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In the definition of the convex hull of a set $A$, does it matter if each element of the convex hull is a convex combination of finite number of elements of $A$ or an infinite number?

I am trying to prove that the convex hull of a bounded set is bounded. So, I take an element of $co(A)$ (the notation I use for convex hull) and apply the Triangle Inequality.

$$\|x\|=\|\sum_{i=1}^n\alpha_iy_i\|$$ where $y_i\in A$ and $\alpha_i\in[0,1]$ and $\sum_{i=1}^n\alpha_i=1$ for some finite $n$. $$\leq \sum_{i=1}^n\alpha_i \| y_i\| < r$$ where $r$ is the radius of the ball that contains the bounded set $A$.

Now, if $n$ is infinite then I'm not sure if I can still apply the Triangle Inequality and, hence, my question.

P.S. I'm using the following definition of convex hull: $$co(A)=\{ x\in X| x=\sum_{i=1}^n t_iy_i \textrm{ for some } n\geq 1, \textrm{ where } t_i\in[0,1], \sum_{i=1}^n t_i=1, \ y_i\in A \}$$ where $X$ is a normed vector space.

Teodorism
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    Is this calculus, real analysis, linear algebra, functional analysis...? I ask just because we have to be careful about series/infinite sums... – Luiz Cordeiro Sep 23 '19 at 01:01
  • @LuizCordeiro Functional analysis! – Teodorism Sep 23 '19 at 01:04
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    So are you taking the closed convex hull? The convex hull of $A$ is the intersection of all convex sets containing $A$, and it coincides with the set of finite affine combinations of $A$. The closed convex hull is the intersection of closed convex subsets of $A$, and it coincides with the set of affine (possibly infinite, as you're considering) combinations. These two may be different – Luiz Cordeiro Sep 23 '19 at 01:06
  • @LuizCordeiro Thank you. I'd never looked at it that way. I just added the definition of convex hull I'm using to the question. – Teodorism Sep 23 '19 at 01:10
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    This is true for general normed spaces, so you should not use Cauchy-Schwarz. Try using the triangle inequality instead. Also note that the definition of convex hull uses finite $n$. – Hans Engler Sep 23 '19 at 01:12
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    Oh, ok. So in your definition no infinite sums are considered, and your proof is already complete. The triangle inequality (which is what you're really using) is still valid for infinite sums. – Luiz Cordeiro Sep 23 '19 at 01:12
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    @LuizCordeiro The closed convex hull does not necessarily correspond to the set of infinite convex combinations. For example, any convex open set is closed under the taking of infinite convex combinations. – Theo Bendit Sep 23 '19 at 01:15
  • @TheoBendit I just noticed that. I should've been more careful, thanks. – Luiz Cordeiro Sep 23 '19 at 01:17
  • It would improve the Question to describe what kind of set $A$ is. If $A\subset \mathbb R^n$, then at most $n+1$ terms are needed to express any particular point in the convex "hull". – hardmath Sep 23 '19 at 01:32
  • $A\subset X$ a normed vector space. – Teodorism Sep 23 '19 at 01:33
  • To solve your problem, $A$ being bounded means that $A$ is contained in a (convex) ball. This ball is closed under the taking of finite convex combinations, so $\operatorname{cl} A$ must also be contained in this same ball, making it bounded. – Theo Bendit Sep 23 '19 at 01:47

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