8.1 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. Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By Lemma 5.1.9, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.$\square$

Proof. (2): Firstly, for any $V \in \fB_{0}$, $U_{V}$ is translation-invariant.

1. For any $V, V' \in \fB_{0}$, there exists $W \in \fB_{0}$ such that $W \subset V \cap V'$. In which case, $U_{V} \cap U_{V'}\supset U_{W} \in \fB$.

2. For any $V \in \fB_{0}$, $0 \in V$, so $\Delta \subset U_{V}$.

3. Let $V \in \fB_{0}$, then by (TVS1), there exists $W \in \fB_{0}$ such that $W + W \subset V$. For any $(x, y), (y, z) \in U_{W}$, $(x - y), (y - z) \in W$, so $(x - z) \in V$. Thus $U_{W} \circ U_{W} \subset U_{V}$.

By Proposition 5.1.8, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.

(1): Let $\mathfrak{V}$ be a translation-invariant uniformity on $E$ inducing the topology. For any symmetric, translation-invariant entourage $V \in \mathfrak{V}$, $V(x) = V(0) + x$ for all $x \in E$, and $(x, y) \in V$ if and only if $y - x \in V$, if and only if $x - y \in V$. Thus $V = U_{V(0)}$.

Let $W \in \cn(0)$, then by Lemma 8.1.5, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_{W}$. Thus $\mathfrak{V}\supset \fU$.

Let $V \in \mathfrak{V}$. Using Lemma 8.1.5, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_{0}$ with $W \subset V(0)$. In which case, $U_{W} \subset V$, and $\fU \supset \mathfrak{V}$.$\square$

Proof. Let $V \in \cn(0)$ be balanced and $U_{V} = \bracs{(x, y) \in E \times E| x - y \in V}$, then $y \in U_{V}(A)$ if and only if there exists $x \in A$ such that $(x, y) \in U_{V}$. This is equivalent to $x - y \in V$ and $y - x \in V$, so $U_{V}(A) = A + V$.

Assume without loss of generality that $\fB$ consists of symmetric entourages. By Proposition 8.1.6, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and Proposition 5.1.13 implies that

Proof. $(1)$: For every $x \in B$, $A + x$ is open by Definition 8.1.2, so

is open.

$(2)$: Let $x \in \overline{A + B}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap (A + B) \ne \emptyset$, so $(U - B) \cap A \ne \emptyset$, and $\fB = \bracs{U - B| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that

By Proposition 8.1.7, $\overline{(-B) + U}\subset (-B) + U + U$, so

Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus

so $x \in y + B \subset A + B$.$\square$

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

By Proposition 5.1.16, $E$ 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. Forward: By Proposition 8.1.11, 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. 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. For any $x \in E$ and $V \in \fB$, $x - x = 0 \in V$, so $\Delta \subset U_{V}$.

3. 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 5.1.8, 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. 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. 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 4.4.4, $\mathcal{S}= \mathcal{T}$.$\square$

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 8.1.12, the radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods, so the path-connected neighbourhoods of $0$ forms a fundamental system as well.$\square$