# Intuition Behind Balanced Sets

Suppose $B \subset X$ where $X$ is a vector space. $B$ is called balanced if $\alpha B \subset B$ for every $\alpha \in \Phi$ with $|\alpha| \leq 1$. Note that $\Phi = \textbf{R}$ or $\Phi = \textbf{C}$. What is the intuition behind balanced sets? Why do we define them? Is it "good" for a vector space to have more balanced sets? Why do we require $| \alpha| \leq 1$?

-
Think about this: Look at $\mathbb{R}^2$ as a vector space over $\mathbb{R}$ ($\mathbb{C}$), what is the smallest balanced set containing $A=\{(1,1)\}$, $B=\{(1,1),(0,1)\}$, $C=\{(x,y):x^2+y^2\le 1\}$? –  AD. Oct 27 '10 at 18:11
The unit circle? –  PEV Oct 28 '10 at 1:44
Do you adopt $|\alpha|\le 1$ (KennyTM edited the text)? Think it over again (using pen+paper), $\alpha$ is real over $\mathbb{R}$ and complex over $\mathbb{C}$. In one occasion you WILL get the unit disc (usually unit circle is the boundary of the unit disc)! –  AD. Oct 28 '10 at 4:37

[Added in response to the comments below:] Despite what is written below, balanced needn't imply convex; it simply doesn't follow from the definition. So one should take the following discussion as rather an explanation about convex balanced sets, for whatever that is worth.

Balanced means that the set is symmetric around the origin (i.e. invariant under $v \mapsto -v$) and is convex. These are nice properties, generalizing some useful features of balls around the origin in $\mathbb R^n$. (Incidentally, if you didn't have the condition $|\alpha| = 1$, then any balanced subset with non-empty interior (in $\mathbb R^n$ say) would be all of $\mathbb R^n$, so that wouldn't be such an interesting condition.)

Also, any vector space will have plenty of such sets; you can form them just by taking the convex hull of a collection of points invariant under $v \mapsto -v$. What is more relevant in functional analysis is whether the origin of a topological vector space contains a basis of balaced neighbourhoods. The topological vector space is then called locally convex and this has important consequences for the space (especially for its duality theory).

Note that balls around the origin in a normed space are automatically balanced, so normed spaces are locally convex; so again, in the topological setting, one sees that this is a way of capturing some the geometric aspects of normed vector spaces, without insisting on the existence of a norm. (So, while normed spaces are not closed under certain operations, such as forming infinite direct sums, locally convex spaces are --- any direct sum (just to take one example) of a collection of locally convex spaces has a natural locally convex topology of its own.)

-
Is a balanced set necessary convex? If over a real vector space, I thought it suffices that the set is star-shaped w.r.t. 0. –  Willie Wong Oct 27 '10 at 18:49
I also don't see why if you didn't have $\lvert\alpha\rvert \le 1$, a balanced subset couldn't be an infinite star-shaped region, like a double cone, and not all of $\mathbb{R}^n$. –  Rahul Oct 27 '10 at 19:17
@Willie, @Rahul: Dear Willie and Rahul, Thanks a lot, you are both correct; I was thinking too much about the locally convex set-up, rather than what was literally under discussion. I have added a big caveat at the start of my answer. –  Matt E Oct 27 '10 at 20:24

It's like balancing a tire, you spin the set around! (A bit tongue in cheek, but that's how I've always felt about it...)

-