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By the lemma you cited, you can lift $W_3$ to $X_{n+1}$ (maintaining convexity, balancedness), and then to $X_{n+2}$, to $X_{n+3}$, etc... If you let $\widetilde{W}^\infty_3$ be the union of all those ''lifts'', then $\widetilde{W}^\infty_3\in\mathcal{U}$, so also $\widetilde{W}_3:=\widetilde{W}^\infty_3\cap W_1\cap W_2\in\mathcal{U}$, and $\widetilde{W}_3$ ...


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Starting with step 3 in the question above, we need to modify things slightly. First, change the definition of $Z$ slightly from the one above. $$Z = \{\alpha x + \beta y : x \in O_1, y \in V, |\alpha| + |\beta| \le 1\} $$ So we are using the condition $|\alpha| + |\beta| \le 1$ whereas, above we use $|\alpha| + |\beta| = 1.$ It's easy to see: $$Z = ...


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


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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 ...


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).


-1

Proof: Let $V$ be a finite dimensional Hausdorff topological vector space and $E$ an Euclidean vector space of the same dimension. Since both dimensions agree there is a linear bijection $L:E\to V$. Now since the Euclidean space carries the product topology the projections $\pi_k:E\to\mathbb{K}$ are continuous. Similar vector addition and scalar ...


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 ...


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Uniform Structure Firstly, $\Phi$ is upward closed by construction. Next, note that $U_M\cap U_N=U_{M\cap N}$. Thus if $A,B\in\Phi$ then so also $A\cap B\in\Phi$. Now, $\Phi$ is nonempty since so is $\mathcal{N}_0$ Also, $\Delta\subseteq U$ for $U\in\Phi$ since $0\in N$ for any $N\in\mathcal{N}_0$. Moreover, inversion (dilation by the factor minus one) ...


2

I think this works: Assume without loss of generality that $V = \mathbb{F}^n$. Let $\tau$ be the Euclidean topology and $\tau_1$ be any other (Hausdorff) linear topology on $V$. a) Suppose $x_{\alpha} \to 0$ in $(V,\tau)$, then each component $x_{\alpha}^i \to 0$ in $\mathbb{F}$. Since addition and scalar multiplication are continuous in $(V,\tau_1)$, it ...



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