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5

I'll ignore the missing point at the north pole for now as it doesn't really add anything to the picture as far as I can tell. I think this space could be considered some kind of bastard demon-spawn of the line with two origins. You could so something very similar to you example, except start with a copy of the circle in the plane, glue together the east ...


4

For $0 < p < 1$, let $$\ell^p(\mathbb{N}) = \Biggl\{ x \colon \mathbb{N}\to \mathbb{C} : \sum_{n = 0}^\infty \lvert x(n)\rvert^p < +\infty\Biggr\}.$$ We endow it with the $p$-seminorm $$s_p(x) = \sum_{n = 0}^\infty \lvert x(n)\rvert^p$$ and the metric $d_p(x,y) = s_p(x-y)$ derived from it. Note the difference from the case $p \geqslant 1$. If ...


4

Scalar multiplication will not be continuous unless the topology on the scalar field $K$ is also discrete (or unless $X=0$). Indeed, if $x\in X$ is any nonzero vector, $a\mapsto a\cdot x$ is a continuous injection $K\to X$, as the restriction of the scalar multiplication map $K\times X\to X$ to the subspace $K\times\{x\}\cong K$. If $X$ has the discrete ...


2

Since the $\inf$ is $0$ you find a sequence $(x_n)_{n\in\mathbb N}$ in $K$ and a sequence $(y_n)_{n\in\mathbb N}$ in $E$ such that $$\lim_{n\to\infty} d(x_n,y_n)=0.$$ Since $K$ as compact set is bounded the sequence $y_n$ is bounded and hence lies in a compact subset of $E$. This implies that both sequences have a convergent subsequence. The limit of both ...


2

The topology $T$ is usually called the norm topology (when we view $V,W$ as normed spaces, and not only as normable spaces), or the topology of uniform convergence on bounded subsets of $V$. To avoid trivialities, I assume that neither $V$ nor $W$ is the trivial vector space in the following (for then $B(V,W) = \{0\}$ and all topologies coincide). Since ...


1

If $x\ne 0$, the map $v\mapsto xv$ is a homeomorphism since its reciprocal is multiplication by $x^{-1}$. It is even an isomorphism of topological vector spaces. The map $\;\begin{aligned}[t]\mathbf C&\to\langle v\rangle\\[-1ex]x&\mapsto xv\end{aligned}$ is also an isomorphism of topological vector spaces, if $V$ is Hausdorff, because on the finite ...


1

The same thing holds for the spaces $L^p$, $0 < p < 1$, and Rudin gives an elegant proof for this in his book (section 1.47). Pretty much the same argument works for $C[0,1]$ with the given metric (some boring details need to be added). Assume that $\emptyset \neq V \subset C$ is open and convex and $0 \in V$. From this it follows that the open ball ...



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