Subsets, Scalar Mults., Fin. Uns., and Arb. Ints. of Bounded Sets

Table of Contents

Proposition 1: Let

$E$

be a locally convex topological vector space and let

A \subseteq E

. If

$A$

is a bounded set then so it any subset or scalar multiple of

$A$

Proof: Let $\mathcal U$ be a base of absolutely convex neighbourhoods of the origin. Since $$A$$ is bounded, for each $U \in \mathcal U$ there exists a $\lambda_U > 0$ such that $A \subseteq \lambda_U U$ .

Thus, if $B \subseteq A$ then $B \subseteq \lambda_U U$ , so that $$B$$ is bounded too.

Furthermore, $\mu A \subseteq (\mu \lambda_U) U$ , so that $\mu A$ is bounded too, for each $\mu \in \mathbf{F}$ . $\blacksquare$

Proposition 2: Let

$E$

be a locally convex topological vector space.
(1) If

$A_1, A_2, ..., A_n$

is a finite collection of bounded sets, then

\displaystyle{\bigcup_{i=1}^{n} A_i}

is bounded.
(2) If

\{ A_{\alpha} \}_{\alpha}

is an arbitrary collection of bounded sets, then

\displaystyle{\bigcap_{\alpha} A_{\alpha}}

is bounded.

Proof of (1): Let $\mathcal U$ be a base of absolutely convex neighbourhoods of the origin. Since each $$A_i$$ is bounded for each $U \in \mathcal U$ there exists a $\lambda_{U, i} > 0$ such that $A_i \subseteq \lambda_{U, i} i$ for $1 \leq i \leq n$ . Let $\lambda_U = \max_{1 \leq i \leq n} \{ \lambda_{U_i} \}$ . Then $\lambda_{U, i} \leq \lambda_U$ for each $1 \leq i \leq n$ , and by the absolute convexity of $$U$$ :

(1)

\begin{align} \quad A_i \subseteq \lambda_{U, i} U \subseteq \lambda_U U \end{align}

(2)

\begin{align} \quad \bigcup_{i=1}^{n} A_i \subseteq \lambda_U U \end{align}

Proof of (2): Let $\mathcal U$ be a base of absolutely convex neighbourhoods of the origin. Take any $\alpha'$ . Then $A_{\alpha'}$ is bounded and:

(3)

\begin{align} \quad \bigcap_{\alpha} A_{\alpha} \subseteq A_{\alpha'} \end{align}

But by Proposition 1, subsets of bounded sets are bounded. Thus $\displaystyle{\bigcap_{\alpha} A_{\alpha}}$ is bounded. $\blacksquare$

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