Standard Topology of Locally Convex Space has Local Basis of Balanced Convex Absorbing Sets

Theorem

Let $\GF \in \set {\R, \C}$.

Let $\struct {X, \PP}$ be a locally convex space over $\GF$ with standard topology $\tau$.

Then there exists a local basis $\BB$ for ${\mathbf 0}_X$ in $\struct {X, \tau}$ such that:

each $A \in \BB$ is balanced, convex and absorbing.

Proof

For each $\epsilon > 0$ and $S$ a finite subset of $\PP$, set:

$U_{\epsilon, S} = \set {x \in X : \map p x < \epsilon \text { for each } p \in S}$

and:

$\BB = \set {U_{\epsilon, S} : \epsilon > 0 \text { and } S \subseteq \PP \text { finite} }$

From Open Sets in Standard Topology of Locally Convex Space, $\BB'$ is a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$.

From Convex Open Neighborhood of Origin in Topological Vector Space contains Balanced Convex Open Neighborhood:

for each $A \in \BB'$ there exists a balanced convex open neighborhood $V_A$ for ${\mathbf 0}_X$ such that $U_A \subseteq A$.

Then:

$\BB = \set {V_A : A \in \BB'}$

forms a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$, consisting of balanced convex open neighborhoods of ${\mathbf 0}_X$.

From Locally Convex Space is Topological Vector Space, $\struct {X, \tau}$ is a topological vector space.

From Convex Subset of Topological Vector Space containing Zero Vector in Interior is Absorbing Set, we therefore obtain that $V_A$ is absorbing for each $A \in \BB'$.

Hence $\BB$ is a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$, consisting of balanced convex absorbing sets.

$\blacksquare$