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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 ...


2

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 ...


1

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 ...


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 ...


1

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} ...


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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