Local Basis of Topological Vector Space

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Theorem

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

Let $0_\mathcal X$ denote the zero vector of $\mathcal X$.


Then there exists a local basis $\mathcal B$ of $0_\mathcal X$ with the following properties:

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


Proof

The proof will be carried out in various steps.

We will construct a collection of star-shaped neighborhoods of $0_\mathcal X$.

Then we will show that it is indeed a local basis with the required properties.


Firstly we define the following set:

$\mathcal B_0 := \left\{{W \in \tau: 0 \in W, W \text{ is star-shaped} }\right\}$


$\mathcal B_0$ is a local basis for $0_\mathcal X$

Let $U \ni 0_\mathcal X$ be an open set.

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:

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

We have that:

$(1): \quad$ For every $\lambda \ne 0$, the set $\lambda \cdot W$ is open.

Hence $G$ is open as union of open sets.

$(2): \quad 0_\mathcal X \in G$ since $0 \in W$.
$(3): \quad$ If $x \in G$ then $-x \in G$.

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

$\Box$


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:

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

$\Box$


$\mathcal B_0$ 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 amounts to the same thing:

$x \in n V$

$\Box$


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.

$\Box$


The proof is now complete.

$\blacksquare$