Properties of Balanced Sets of Vectors

Proposition 1: Let $E$ be a vector space. If $A \subseteq E$ is a balanced set then for all $\lambda \in \mathbf{F}$, $\lambda A = |\lambda| A$.

• Proof: Clearly, if $\lambda = 0$ then the conclusion follows immediately.

• Suppose that $\lambda \neq 0$. Then observe that $\displaystyle{\frac{\lambda}{|\lambda|}, \frac{|\lambda|}{\lambda} \in \mathbf{F}}$ are such that $\displaystyle{\left | \frac{\lambda}{|\lambda|} \right | = 1}$ and $\displaystyle{\left | \frac{|\lambda|}{\lambda} \right | = 1}$. Since $A$ is balanced, we have that:

(1)

\begin{align} \quad \frac{\lambda}{|\lambda|} A \subseteq A \quad \mathrm{and} \quad \frac{|\lambda|}{\lambda} A \subseteq A \end{align}

• Therefore:

(2)

\begin{align} \lambda A \subseteq |\lambda| A \quad \mathrm{and} \quad |\lambda| A \subseteq \lambda A \end{align}

• So $\lambda A = |\lambda| A$. $\blacksquare$

Proposition 2: Let $E$ be a vector space. If $A \subseteq E$ is balanced and if $\lambda, \mu \in \mathbf{F}$ are such that $|\lambda| \leq |\mu|$ then $\lambda A \subseteq \mu A$.

• The conclusion is trivially true when either $\lambda = 0$ or $\mu = 0$.

• So suppose that $\lambda, \mu \neq 0$. Since $\mu A$ is balanced and since $\left | \frac{\lambda}{\mu} \right | \leq 1$, we have that:

(3)

\begin{align} \lambda A = \frac{\lambda}{\mu} (\mu A) \subseteq \mu A \quad \blacksquare \end{align}