Bounded Sets in a LCTVS

# Bounded Sets in a LCTVS

## Absorbing Sets in a Vector Space

Definition: Let $E$ be a vector space and let $A, B \subseteq E$. Then $A$ Absorbs $B$ if there exists a $\lambda > 0$ such that $B \subseteq \mu A$ for all $\mu \in \mathbf{F}$ with $|\mu| \geq \lambda$. |

Recall from the Absorbent Sets of Vectors page that if $E$ is a vector space and $A \subseteq E$ then $A$ is said to be **Absorbent** if for each $x \in E$ there exists a $\lambda > 0$ such that $x \in \mu A$ for all $\mu \in \mathbf{F}$ with $|\mu| \geq \lambda$. Thus $A$ is absorbent if for each $x \in E$, $A$ absorbs $\{ x \}$.

## Bounded Sets in a Locally Convex Topological Vector Space

Definition: Let $E$ be a locally convex topological vector space and let $A \subseteq E$. Then $A$ is said to be Bounded if for every neighbourhood $U$ of the origin, $U$ absorbs $A$. |

Equivalently, $A$ is bounded if for every neighbourhood $U$ of the origin, there exists a $\lambda_U > 0$ such that:

(1)\begin{align} \quad A \subseteq \mu U \end{align}

for all $\mu \in \mathbf{F}$ with $|\mu| \geq \lambda_U$.

Proposition 1: Let $E$ be a locally convex topological vector space, $A \subseteq E$, and let $\mathcal U$ be a base of absolutely convex neighbourhoods of the origin. Then $A$ is bounded if and only if for every $U \in \mathcal U$ there exists a $\lambda_U > 0$ such that $A \subseteq \lambda_U U$. |

**Proof:**$\Rightarrow$ If $A$ is bounded, then by definition, for each $U \in \mathcal U$ there exists a $\lambda_U > 0$ such that $A \subseteq \mu U$ for all $\mu \in \mathbf{F}$ with $|\mu| \geq \lambda_U$. In particular, set $\mu = \lambda_U$. Then $A \subseteq \lambda_U U$.

- $\Leftarrow$ Suppose that for every $V \in \mathcal U$ there exists a $\lambda_V > 0$ such that $A \subseteq \lambda_V V$. Let $U$ be ANY neighbourhood of the origin and let $V \in \mathcal U$ be such that $V \subseteq U$.

- Let $\mu \in \mathbf{F}$ be such that $|\mu| \geq \lambda_V$. Then:

\begin{align} \quad A \subseteq \lambda_V V \subseteq \mu V \subseteq \mu U \end{align}

- where the inclusion $\lambda_V V \subseteq \mu V$ comes from $V$ being absolutely convex and one of the propositions on the Absolutely Convex Sets of Vectors page. Thus $A$ is bounded. $\blacksquare$

Proposition 2: Let $E$ be a vector space, $Q$ a set of seminorms on $E$, and equip $E$ with the locally convex topology determined by $Q$ so that the sets of the form $\displaystyle{\{ x : \sup_{1 \leq i \leq n} p_i(x) \leq \epsilon \}}$ with $p_1, p_2, ..., p_n \in Q$ form a base of neighbourhoods of the origin, and let $A \subseteq E$. Then $A$ is bounded if and only if $p(A)$ is a bounded set of nonnegative real numbers for each $p \in Q$. |

**Proof:**$\Rightarrow$ Suppose that $A$ is bounded. Let $p \in Q$ and let $U$ be an absolutely convex and absorbent neighbourhood of the origin for which $p = p_U$ where $p_U$ is the gauge of $U$. Then there exists a $\lambda_U > 0$ such that $A \subseteq \lambda_U U$. Thus, for each $a \in A$ we have that $a = \lambda_U u$ for some $u \in U$, and so:

\begin{align} \quad p(a) = p_U(a) = p_U(\lambda_U u) = \lambda_U p_U(u) \leq \lambda_U \cdot 1 = \lambda_U \end{align}

- So $p(A) \subseteq [0, \lambda_U]$, and so $p(A)$ is a bounded set of nonnegative real numbers.

- $\Leftarrow$ Suppose that $p(A)$ is a bounded set of nonnegative real numbers for each $p \in Q$. Let $U$ be any absolutely convex and absorbent set. Then there exists $\epsilon_U > 0$ and $p_1, p_2, ..., p_n \in Q$ such that:

\begin{align} \quad V := \{ x : \sup_{1 \leq i \leq n} p_i(x) \leq \epsilon_U \} \subseteq U \end{align}

- For each $1 \leq i \leq n$, let $\mu_i > 0$ be such that $p_i(a) \leq \mu_i$ for all $a \in A$. If $\mu_U := \max \{ \mu_1, \mu_2, ..., \mu_n \}$ then $p_i(A) \subseteq [0, \mu_U]$ for all $1 \leq i \leq n$. Therefore:

\begin{align} \quad A \subseteq \epsilon_U \mu_U^{-1} V \subseteq \epsilon_U \mu_U^{-1} U \end{align}

- By setting $\lambda_U := \epsilon_U \mu_U^{-1}$ we see that $A \subseteq \lambda_U U$. So by the previous proposition, $A$ is bounded. $\blacksquare$