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Let $D=(x_n)$ dense in $X$ and let $\sigma(X^*,D)$ the locally convex topology in $X^*$ with countable local basis at $0\in X^*$ given by: $$U_n=\{x^*\in X^*: \sup_{1\leq j\leq n}|x^*(x_j)|< 1/n \}$$ Then $(X^*,\sigma(X^*,D))$ is a Hausdorff TVS. In fact, if $x^*\neq 0$, as $x^*$ is continuous, there is $j_0$ such that $x^*(x_{j_0})\neq 0$, so we have ...


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It seems the following. It depends. If you consider on $E=E_0\oplus E_1$ the product topology from $E_0\times E_1$ then the answer is positive, because the set $A\times E_1$ is open in $E_0\times E_1$. But in the opposite cases the answer may be negative. For instance, suppose that $E_1$ is dense in $E$, $A$ is non-empty and $E_0\ne A-A$. If the set ...


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I don't immediately see if your approach leads to a solution (in particular I think you need "normed" for this to be true). This is the way I would do it: assume that $\|\phi\|\leq c$ for all $\phi\in K$. Let $D=(x_n)$ be dense in the unit ball of $X$. Now define $$ d(\phi,\psi)=\sum_{k=1}^\infty2^{-k}|\phi(x_k)-\psi(x_k)|. $$ It is straighforward to check ...


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OK, I think I got it. First we can assume w.l.o.g. that $0\in A$, I won't need $A$ to be balanced. Since $f$ is an embedding we can find $B'\subseteq E'$ open with $f(A)=f(E)\cap B'$ but this $B'$ doesn't have to be convex. Now choose a convex, open neighbourhood $U'\in{\cal U}(0)$ in $E'$ such that $U'\subseteq B'$. For each $x\in A$ we find some ...


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OK, I didn't thought about it the way Tom Cruise mentioned, the supremum as the topology generated by the union. I guess it's pretty much straight forward now, let's say we have the linear topologies $\tau_\alpha$ on the vector space $X$ and $\tau$ is the supremum of the $\tau_\alpha$, i.e. the topology generated by sets of the form $\cap_{i<n} ...


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The space $\mathbb{R}^2 \setminus \{(0,0)\}$ is not homeomorphic to $(a)$, $(c)$, or $(d)$, but it is homeomorphic to $(b)$. To prove that $\mathbb{R}^2 \setminus \{(0,0)\}$ is homeomorphic to the cylinder $(b)$, define the following map. For the circle of radius $r$ around the origin in $\mathbb{R}^2 \setminus \{(0,0)\}$, map this circle homeomorphically ...


4

It's going to be hard to find an invariant that distinguishes these spaces from $ \Bbb R^2 \setminus \{(0,0)\} $, if you only know compactness, connectedness, and path-connectedness. If you knew what the fundamental group of a space was, this would be rather straightforward. As is, you best bet would be to show that these spaces are homeomorphic to a space ...


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OK, I think I got a counterexample now, i.e. the final topology w.r.t. to the inclusion maps is not linear. I only worked it through on ${\bf R}$ though. The only assumption that I need is that for given $\epsilon>0$ and $k,n\in{\bf N}$ we find some $f\in C^\infty([-1,1])$ such that $\|f^{(j)}\|_\infty<\epsilon$ for $j<k$ and $f^{(k)}(0)>n$, ...


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The weak* topology is the topology of pointwise convergence: $f_n \to f$ in the weak* topology if and only if $f_n(x) \to f(x)$, which I think is a pretty good reason to care about it. The weak topology on a topological vector space $V$ gives the convergence condition $v_n \to v$ if and only if for all $\lambda \in V^*$ we have $\lambda(v_n) \to ...


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Let $K_n$ be the unit ball of radius $n$, so your $C_K^\infty = \bigcup_n C^\infty(K_n)$ and let $f_n:C^\infty(K_n)\longmapsto C_K^\infty$ be the natural embeddings. The system $$ \bigl\{ \bigcup_n f_n[U_n]\, ;\, U_n \text{ is a $0$-neighbourhood in } C^\infty(K_n) \bigr\} $$ is a filter base in $C_K^\infty$, immediately to verify (using $U_n = ...


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The nub of it is that total boundedness is an intrinsic property of a metric [more generally, uniform] space. It does not depend on whether the space is a subspace of some larger space, and if so, what that larger space is, all that matters is the metric. We have two metric spaces, $X = A^\ast(S^\ast)$, with the metric induced by the norm on $E^\ast$, and ...


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Wikipedia answers both of your questions. Specifically, let $X$ be a topological vector space and $M \subset X$ be a subspace. Then $X/M$ in its usual topology is Hausdorff if and only if $M$ is closed. This is essentially because a topological vector space is Hausdorff if and only if $\{ 0 \}$ is closed, and $\{ 0 \}$ is closed in $X/M$ if and only if $M$ ...


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Suppose $f$ is as stated with exponential bound $Ce^{-\delta|x|}$, and suppose that $\overline{g} \in L^{2}(\mathbb{R})$ is orthogonal to the functions $\{ x^{n}f(x) \}_{n=0}^{\infty}$ in the $L^{2}$ inner-product. Define $$ H(\lambda) = \int_{-\infty}^{\infty}g(x)f(x)e^{-i\lambda x}\,dx. $$ Suppose that $|\Im\lambda| \le \delta' < ...



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