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New answers tagged topological-vector-spaces

0

No, it is not first countable as it is not metrizable. Some arguments are mentioned here, observing that $C_b(Compact)$ has an uncountable Hamel basis.

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I think I found the answer.It's a modification of the proof in "Introduction to Tensor Products of Banach spaces, R.A.Ryan" Prop4.6. Suppose $x^{**}$ is an element of the dual of $W^*$ endowed with the given topology. It's easy to see that there exsist a compact subset $K\subset X$, s.t. $$|x^{**}(f)|\leq \sup_{x\in K}|f(x)|,\forall f\in W^*.$$ Use that ...

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. ...

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 ...

3

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 ... 0 The dual space is a concept that shows up in greater generality in Linear Algebra. If$V$is a vector space then its dual space$V^*$is the set of all linear functionals from$V$to its base field$F$. When$V$is a finite dimensional vector space, then so is$V^*$and$\text{dim } V^*=\text{dim }V$. Thus there is a linear isomorphism between them when ... 2 More generally, if$X$and$Y$are sets (resp. vector spaces over a common field$k$), we can consider the set$Y^X$of all mappings (resp. the vector space$\mathrm{Hom}(X,Y)$of all linear mappings) from$X$to$Y$. These are obviously interesting to study if you think in terms of structures and structure preserving maps; and dual spaces are simply a ... 0 There is a textbook by Halmos, Measure Theory, in which he does as much as possible in the setting of$\sigma$-rings, where complements are not assumed. But subsequent mathematicians have not adopted that point of view. (1) In the Halmos definitions, we can postulate that every measurable set is$\sigma$-finite; then when (rarely) it is necessary, extend ... 1 While not an answer per se, here is an example to get you thinking about why you might want more than countable disjoint unions. Consider the space$[0,\infty)\subset \mathbb R$, and take as your "algebra" the sets of the form$[0,x)\$. The only step functions on this space contain only one step, and the sum of two step functions will not in general be ...

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