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I have a question about proposition in this set of notes (https://www.math.ubc.ca/~cass/research/pdf/TVS.pdf) on topological vector spaces.

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We are in a complex vector space $V$. A subset $S$ is said to be convex if for any $v, w \in S$, $tv +(1-t)w \in S$ for all $t \in [0,1]$, balanced if $|c| = 1, v \in S$ implies $cv \in S$.

In the proof, it says we replace a convex neighborhood $U$ of $0$ with the union $$\bigcup\limits_{|c| \leq 1} cU$$ It is clear that this union is balanced, but I don't understand why it should remain convex.

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    Since $U$ is convex and contains 0, you can show that $U = \bigcup_{|c| \le 1} cU$. – 2017-02-27
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    I don't think that's true. Otherwise, why would anyone bother with the definition of balanced? – 2017-02-27
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    I don't see why the equality would render useless the definition of balanced. Note that the assumptions that $U$ is convex and contains 0 are important here. Which of the containments $U \subseteq \bigcup_{|c|\le 1} cU$ and $\bigcup_{|c|\le 1} cU \subseteq U$ do you disagree with? – 2017-02-27
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    The second one. – 2017-02-27
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    But convexity only tells you that $|c|u + (1-|c|)0 = |c|u \in U$, not $cu$. – 2017-02-27
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    I see your point. – 2017-02-27
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    For a counterexample with real scalars, take $U$ to be the interior of a triangle in $\mathbb{R}^2$ containing the origin. The union $\bigcup_{|c| \leq 1} cU$ looks like a bowtie and certainly isn't convex. – 2017-03-03

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