Kreyszig's Functional Analysis Section 2.8: How is the canonical embedding map injective? Let $X$ be a vector space over the field $K$ of the real or complex numbers. Let $X^*$ denote the vector space of all linear functionals defined on $X$, and let $X^{**}$ denote the vector space of all linear functionals defined on $X^*$. 
Let $x \in X$ be fixed, and let $g_x \colon X^* \to X^{**}$ be defined as 
$$g_x(f) \colon= f(x) \; \; \; \forall f \in X^*.$$ 
Then $g_x$ is a linear functional on $X^*$ so that $g_x \in X^{**}$. 
Then the mapping $C \colon X \to X^{**}$ defined by 
$$Cx \colon= g_x \; \; \; \forall x \in X$$ 
is linear. 
But Kryszeg states that this map $C$ is also injective. How to show this? 
My work: 
Suppose that, for some $x$, $y \in X$, we have the equality
$$Cx = Cy.$$
Then 
$$g_x = g_y.$$
So, for all $f \in X^*$, we have 
$$g_x(f) = g_y(f).$$ 
This implies that $f(x) = f(y)$ and so $x-y \in N(f)$ for all linear functionals $f$ defined on $X$, where $N(f)$ denotes the null space of $f$. 
What next? How to show that $x = y$, especially in the case when $X$ is not finite-dimensional? 
 A: This has nothing to do with functional analysis.
If $ v\neq 0$ there always exist a functional $ f:X\to K $ such that $ f (v)\neq 0$: complete $ v $ to a basis $\mathcal{B}$ of $ X $ (as a vector space) and define $ f $ by declaring that $ f (v):=1$ and, for any $w\in \mathcal{B}\setminus\{v\} $, $f(w):=0$.
Choosing in particular $ v:=x-y $ this shows that $ v=0$, i.e. $ x=y $.
A: Assume that $X$ has infinite dimension and let $x\in X\setminus \{0\}$ ; let $E=(e_i)_i$ be a basis of $X$ that contains $x$. An element of $X$ is a function $E\rightarrow K$ that is zero almost everywhere.  Every function $E\rightarrow K$ gives birth (by linearity) to an element of $X^*$ and the converse is true. In particular, there is $f\in X^*$ s.t. $f(x)=1$. Thus $g_x\not= 0$ and we are done. Note that $\dim(X)<\dim(X^*)<\dim(X^{**})$, and consequently, $C$ is never a bijection.
EDIT. Answer to Saaqib. When $X$ has infinite dimension, $\dim(X)$ is a cardinal number (the cardinality of $E$) and not an integer. Then $\dim(X)<\dim(X^*)$ iff there is no surjective function $X\rightarrow X^*$. There is a pretty proof in MO 
https://mathoverflow.net/questions/13322/slick-proof-a-vector-space-has-the-same-dimension-as-its-dual-if-and-only-if-i
