# Bases of Neighbourhoods for a Point in a Topological Vector Space

On the Topological Vector Spaces over the Field of Real or Complex Numbers page we saw that if $E$ is a topological vector space then for each $a \in E$ and for each $\lambda \in \mathbf{F}$ with $\lambda \neq 0$, the functions $f_a :E \to E$ and $f_{\lambda} : E \to E$ defined by:

(1)are homeomorphisms of $E$ onto $E$. If $o \in E$ denotes the zero vector in $E$ and if $\mathcal U_o$ is a base of neighbourhoods of $o$. In particular, for each $a \in E$, $\mathcal U_o + a = \{ U + a : U \in \mathcal U_o \}$ is a base of neighbourhoods of $a$. Thus the topology on a topological vector space can be described entirely by a base of neighbourhoods of the origin $o$.

Proposition 1: Let $E$ be a topological vector space and let $\mathcal U$ be a base of neighbourhoods of the origin. Then:(1) For each $U \in \mathcal U$, $U$ is an absorbent set.(2) For each $U \in \mathcal U$, there exists a $V \in \mathcal U$ with $V + V \subseteq U$.(3) For each $U \in \mathcal U$, there exists a balanced neighbourhood $W$ of the origin such that $W \subseteq U$.(4) For each $U \in \mathcal U$, there exists a balanced neighbourhood $V$ of the origin such that $V + V \subseteq U$.(5) For every convex $U \in \mathcal U$, there exists a convex and balanced neighbourhood $W$ of the origin such that $W \subseteq U$. |

*Consequently, if $E$ is a topological vector space then for every $x \in E$, if $\mathcal U_x$ is a base of neighbourhoods of $x$ then each $U \in \mathcal U_x$ is absorbent, for each $U \in \mathcal U_x$ there exists a $V \in \mathcal U_x$ with $V + V \subseteq U$, and for each $U \in \mathcal U_x$ there is a balanced neighbourhood $W$ of $x$ with $W \subseteq U$ (and moreover if $U$ is convex, then $W$ can be chosen to be convex).*

**Proof of (1):**Let $x \in E$. Let $f_x : \mathbf{F} \to E$ be defined for all $\lambda \in \mathbf{F}$ by:

- The function $f_x$ is a homeomorphism of $\mathbf{F}$ onto $\mathbf{F} x$, and is thus $f_x : \mathbf{F} \to \mathbf{F}x$ is continuous. Since every open subset of $\mathbf{F}x$ is the intersection of an open subset of $E$ with $\mathbf{F}x$, we see that if $V$ is any open subset of $E$, then $V \cap \mathbf{F}x$ is open in $\mathbf{F}x$, and consequently, $f^{-1}(V) = f^{-1}(V \cap \mathbf{F}_x)$ is open in $\mathbf{F}$.

- In particular, since $f_x$ is continuous at $0$ and since $U \in \mathcal U$ is a neighbourhood of $o = f_x(0)$ and since $f$ is continuous at $0$, there exists an $\epsilon > 0$ such that:

- Or equivalently, $\lambda U \subseteq U$ whenever $|\lambda| \leq \epsilon$. So $\displaystyle{U \subseteq \frac{1}{\lambda} U}$ whenever $|\lambda| \leq \epsilon$, i.e., whenever $\frac{1}{\epsilon} \leq \frac{1}{|\lambda|}$. This is equivalent to saying that:

- whenever $|\mu| \geq \frac{1}{\epsilon}$. Thus for each $x \in E$ the number $\frac{1}{\epsilon} > 0$ (where $\epsilon$ depends on $x$) is such that for all $\mu \in \mathbf{F}$ with $\displaystyle{|\mu| \geq \frac{1}{\epsilon}}$ we have that $U \subseteq \mu U$. So $U$ is an absorbent set. $\blacksquare$

**Proof of (2):**Since $E$ is a topological vector space, the operation $+ : E \times E \to E$ defined for all $x, y \in E$ by $(x, y) \mapsto x + y$ is continuous on all of $E \times E$. In particular, $+$ is continuous at $(o, o)$. Since $U$ is a neighbourhood of $o + o = o \in E$, by continuity, there exists two neighbourhoods $V_1$ and $V_2$ of $o$, such that $V_1 + V_2 \subseteq U$.

