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

# Subsets, Scalar Multiples, Finite Unions, and Arbitrary Intersections of Bounded Sets in a LCTVS

 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}
• for all $1 \leq i \leq n$. Therefore:
(2)
\begin{align} \quad \bigcup_{i=1}^{n} A_i \subseteq \lambda_U U \end{align}
• and hence $\displaystyle{\bigcup_{i=1}^{n} A_i}$ is bounded. $\blacksquare$
• 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. $\blacksquare$