Local Basis of Topological Vector Space

Theorem
Let $\left(\mathcal{X},\tau\right)$ be a topological vector space.

Then, there is a local basis $\mathcal{B}$ of $0_\mathcal{X}$ (where $0_\mathcal{X}$ denotes the zero vector of $\mathcal{X}$) with the following properties:


 * 1) $\forall W\in\mathcal{B},\ \exists V\in\mathcal{B}$ such that $V+V\subseteq W$ (where the addition $V+V$ is meant in the sense of the minkowski sum).
 * 2) Every $W\in\mathcal{B}$ is star-shaped (balanced).
 * 3) Every $W\in\mathcal{B}$ is absorbent.
 * 4) $\bigcap_\mathcal{B}W=\left\{0_\mathcal{X}\right\}$.

Proof
The proof will be carried out in various steps. We will construct a collection of star-shaped neighborhoods of $0_\mathcal{X}$ and we will show that it is indeed a local basis with the required properties. Firstly we define the following class:


 * $\mathcal{B}_0 := \left\{ W\in \tau,\ 0\in W,\ W \text{ is star-shaped}\right\}$

$\mathcal{B}_0$ is a local basis for $0_\mathcal{X}$
Let $U\ni 0_\mathcal{X}$ be an open set and let us notice that $0\cdot 0_\mathcal{X}=0_\mathcal{X}$.

This way we have proved that $\mathcal{B}_0$ is a local basis for $0_\mathcal{X}$.

Since $\cdot:K\times\mathcal{X}\to\mathcal{X}$ is a continuous mapping, there is a neighborhood of $\left(0,0_\mathcal{X}\right)\in K\times \mathcal{X}$ in the form


 * $\left(-\epsilon,\epsilon\right)\times W$

such that:


 * $\cdot\left( \left(-\epsilon,\epsilon\right)\times W\right) \subseteq U$

which means that


 * $\left(-\epsilon,\epsilon\right)\cdot W \subseteq U$

Let


 * $G:= \displaystyle\bigcup_{\lambda\in\left( -\epsilon,\epsilon\right),\lambda\neq 0}\lambda W \subseteq U$

We have that:


 * 1) For every $\lambda\neq 0$, the set $\lambda \cdot W$ is open, hence $G$ is open as union of open sets.
 * 2) $0_\mathcal{X}\in G$ since $0\in W$.
 * 3) If $x\in G$ then $-x\in G$.

Therefore, $G$ is open, star-shaped, contains $0_\mathcal{X}$ and $G\subseteq U$.

Condition #1 is satisfied by $\mathcal{B}_0$
Let $V\in\mathcal{B}_0$. We have that $+\left( 0_\mathcal{X},0_\mathcal{X} \right)=0_\mathcal{X}\in W$.

Since $+$ is a continuous mapping and $\mathcal{B}_0$ is a local basis of $0_\mathcal{X}$, there exists a $V\in\mathcal{B}_0$ such that:


 * $+\left(V,V\right)=V+V\subseteq W$

$\mathcal{B}_0$ consists of absorbent sets
We notice that:


 * $\left(\dfrac{1}n, x\right) \overset{n}{\to} \left(0,x\right)$

and


 * $\cdot\left(\left(\dfrac{1}n, x\right)\right) \overset{n}{\to} \cdot\left(0,x\right)=0_\mathcal{X}$

Thus, there is a $n\in\N$ such that $\frac{1}{n}x\in V$ or what is the same $x\in nV$

Condition #4 is satisfied by $\mathcal{B}_0$
It suffices that we find a $V\in\mathcal{B}_0$ such that $x\notin V$. This is possible since $\tau$ is a Hausdorff topology.

The proof is now complete.