Tag Info

New answers tagged

0

No, there isn't in general. If $J$ were such a set then $J$ would span the whole vector space. So we would have $J=I$. So your question can be reduced to the question if there exists vector spaces with an uncountable basis. Each infinite dimensional Banach space is such a space. For example take $C_0([0,1])$ with supremum norm.


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 ...


0

Every Hausdorff topology on a real or complex finite-dimensional topological vector space is equivalent to the usual topology. So, without doing any checking: No, your topology cannot result in a TVS.


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 ...


0

The map $f \colon \mathbb{C} \to V$ given by $x \mapsto xv$ for $v \neq 0$ is only a homeomorphism onto its image if $V$ is assumed to be a Hausdorff topological vector space. Note that $f(\mathbb{C})$ inherits a Hausdorff topology from $V$ and use that there is only one Hausdorff vector space topology on $\mathbb{C}^n$. Perhaps an easier way to see this is ...


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 ...


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 ...


0

If $x\notin (A+B)$, then $A\cap(x-B)=\varnothing$. Since $(x-B)$ is closed, it follows from Theorem 1.10 in Rudin's book that there exists a neighborhood $V$ of $0$ such that $(A+V)\cap(x-B+V)=\varnothing$. Therefore $(A+B+V)\cap(x+V)=\varnothing$ and, in particular, $(A+B)\cap(x+V)=\varnothing.$ As $(x+V)$ is a neighborhood of $x$, this shows that $x\notin ...


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 ...


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 ...


0

The main difference is that an inner product is a binary operation but norm acts on one element.


0

Yes the product of tow complete topological vector space is complete. The very simple example is considering $\mathbb R $ and $\mathbb R^2$



Top 50 recent answers are included