Codimension of a subspace I'm studying the book "Topologies on closed and closed convex sets" by myself. I have searched a lot for this but couldn't get an answer. 

If $X$ is a normed linear space and $ f : X\to \mathbb R$ is a linear functional, then what is the co-dimension of $\ker(f)$?  

 A: the kernel of $f$, denoted $\text{ker}(f)$ is the set of all vectors that are mapped to 0 by $f$, i.e. $\text{ker}(f) = \{x \in X: f(x) = 0\}$. This is a subspace of $X$, hence it makes sense to talk about the codimension of $\text{ker}(f)$ in $X$. See wikipedia. If you're looking for a more detailed discussion of the notion of codimension, I suggest you consult some basic linear algebra books.
The range of a linear functional is one-dimensional, except for the zero functional, in which case it is zero-dimensional. Hence the codimension of the kernel of a linear functional is $1$ if the functional is not identically $0$, and $0$ if it is the zero functional.
A: Let $f \in X^*$. We know that $Ker(f)$ is a subspace of $X$.
Now let $x_0 \in X$ such that $f(x_0) \neq 0$ and $x \in X$.
We have that $$f(x- \frac{f(x)}{f(x_0)} x_0)=0$$ thus $x- \frac{f(x)}{f(x_0)} x_0 \in Ker(f)$ thus $[x]_{X/Ker(f)}= \frac{f(x)}{f(x_0)}[x_0]_{X/Ker(f)}$ in the quotient space $X/Ker(f)$.
Also we $f(x_0) \neq 0$ form our assumption thus $x_0 \notin Ker(f)$ thus $[x_0]_{X/Ker(f)} \neq [0]_{X/Ker(f)}$.
We proved that every element of $X/Ker(f)$ is a multiple of $[x_0]_{X/Ker(f)}$ therefore the set $\{[x_0]_{X/Ker(f)}\}$ is a basis of $X/Ker(f)$ 
Thus $codim(Ker(f))=1$
