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The Answer to question 1 is yes: Let $V = \mathbb R^n$ and $I = \mathbb N$. It is clear that $\operatorname{conv}(\bigcap_{i \in \mathbb N} C_i) \subset \operatorname{conv}(C_j)$ for every $j \in \mathbb N$ since $\bigcap_{i \in \mathbb N} C_i \subset C_j$ for every $j \in \mathbb N$. Thus $\operatorname{conv}(\bigcap_{i \in \mathbb N} C_i) \subset ... 2 In general, if$X$is a locally convex topological vector space of uncountable dimension (as a linear space), then the weak$^*$topology on$X^*$is not first countable. Proof. In the weak$^*$topology a sub-base of the neighborhoods of$0$is obtained by sets of the form$$W_{x,\varepsilon}=\{x^*\in X^*: |x^*(x)|<\varepsilon\}, \quad ... 1 You seem to be slighly confuzed when you are writing about the vectors. You say that$j$can be derived from$i$by the process of differentiation, which is not true. In fact,$i$is a constant vector, not a function, you can only have the differential of a function. As far as the definition of independency goes, you have$2$equivalent versions: The ... 1 The starting idea to take a sequence$x_n$in$S$that converges to some$x$and see whether$x\in S$is OK, the rest is nonsense, e.g. nobody said that$x=0$neither that$x_i^{(n)}=0$, or$S$may have more than countably infinite dimensions, whence the sequence setting$x=(x_1,x_2,\dots)$is not that much justified. Probably the fastest argument to ... 1 Suppose$p \notin S$, and let$v_1,...v_n$span$S$. Define the functional$f$on the space spanned by$p,v_1,...v_n$by$f(v_k) = 0$,$f(p) = 1$. Use Hahn Banach to extend this functional to the whole space. Let$U = \{ x | f(x) > {1 \over 2} \}$which is open, and since$f(s) = 0$when$s \in S$, we see that$S \cap U = \emptyset$, hence$S^c\$ is open. ...