Convex and Balanced Sets of Vectors

# Convex and Balanced Sets of Vectors

## Convex Sets of Vectors

 Definition: Let $E$ be a vector space. A subset $A \subseteq E$ is a Convex Set if for all vectors $x, y \in A$ and for all $\lambda, \mu \in \mathbf{F}$ with $\lambda + \mu = 1$, we have that $\lambda x + \mu y \in A$.

In $\mathbb{R}^n$, convex sets are fairly easy to visualize. A subset $A \subseteq \mathbb{R}^n$ is convex if, when given any pair of points $x, y \in \mathbb{R}^n$, the line segment connecting $x$ and $y$ is contained in $A$.

 Proposition 1: Let $E$ be a vector space. If $\{ M_i : i \in I \}$ is a collection of convex sets then $\bigcap_{i \in I} M_i$ is a convex set.

It should be remarked that in general, a union of a collection of convex sets is not convex.

• Proof: Let $M := \bigcap_{i \in I} M_i$. Let $x, y \in M$ and let $\lambda \in [0, 1]$. Then $x, y \in M_i$ for each $i \in I$ and so $\lambda x + (1 - \lambda) y \in M_i$ for each $i \in I$, by the convexity of $M_i$. Hence $\lambda x + (1 - \lambda)y \in M$. So $M$ is convex. $\blacksquare$.
 Proposition 2: Let $E$ be a vector space. Then: (1) If $A$ is convex then for all $x \in E$, $x + A$ is convex. (2) If $A$ is convex then for all $\lambda \in \mathbf{F}$, $\lambda A$ is convex.
• Proof of (1): Suppose that $A$ is convex and let $x \in E$. Let $x + y, x + z \in x + A$, with $y, z \in A$, and let $\lambda, \mu \in \mathbf{F}$ with $\lambda + \mu = 1$. Since $A$ is convex we have that $\lambda y + \mu z \in A$. Therefore:
(1)
\begin{align} \lambda (x + y) + \mu (x + z) = x + \lambda y + \mu z \in x + A \end{align}
• So $x + A$ is convex. $\blacksquare$
• Proof of (2): Suppose that $A$ is convex and let $\lambda \in \mathbf{F}$. Let $\lambda x, \lambda y \in \lambda A$, with $x, y \in A$, and let $\mu, \delta \in \mathbf{F}$ with $\mu + \delta = 1$. Since $A$ is convex we have that $\mu x + \delta y \in A$. Thus:
(2)
\begin{align} \mu \lambda x + \delta \lambda y = \lambda [\mu x + \delta y] \in \lambda A \end{align}
• So $\lambda A$ is convex. $\blacksquare$
 Proposition 3: Let $E$ be a vector space and let $A, B \subseteq E$. If $A$ and $B$ are convex then $A + B$ is convex.

## Balanced Sets of Vectors

 Definition: Let $E$ be a vector space. A subset $A \subseteq E$ is a Balanced Set if for all $\lambda \in \mathbf{F}$ with $|\lambda| \leq 1$, $\lambda x \in A$.

Equivalently, $A \subseteq E$ is a balanced set if $\lambda A \subseteq A$ for all $\lambda \in \mathbf{F}$ with $|\lambda| \leq 1$.

 Proposition 4: Let $E$ be a vector space. Then: (1) If $M$ is a nonempty balanced subset of $E$ then $o \in M$. (2) If $\{ M_i : i \in I \}$ is a collection of balanced subsets of $E$, then $\bigcup_{i \in I} M_i$ is a balanced set. (3) If $\{ M_i : i \in I \}$ is a collection of balanced subsets of $E$, then $\bigcap_{i \in I} M_i$ is a balanced set.
• Proof of (1): If $M$ is a nonempty, take $x \in M$. Then, since $M$ is balanced, for $\lambda = 0$, $\lambda x = 0x = o \in M$. $\blacksquare$
• Proof of (2): Let $\{ M_i : i \in I \}$ be a collection of balanced subsets of $E$ and let $M := \bigcup_{i \in I} M_i$. Let $\lambda \in \mathbf{F}$ be such that $|\lambda| \leq 1$. If $x \in M$ then $x \in M_i$ for some $i \in I$. Since $M_i$ is balanced, $\lambda x \in M_i$. But $M_i \subseteq M$, so $\lambda x \in M$. Thus $\bigcup_{i \in I} M_i$ is a balanced set. $\blacksquare$
• Proof of (3): Let $\{ M_i : i \in I \}$ be a collection of balanced subsets of $E$ and let $M := \bigcap_{i \in I} M_i$. Let $\lambda \in \mathbf{F}$ be such that $|\lambda| \leq 1$. If $x \in M$ then $x \in M_i$ for all $i \in I$. Since each $M_i$ is balanced, $\lambda x \in M_i$ for all $i \in I$. So $\lambda x \in M$. Thus $\bigcap_{i \in I} M_i$ is a balanced set. $\blacksquare$
 Proposition 5: Let $E$ be a vector space and let $A, B \subseteq E$. If $A$ and $B$ are balanced then $A + B$ is balanced.