In a Banach space, is the convex hull of finite set compact?
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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 $\mathrm{conv}\{x_1,\dots,x_n\}$. 


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. 

