Hyperplane $V=[V]\neq X$ or $[V]=X$

Question

$(X,\tau)$ - linear topological space

$ V \subset X $ - hyperplane
$ V = \left \{ x \in X \ | \ f(x) = \alpha \right \}, \alpha \in K = \mathbb{R}(\mathbb{C}), f $ - linear functional on $X $

I should prove that hyperplane $V=[V]\neq X$ or $[V]=X$

I know:

In a linear topological space a hyperplane is closed if and only if it is the kernel of a continuous linear functional.

How can I prove this statement?

Answer 1

Definition 1: A hyperspace in $X$ is a maximal proper subspace of $X$. Hence, a subspace $L$ of $X$ is a hyperspace of $X$ if and only if $[L\cup\{a\}]=X$ for any $a\in X\setminus L$.

Definition 2: A hyperplane in $X$ is a set of the form $L+a$ where $a\in X$ and $L$ is a hyperspace.

Theorem: A subspace of $X$ is a hyperspace if and only if it is the kernel of a non-zero functional $f:X\to\mathbb{R}$ (or $\mathbb{C}$).

Corollary: If $V=\{x\in X:f(x)=\alpha\}$, for $\alpha\in\mathbb{R}$ and $f\in X'\setminus\{0\}$, then $[V]=V$ or $[V]=X$.

Proof: Let $L=ker(f)$ and $a\in X$ such that $f(a)=\alpha$. Then $V=L+a$.

Case 1: $a\in L$. In this case, $V$ is a subspace of $X$ and hence $[V]=V$.

Case 2: $a\not\in L$. By Theorem, $L$ is a hyperspace of $X$ and hence $$X=[L\cup\{a\}]=\{l+\lambda a:l\in L,\, \lambda\in\mathbb{R}\}=[L+a]=[V].$$