11.1 Vector Space Topologies

Proof. Let $U \subset E$ open and $y \in E$, then $U + y$ is the preimage of $U$ by the map $x \mapsto x - y$. By (TVS1), $U + y$ is open.$\square$

Proof. By Definition 9.1.5.$\square$

Proof. Firstly, (TVS2) implies that every neighbourhood of $0$ is circled.

By Proposition 6.1.16, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets.

Let $U \in \cn^{o}(0)$ be open. By (TVS2), there exists $r > 0$ such that $\lambda U \subset U$ for all $\lambda \in K$ with $\abs{\lambda}\le r$. Define

then for any $x \in V$, there exists $\lambda \in K$ with $\abs{\lambda}\le r$ and $y \in U$ such that $x = \lambda y$. In which case, for any $\mu \in K$ with $\abs{\mu}\le 1$, $\mu(\lambda y) = (\mu \lambda) y \in \mu\lambda U$. Since $\abs{\mu \lambda}\le r$, $\mu \lambda U \subset V$. Thus $V \subset U$ is balanced.

Let $U \in \cn(0)$ be closed, then there exists a balanced neighbourhood $V \in \cn^{o}(0)$ such that $V \subset U$. In which case, for any $\lambda \in K$ with $0 < \abs{\lambda}\le 1$, $\lambda \overline{V}= \overline{\lambda V}\subset \overline{V}$ by (TVS2). Therefore $\overline{V}\subset U$ is balanced as well.$\square$

Proof, [I.1.2, SW99]. Forward: By Proposition 11.1.6, there exists a fundamental system of neighbourhoods $\fB \subset \cn_{E}(0)$ consisting of circled and radial sets. By (TVS1), $\fB$ satisfies (TVB1).

Converse: For each $V \in \fB$, let $U_{V} = \bracs{(x, y) \in E|x - y \in V}$, then $U_{V}$ is symmetric and translation-invariant by (TVB1). Let

then

1. (FB1)

   For any $V, V' \in \fB$, there exists $W \in \fB$ with $W \subset V \cap V'$. In which case, $U_{V}\cap U_{V'}\supset U_{W} \in \mathfrak{V}$.

2. (UB1)

   For any $x \in E$ and $V \in \fB$, $x - x = 0 \in V$, so $\Delta \subset U_{V}$.

3. (UB3)

   For any $V \in \fB$, by (TVB1), there exists $W \in \fB$ such that $W + W \subset V$. In which case, for any $x, y, z \in E$ with $x - y, y - z \in W$, $x - z \in V$. Therefore $U_{W} \circ U_{W} \subset U_{V}$.

By Proposition 6.1.8, there exists a unique uniformity $\fU$ on $E$ for which $\mathfrak{V}$ is a fundamental system of entourages.

(1): Since $\mathfrak{V}$ is translation-invariant, so is $\fU$.

(2): By definition of the uniform topology, $\fB = \bracs{U_V(0)|V \in \fB}$ is a fundamental system of neighbourhoods at $0$.

(3):

1. (TVS1)

   Let $V \in \fB$, then there exists $W \in \fB$ such that $W + W \subset V$ by (TVB1). In which case, for any $x, x', y, y'$ with $x - x' \in W$ and $y - y' \in W$, $(x + y) - (x' + y') \in W + W \subset V$.

2. (TVS2)

   Let $V \in \fB$, $\eps > 0$, $x, x' \in E$ with $x - x' \in V$, and $\lambda, \lambda' \in K$ with $\abs{\lambda - \lambda'}< \eps$, then

   \begin{align*}\lambda x - \lambda' x'&= \lambda x - \lambda x' + \lambda x' - \lambda' x' \\&= \lambda(x - x') + (\lambda - \lambda')x' \in \lambda V + (\lambda - \lambda')x'\end{align*}

   
   

   Since $V$ is radial, there exists $\mu \in K$ such that $x \in \mu V$. Given that $V$ is circled, $x' = x + (x - x') \in \mu V + V \subset (\abs{\mu}+ 1)V$, and

   \[\lambda x - \lambda' x' \in \lambda V + \eps(\abs \mu + 1)V \subset \abs{\lambda}V + \eps(\abs \mu + 1)V\]

   
   

   Let $W \in \fB$ and $\eps > 0$ such that $\eps(\abs{\mu}+1) \le 1$, then by repeated application of (TVB1), there exists $V \in \fB$ such that $\abs{\lambda}V + V \subset W$. Therefore scalar multiplication is jointly continuous.

(Uniqueness): Let $\mathcal{S}\subset 2^{E}$ be a topology on $E$ satisfying (1) and (2), then for each $x \in E$, $\cn_{(E, \mathcal{S})}(x) = \cn_{(E, \mathcal{T})}(x)$. By Proposition 5.4.4, $\mathcal{S}= \mathcal{T}$.$\square$

Proof. Let $U \in \cn(0)$ be radial, then for any $y \in U$, the mapping $t \mapsto ty$ is a path from $0$ to $y$ contained in $U$. Thus $U$ is path-connected. By Proposition 11.1.7, the radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods, so the path-connected neighbourhoods of $0$ forms a fundamental system as well.$\square$