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 $| \alpha| \leq 1$?  
 A: [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.
[Original answer follows:]
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.)
A: In Complex analysis in locally convex spaces by Seán Dineen, it states that:
Balanced open sets arise as the natural domain of convergence of Taylor series expansion at the origin of holomorphic functions.
A: With regard to the above answer made by (now) @anonymous, the "natural domains of convergence of Taylor series" are logarithmically-convex complete Reinhardt domains. Two questions arise: 
Q1. What do these words mean?
Q2. What does natural domain of convergence mean?
To answer these questions, let us first note by "domain", we mean a connected open set in $\mathbb{C}^n$. I do not want to touch on anything outside of domains in $\mathbb{C}^n$. 
A domain $\Omega \subseteq \mathbb{C}^n$ is said to be a Reinhardt domain centred at $w \in \Omega$ if, for any point $z = (z_1, ..., z_n) \in \Omega$, the rotations $(w_1 + z_1 e^{i\vartheta_1}, ..., w_n + z_n e^{i\vartheta_n})$ lie in $\Omega$ also, where $\vartheta_j \in \mathbb{R}$ for all $j \in \{ 1, ..., n \}$. If $w = (w_1, ..., _n)$ is the origin in $\mathbb{C}^n$, I believe this is the definition of balanced. 
A Reinhardt domain (centred at $w \in \mathbb{C}^n$) $\Omega \subseteq \mathbb{C}^n$ is said to be complete if for any point $\zeta \in \Omega$, the polydisk $$\Delta(w, \zeta - w) := \left \{ z \in \mathbb{C}^n : \left| z_j - w_j \right| \leq \left| \zeta_j - w_j \right|, 1 \leq j \leq n \right \}$$ is contained in $\Omega$. In other words, a Reinhardt domain is complete if it can be "filled up by polydisks".
To describe logarithmic convexity, let us denote by $\mathcal{D}$ the complement of all coordinate hyperplanes in $\mathbb{C}^n$, i.e., $$\mathcal{D} = \{ z= (z_1, ..., z_n) \in \mathbb{C}^n : z_j \neq 0, \ \forall \ 1 \leq j \leq n \}.$$ Then a set $\Omega \subseteq \mathbb{C}^n$ is logarithmically-convex if $$\lambda(\Omega \cap \mathcal{D}) = \left \{ \zeta = (\zeta_1, ..., \zeta_n) : \zeta_j = \log | z_j |, \ 1 \leq j \leq n \right \}$$ is a convex set. 
One may show without too much difficulty that the domain of convergence of a power series is a complete Reinhardt domain. If one asks the converse question, however, i.e., whether, given a complete Reinhardt domain $\Omega \subseteq \mathbb{C}^n$, does there exist a power series whose domain of convergence is exactly $\Omega$, the answer is negative. If $\Omega$ is further assumed to be logarithmically-convex however, then it will be true that we can find a power series with $\Omega$ as the domain of convergence. 
Correct me if I am wrong, but I believe that complete in the sense I have just described, corresponds to absorbent in functional analysis? 
Hope this helps. 
