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In a topological vector space $X$, a subset $S$ is convex if \begin{equation}tS+(1-t)S\subset S\end{equation} for all $t\in (0,1)$.

$S$ is balanced if \begin{equation}\alpha S\subset S\end{equation} for all $|\alpha|\le 1$.

So if $S$ is balanced then $0\in S$, $S$ is uniform in all directions and $S$ contains the line segment connecting 0 to another point in $S$.

Due to the last condition it seems to me that balanced sets are convex. However I cannot prove this, and there are also evidence suggesting the opposite.

I wonder whether there is an example of a set that is balanced but not convex.

Thanks!

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    What "$S$ is balanced" tells you is that the intersection of $S$ with every one-dimensional subspace is either the subspace itself, a disk centered at the origin or $0$. It doesn't tell you anything about affine segments that don't lie in a one-dimensional linear subspace, which is what convexity does. It is somewhat similar to *starshaped* versus convex.2012-09-16

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Consider the vector space $\mathbb{R}^2$ and the set $S=\{(x,y)|x=0 \text{ or } y=0\}$, i.e. the union of the two axes. Then $S$ is balanced but not convex.

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    @ChrisEagle I thought you consider $\mathbb{R}^2$ as $\mathbb{C}$2012-09-16
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Astroid: $|x|^{2/3}+|y|^{2/3} \le 1$.

http://en.wikipedia.org/wiki/Astroid

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The interior of a regular pentagram centered at the origin is balanced but not convex.

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    That's not balanced: If $x$ is near one of the corners then $-x$ is not in the pentagram. Similar stars work if you take an even number of corners.2012-09-16