In a Banach space, is the convex hull of finite set compact?

closed as too localized by George Lowther, Matthew Daws, Andreas Blass, Bill Johnson, Alain Valette Nov 4 '11 at 15:26
This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.
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 $\mathrm{conv}\{x_1,\dots,x_n\}$. 


Of course yes: the n points lie in the finitedimentional linear subspace generated by themselves (remember that any norm, when restricted to a finitedimensional linear subspace, gives rise to the same topology on that space). 


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. 

