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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 $\cap_{i<n} ... 11 The space$\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)\} $, if you only know compactness, connectedness, and path-connectedness. If you knew what the fundamental group of a space was, this would be rather straightforward. As is, you best bet would be to show that these spaces are homeomorphic to a space ... 0 OK, I think I got a counterexample now, i.e. the final topology w.r.t. to the inclusion maps is not linear. I only worked it through on${\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$, ... 1 The weak* topology is the topology of pointwise convergence:$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 $U_n = ... 1 The nub of it is that total boundedness is an intrinsic property of a metric [more generally, uniform] space. It does not depend on whether the space is a subspace of some larger space, and if so, what that larger space is, all that matters is the metric. We have two metric spaces,$X = A^\ast(S^\ast)$, with the metric induced by the norm on$E^\ast$, and ... 0 Wikipedia answers both of your questions. Specifically, let$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$... 2 Suppose$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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