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Let $(X,\|\;\|)$ be a normed vector space over $K\;(\Bbb R\text{ or}\;\Bbb C)$. Lets define two functions: $$\oplus:X\times X\to X\;\text{s.t.}\;\oplus((x,y))=x+y\;\;\forall (x,y)\in X\times X\\ \odot:K\setminus\{0\}\times X\to X\;\text{s.t.}\;\odot((a,x))=ax\;\;\forall (a,x)\in K\setminus\{0\}\times X$$

I want to prove that $\oplus\;\text{and}\;\odot$ are continuous and open functions:

I've already proved that both functions are continuous by taking any sequence that converge to an element in their respective domain and showing that the sequence-function converges to the function of the limit. Then I defined $\|\;\|_{X^2}:X\times X\to X$ s.t. $\|(x,y)\|_{X^2}=\|x\|+\|y\|$, which is clearly a norm in $X\times X$ and proved that $\oplus$ is open.

Where I'm stucked is trying to prove that $\odot$ is open, since I couldn't define a norm in $K\setminus\{0\}\times X$ nor prove is even a vector space, I started by taking any open set $A\subseteq X\times X$ s.t. $A\neq\emptyset$ so $$\forall (x,y)\in A\;\exists B_{xy}\subset A\;\text{s.t.}\;\ (x,y)\in B_{xy}\;\text{and}\;\ B_{xy}\;\text{is open}\\ \Rightarrow A=\bigcup_{(x,y)\in A}B_{xy}\\ \Rightarrow \odot(A)=\bigcup_{(x,y)\in A}\odot(B_{xy})$$ But got stucked here since I don't know much about $\odot(B_{xy})$. Any ideas would be appreciated.

Arnulf
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  • Presumably the topology on $K\backslash {0} $ is the subspace topology as a subspace of $K.$ So the usual metric on $K,$ restricted to $K\backslash {0},$ generates the topology on $K\backslash {0}.$ Note that $(K\backslash {0})\times X$ is not a vector space. – DanielWainfleet Mar 06 '16 at 00:44
  • I was trying defining a metric $d:K\setminus {0}\times X\to K$ as $$d((a,x),(b,y))=|a-b|+|x-y|$$ With $(K\setminus {0}\times X, d)$, $;\forall A\subset K\setminus{0}\times X$ s.t. $A\neq\emptyset$ is open we get that: $$\forall (a,x)\in A;; \exists r_{ax};\text{s.t.};\ (a,x)\in B_{r_{ax}}((a,x))\subset A\ \Rightarrow A=\bigcup_{(a,x)\in A} B_{r_{ax}}((a,x))\ \Rightarrow \odot (A)=\bigcup_{(a,x)\in A}\odot (B_{r_{ax}}((a,x)))$$ And from here I was trying to prove that: $$\odot (B_{r_{ax}}((a,x)))=B_{r_{ax}}^X (ax)$$ But I have only achieved the $\subseteq$ part – Arnulf Mar 06 '16 at 01:14
  • I think I got it: Let $$y\in B_{r_{ax}-|a-1|(1+|x|)}((a,x))\Rightarrow d(ax,y)<r_{ax}-|a-1|(1+|x|)\Leftrightarrow |ax-y|<r_{ax}-|a-1|(1+|x|)$$ So, lets take $;(1,y)\in K\setminus {0}\times X\Rightarrow \odot (1,y)=y)$, and $$d((a,x),(1,y))=|a-1|+|x-y|=|a-1|+|(1-a)x+ax-y|\le |a-1|+|(1-a)x|+|ax-y|=|a-1|+|a-1||x|+|ax-y|<|a-1|(1+|x|)+r_{ax}-|a-1|(1+|x|)=r_{ax}$$ thus, $$(1,y)\in B_{r_{ax}}((a,x))$$ thereby $y\in \odot (B_{r_{ax}}((a,x)))$ and thus $\odot (B_{r_{ax}}((a,x)))=B_{r_{ax}}^X (ax)$ – Arnulf Mar 06 '16 at 01:37
  • I was typing an answer when your last comment appeared But I have a computer glich. – DanielWainfleet Mar 06 '16 at 02:03
  • Sorry I was badly wrong in my last comments, but now I think I really got it. Will post it as an answer. – Arnulf Mar 06 '16 at 02:29

1 Answers1

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Defining $d_{KX}:K\setminus\{0\}\times X\to K$ as $$d_{KX}((a,x),(b,y))=|a-b|+\|x-y\|_X\;\;\forall (a,x),(a,y)\in K\setminus\{0\}\times X$$ is clearly that $\ d_{KX}$ is a metric. So a neigborhood of $(a_0,x_0)$ in $\;K\setminus\{0\}\times X\;$ will be $B_r(a_0,x_0)=\{(b,y)\in K\setminus\{0\}\times X:d_{KX}((b,y),(a_0,x_0))<r\}$.

So, let $A$ be any open and not-empty subset of $K\setminus\{0\}\times X$. I want to prove that $\odot (A)$ is open (in $X$).

Let $y\in\odot (A)\Rightarrow (1,y)\in A$, since $\odot (1,y)=y$. And since $A$ is open we have that $\exists r_0>0$ s.t.$B_{r_0}(1,y)\subset A$.

Thus, taking $r_y=r_0$ we get that: $$\;\forall x\in B_{r_y}^X(y),\;\;d_{KX}((1,x),(1,y))=\|x-y\|_X<r_y=r_0$$ $$\Rightarrow (1,x)\in B_{r_0}(1,y)\Rightarrow (1,x)\in A\;\text{and}\;\odot (1,x)=x\Rightarrow x\in\odot (A)$$ thus $$B_{r_y}^X(y)\subset\odot (A)$$ and, since this happens for every $y\in\odot (A)$, thereby $\odot (A)$ is open and thus $\odot$ is open.

Arnulf
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