- But since $V_1$ and $V_2$ are neighbourhoods of $o$, so is $V_1 \cap V_2$. Set $V := V_1 \cap V_2$. Then:

**Proof of (3):**Since $E$ is a topological vector space, the operation $\cdot : \mathbf{F} \times E \to E$ is continuous on $\mathbf{F} \times E$ and in particular, continuous at $(0, o) \in \mathbf{F} \times E$. Since $U$ is a neighbourhood of $0 \cdot o = o$, by continuity, there exists an open neighbourhoods $0 \in A \subseteq \mathbf{F}$ and $o \in V \subseteq E$ such that $A \cdot V \subseteq U$. Since $A$ is an neighbourhood of numbers in $\mathbf{F}$ containing $0$, there exists an $\epsilon > 0$ such that the closed ball centered at $0$ with radius $\epsilon$ is contained in $A$. Thus:

- for all $\lambda \in \mathbf{F}$ with $|\lambda| \leq \epsilon$. Or equivalently: $V \subseteq \frac{1}{\lambda} U$ for all $\lambda \in \mathbf{F}$ with $|\lambda| \leq \epsilon$, and $\epsilon V \subseteq \frac{\epsilon}{\lambda} U$ for all $\lambda \in \mathbf{F}$ with $|\lambda| \leq \epsilon$. But this is equivalent to saying that:

- for all $\mu \in \mathbf{F}$ with $|\mu| \geq 1$. Let:

- We claim that $W \subseteq U$. To see this, let $w \in W$. By the definition of $W$, this means that $w \in \mu U$ for all $\mu \in \mathbf{F}$ with $|\mu| \geq 1$. Thus $\displaystyle{\frac{1}{\mu} w \in U}$ for all $\mu \in \mathbf{F}$ with $|\mu| \geq 1$. But this is equivalent to saying that $\delta w \in U$ for all $\delta \in \mathbf{F}$ with $|\delta| \leq 1$. Hence $\delta W \subseteq U$ for all $\delta \in \mathbf{F}$ with $|\delta| \leq 1$. Taking $\delta = 1$, and we see that $W \subseteq U$.

- Now observe that $W$ is a neighbourhood of $o$. Indeed, since $V$ is a neighbourhood of $o$, so is $\epsilon V$ (since the map $(\lambda, x) \mapsto \epsilon x$ is a homeomorphism). But $\epsilon V \subseteq \mu U$ for all $\mu \in \mathbf{F}$ with $|\mu| \geq 1$, and so $\epsilon V \subseteq W$. So $W$ is a neighbourhood of $o$.

- All that remains to show is that $W$ is a balanced set, i.e., show that $\delta W \subseteq W$ for all $\delta \in \mathbf{F}$ with $|\delta| \leq 1$.

- Clearly $0W = \{ o \} \subseteq W$. Let $\delta \in \mathbf{F}$, $\delta \neq 0$, and $|\delta| \leq 1$. Let $w \in W$. Then for all $\mu \in \mathbf{F}$ with $|\mu| \geq 1$ we have that $\left | \frac{\mu}{\delta} \right | \geq 1$. So if $w \in W$ then $w \in \frac{\mu}{\delta} U$, or equivalently, $\delta w \in \mu U$. Since this holds true for all $\mu \in \mathbf{F}$ with $|\mu| \geq 1$, we see that $\delta w \in W$. So $\delta W \subseteq W$ for all $\delta \in \mathbf{F}$ with $|\delta| \leq 1$. Hence $W$ is balanced.

- Hence, $W$ is a balanced neighbourhood of the origin with $W \subseteq U$. $\blacksquare$

**Proof of (4):**Let $U \in \mathcal U$ . By (2), there exists a neighbourhood $V'$ of the origin such that $V' + V' \subseteq U$. Then for the neighbourhood $V'$ of the origin, by applying (3), there exists a balanced neighbourhood $V$ of the origin such that $V + V \subseteq V'$. Thus $V + V \subseteq W$. $\blacksquare$

**Proof of (5):**If $U$ is convex, then $\mu U$ is convex for every $\mu \in \mathbf{F}$ with $\mu \neq 0$. By (3) there exists a balanced neighbourhood $W$ of the origin such that $W \subseteq U$ and by the construction of $\displaystyle{W = \bigcap_{|\mu| \geq 1} \mu U}$, we see that $W$ is an intersection of convex sets and is thus a convex and balanced neighbourhood of the origin with $W \subseteq U$. $\blacksquare$

Corollary 2: Let $E$ be a topological vector space. Then there exists a base of balanced neighbourhoods of the origin. |

*Consequently, in a topological vector space $E$, there exists a base of balanced neighbourhoods of every $x \in E$.*

**Proof:**Let $\mathcal U$ be a base of neighbourhoods of the origin. Then by (3) in the previous proposition, for each $U \in \mathcal U$ there exists a balanced neighbourhood of thee origin, $W_U$ with $W_U \subseteq U$. Let $\mathcal W = \{ W_U : U \in \mathcal U \}$.

- Given a neighbourhood $V$ of the origin, since $\mathcal U$ is a base of neighbourhoods of the origin there exists a $U \in \mathcal U$ such that $U \subseteq V$. But then $W_U \subseteq V$ too. So $\mathcal W$ is a base of balanced neighbourhoods of the origin.