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3

For example, in an infinite dimensional Banach space, open balls are not open in the weak topology.


2

Yes, the balanced core of an open set is open. Let $x$ be a point in the balanced core $U^\ast$ of $U$. Then $K = \overline{\mathbb{D}}\cdot x = \{\lambda\cdot x : \lvert\lambda\rvert \leqslant 1\}$ is a subset of $U$, and $K$ is balanced, so $K\subset U^\ast$. Furthermore, as the image of the compact set $\overline{\mathbb{D}}$ under the continuous map ...


2

If there is an $0 < \alpha < 1$ such that $\alpha^{-1}x \in A$, then $$x = (1-\alpha)\cdot 0 + \alpha\cdot(\alpha^{-1}x) \in A,$$ since $A$ is convex, and $0\in A$ (because $A$ is absorbing).


2

Check out http://www.math.uwaterloo.ca/~lwmarcou/Preprints/LinearAnalysis.pdf This was the book used for my functional analysis course. It is a functional text more than a topology text book, but has chapters on those 3 bullets you listed and other related topics. I liked it very much.


1

Let me explain this topology in my own words. Let $A,B$ be commutative $k$-algebras. Assume that $N:=\dim_k(B)$ is finite, so that there is an isomorphism of vector spaces $B \cong k^N$. Hence, there is an isomorphism of vector spaces $A \otimes_k B \cong A^N$. If $A$ is a topological vector space, it follows that also $A^n$ and thereby $A \otimes_k B$ is a ...


1

EDIT: Let $x_1+W_1$, $x_2+W_2$ in $\mathcal{B}$ and $x\in (x_1+W_1)\cap(x_2+W_2)$. Let's show that there exists $W_3\in\mathcal{U}$ such that $x+W_3\subseteq (x_1+W+1)\cap(x_2+W_2)$ (so that we are considering $x_3=x$). Let $V=(W_1+x_1-x)\cap (W_2+x_2-x)$. Note that $V$ is convex and contains $0$, so we have to work on balancedness. Let ...



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