Hyperplanes in finite and infinite dimension vector spaces. I know that hyperplanes of a n-dimensional vector space are sub-spaces of dimension n-1, This is in finite dimension spaces. BUT what about infinite dimension spaces what are hyperplanes? are they the same?
 A: I know this is an old question, but it seems to me that no one has answered it in a "correct" fashion yet, concerning "infinite" of course.
The most generalized definition I've seen is the next one:
Let H be a subspace in a vector space X. H is called hyperplane if $H \neq X$  and for every subspace V such that $ H \subseteq V $ only one of the following is satisfied: $ V = X$ or $ V = H $.
Lemma: A subspace H in X is hyperplane iff e is in X \ H such that $$ <\{e,H \}> =  \{ \lambda e + h : \lambda \epsilon \mathbb{R}, h  \epsilon H\} = X  $$
Just for fun:
Theorem: A subspace H in a vector space X is hyperplane iff there is a non-zero linear function $$ l : X \to \mathbb{R}$$ such that $$ H = \{x \epsilon X : l(x) = 0\} = Ker( l ) $$
Good luck proving this.
A: Usually this is how one hears the definition anyway: "a subspace of codimension 1." 
So yes, I believe that any subspace $W$ of a vector space $V$ can be called a hyperplane if $\dim(V/W)=1$. 

(Added) 
This is, of course, equivalent to saying that $W$ is the kernel of a linear functional (if you want to do it that way.) 
If $\dim(V/W)=1$, then picking a basis for $W$ and extending it by one element "$v$" to be a basis for $V$ allows you to project onto the coefficient of $v$, obtaining a linear functional from $V$ to $F$ with kernel $W$.
In the other direction, an easy application of isomorphism theorems tells you that for a nonzero linear functional $f$, $V/\ker(f)\cong F$, so $V/\ker(f)$ has dimension $1$, and $\ker(f)$ has codimension $1$.
A: When you are generalizing an idea to a domain in which it is not, a priori defined, you can do it in any way you want. 
If your definition of hyperplane is that it is a subspace of dimension $n-1$ where $n$ is the dimension of the space, and now you want to extend this for when $n$ is $\infty$, you could say that $\infty-1=\infty$ and therefore you will call hyperplane any subspace of infinite dimension. 
The thing is that often one generalizes when there is a need for it. When you can find something interesting in the generalization. 
For finite dimension a hyperplane is the zero set of a linear form (a linear functional), a linear function from the space to the scalar field. So, another possibility is to say that a hyperplane is the zero set of a linear functional. This is the option that is more often used.
A rough criterion of how good a definition is Halmos': A good definition is the hypothesis of a theorem. The thing is that many theorems involve the zero set of linear functionals. 'any infinite dimensional subspace' are less interesting as a class. 
A: I have seen that this is a very very old question nevertheless I would like to make this little contribution which would be of interest for someone.
I will consider the correct: $H$ is called a hyperplane of a given space $V$ iff $H$ is a subspace of $V$, $H\neq V$ and for any other subspace $W$ of $V$, $H\subset W\implies$ either $H=W$ or $W=X$.
We are then going to prove that: $H=$hyperplane of $V$ iff $dim(V/H)=1$ iff $H=Ker(f)$ for some linear form $f$.
(1) Suppose $H=$hyperplane of $V$ and $dim(V/H)\neq 1$. Sine $H\neq V$, $dim(V/H)\neq 0$. Then $dim(V/H)>1$. Hence there are $\tilde{x},\tilde{y}$ linearly independents in $V/H$ such that $\mathbb{K}\tilde{x}+\mathbb{K}\tilde{y}\subset V/H$. Let $x$ and $y$ be their representers in $V$. It is clear that $x,y\notin H$. The space $W=\mathbb{K}y+H$ is a subspace of $V$ containing $H$. Since $y\neq \overrightarrow{0}$, then $H\neq W$. H being a hyperplane, this condition implies that $W=V\ni x$ therefore $x=k\cdot y+h$ for some $0\neq k\in\mathbb{K}$ and $h\in H$. So $\tilde{x}=k\tilde{y}$ which means that $\tilde{x}$ is proportional to $\tilde{y}$ which is impossible as they are independents.
(2) If $dim(V/H)=1$ then there is a non zero vector $w\in V$ such that $V/H=\mathbb{K}\tilde{w}$. So for all $x\in V$ there is $f(x)\in \mathbb{K}$ such that $\tilde{x}=f(x)\cdot \tilde{w}$. We get the rule $x\overset{f}{\mapsto} f(x)$ from $V$ to $\mathbb{K}$. It is easy to show that $f$ is a linear form and that $H=ker(f)$.
A: They are kernels of linear forms thus their codimension is one.
