In a Banach space, is the convex hull of finite set compact?
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18$\begingroup$ The convex hull of a set of $n$ points is the image of the $n$-simplex in $\mathbb R^n$ under a continuous function, so yes. $\endgroup$– Mariano Suárez-ÁlvarezCommented Apr 26, 2010 at 5:46
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4$\begingroup$ Mariano: do you want to write this as an answer (community wiki if you wish) so I can vote it up? I think it's slightly more to the point than Pete's answer (no offence, Pete!) which works fine but I think is slightly over-elaborate. To be fair, the same idea underlies both. $\endgroup$– Yemon ChoiCommented Apr 26, 2010 at 8:05
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4$\begingroup$ I am also not keen on questions which give little to no indication of (a) why the questioner wants to know (b) what they've tried doing (c) what level they are at. $\endgroup$– Yemon ChoiCommented Apr 26, 2010 at 8:06
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1$\begingroup$ @Yemon: I agree so much that I deleted my answer. $\endgroup$– Pete L. ClarkCommented Apr 26, 2010 at 16:58
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3 Answers
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Suppose $X$ is your Banach space and let $\{x_1,\dots,x_n\}$ be a finite subset of $X$. Let $$S=\{(t_1,\dots,t_n)\in\mathbb R^n:t_1,\dots,t_n\geq0,\\,t_1+\cdots+t_n=1\}$$ be the standard simplex in $\mathbb R^n$. The map $$\phi:(t_1,\dots,t_n)\in S\mapsto t_1x_1+\cdots+t_nx_n\in X$$ is evindently continuous and its image is $\mathrm{conv}\{x_1,\dots,x_n\}$. Since $S$ is compact, so is $\mathrm{conv}\{x_1,\dots,x_n\}$.
answered Apr 26, 2010 at 15:20
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1$\begingroup$ Just to make perfectly clear what is used in the proof: $\phi$ is continuous because $X$ is a topological vector space and $S$ is compact as $X$ is a separated space. $\endgroup$ Commented Apr 26, 2010 at 15:46
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2$\begingroup$ $S$ is compact because it is closed and bounded in $\mathbb R^n$, rather! :) $\endgroup$ Commented Apr 26, 2010 at 15:48
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$\begingroup$ Sorry, misread I thought $S$ was the image. $\endgroup$ Commented Apr 26, 2010 at 16:48
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Of course yes: the n points lie in the finite-dimentional linear subspace generated by themselves (remember that any norm, when restricted to a finite-dimensional linear subspace, gives rise to the same topology on that space).
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Actually, the convex hull of a sequence of points $(x_n)$ is (relatively) compact when $x_n\rightarrow 0$, and this easily gives a positive answer to your question (but is somewhat overkill). In fact, a closed convex set K in a Banach space is compact if and only if it's contained in the closed convex hull of a sequence $(x_n)$ with $x_n\rightarrow 0$. See, for example, Lindenstrauss and Tazfriri, vol I, Proposition 1.e.2.
answered Apr 26, 2010 at 9:52
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1$\begingroup$ Actually, the statement in the question holds for all topological vector spaces, not just Banach spaces. The generalisation here does not apply so widely. I think you need local convexity. $\endgroup$ Commented Nov 4, 2011 at 12:21