# Is the convex hull of a compact set compact?

Let $$V$$ be a normed vector space and $$K$$ a compact subset of $$V$$. Is the convex hull of $$K$$ given by $$\langle K \rangle = \left\{\sum_{i=1}^n t_i x_i\mid x_i\in K, t_i≥0\text{ s.t. }\sum_i t_i=1\right\}$$ again compact?

Consider $$u_n=(\underbrace{0,...,0}_{n-1},1/n,0,...)$$ and $$K=\bigcup_n \{u_n\} \cup \{0\}$$ a compact subset of $$\mathscr l^p(\mathbb N)$$. The convex hull of $$K$$ is given by elements of the form: $$\sum_{n=1}^k a_n u_{n}\qquad\text{s.t.:}\quad \sum_{n=1}^k a_n≤1\qquad a_n≥0$$ So also $$\sum_{n=1}^k 2^{-n}u_n$$ lies in it. But this sequence converges to $$\sum_{n=1}^\infty 2^{-n}u_n$$ which does not lie in it.
For convenience we include the proof of the book, which shows the statement in the setting of completely metrisable locally convex vector spaces. More specifically one shows that for $$K$$ compact the convex hull $$\langle K\rangle$$ is completely bounded.
Let $$\epsilon>0$$, since $$K$$ is compact there is a finite covering of $$K$$ by balls of radius $$\frac\epsilon2$$, it is convenient to write this as: $$K\subseteq F+B_{\epsilon/2}(0)$$ for a finite set $$F$$. It then follows that: $$\langle K\rangle \subseteq \langle F\rangle +B_{\epsilon/2}(0)$$ because $$B_{\epsilon/2}(0)$$ is already convex. Now since $$F$$ is finite one has that $$\langle F\rangle$$ is compact and hence admits a covering by finitely many balls of radius $$\frac\epsilon2$$, write $$\langle F\rangle = \widetilde F + B_{\epsilon/2}(0)$$ for some finite set $$\widetilde F$$, then: $$\langle K \rangle \subseteq \langle F\rangle + B_{\epsilon/2}(0)\subseteq \widetilde F + B_{\epsilon/2}(0)+B_{\epsilon/2}(0)\subseteq \widetilde F + B_{\epsilon}(0)$$ Giving the conclusion that for any $$\epsilon>0$$ you may cover $$\langle K\rangle$$ by finitely many balls of radius $$\epsilon$$, whence $$\langle K \rangle$$ is totally bounded.