ and $V'$ have the same dimension.
A: Let me try and give an answer to the question

In what sense is the dual space of a normed space larger than the original space?

Smallness is not easily phrased for such a thing as an infinite-dimensional normed space. Cardinality certainly doesn't seem to help much (more on that later). One sense in which a space can be small is given by separability (Recall that any subspace of a separable (normed!) space is again separable).
If $X$ is a normed space and $X'$ is its dual, then one can prove the following.


*

*If $X$ is not separable, then neither is $X'$

*If $X$ is separable, then $X'$ need not be.


The spaces $\ell^1$ and $\ell^\infty$ are examples for the second statement.
Another reason to view the transition to the dual as an enlargement is the fact that any normed space can be embedded in a very nice manner into its bidual. 
Let's consider the spaces from earlier again. Since we actually have that $\ell^1$ is the dual of $c_0$, i.e. the space of sequences that converge to zero, $\ell^\infty$ is the bidual of $c_0$. Now, one can prove that $c_0$ can be viewed as the space $C(\alpha \mathbb N)$, i.e. the space of continuous functions on the Alexandroff (or 1-point)-compactification of $\mathbb N$. One can also prove that $\ell^\infty$ can be viewed as $C(\beta \mathbb N)$, i.e. the Stone-Čech compactification of $\mathbb N$.
And that's where cardinality comes into play again. The set $\beta N$ has the cardinality $2^{2^{\mathbb N}}$, whereas $\alpha N$ still has cardinality $\mathbb N$. So in that sense, it's quite the leap from $C(\alpha \mathbb N)$ to $C(\beta \mathbb)$.
