In the Banach space $X$, we define the distance between two sets $A$ and $B$ as $$d(A,B)=\inf \{||a-b||: a\in A,b\in B\}.$$ Let $M$ and $N$ be closed subspaces of $X$. Given $x, y \in X$, assume that $x+M$ and $y+N$ are disjoint. Can we concude that $d(x+M,y+N)>0$?
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I don't remember much of functional analysis, but if you define $T(u,v)=u-v$ on $X\times X$ I think you can apply the closed graph theorem . – ajotatxe Jul 13 '17 at 10:21
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Here is a category of counter examples :
In any infinite dimensional normed vector space $X$, There are two closed subspace, $M$ and $N$ such that $M+N$ is not closed . Then the closed subspace $Y : = \overline{M+N}$ of $X$ contains an element, say $y \in Y$ which is not in $M+N$. Then it is easy to verify that $(y+M )\cap N = \emptyset .$
Observe that $$0=d(y,M+N) = d(y+M ,N)$$
P.S: As an example of such $M$ and $N$ in Hilbert space see: Sum of closed spaces is not closed
Red shoes
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Thank you very much for your answer. So what about the conclusion if we assume that $M$ and $N$ are closed convex cones, instead of closed subspaces? – Math253 Jul 14 '17 at 06:32
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Thanks. Sorry that my question is not clear. Acttually what I mean is that $M$ and $N$ are only closed convex cones but not closed closed subspaces. In this case the equality $d(y,M+N)=d(y+M,N)$ may not true. – Math253 Jul 14 '17 at 06:44
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So your question is that , you have two closed convex cones which are not subspace , then you translate them so that they don't intersect each other!
and you want see if they have positive distance.
– Red shoes Jul 14 '17 at 06:52 -
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@Math253 I am very positive still the answer is NO ! but I don't know any counter example on top of my mind right now ! you better ask it in separate question ! – Red shoes Jul 14 '17 at 07:02
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