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Let $B$ be the unit ball of $\ell^2(\mathbb{N})$, i.e. $B=\lbrace x\in \ell^2(\mathbb{N}): \|x\|\le 1\rbrace.$ For each $x=(x_1,x_2,\cdots)\in B$, let $f(x)=(1-\|x\|,x_1,x_2,\cdots).$ Define $T:B\to 2^B$ by $T(x)=B(f(x),r(x))\cap B, \mbox{ where }r(x)=\frac{1}{2}(\|x-f(x)\|).$ Is it true that $D(Tx,Ty)\le \|x-y\| \mbox{ for all }x,y\in B?$ Here $B(y,r)$ denotes the closed ball with radius $r$ centered at $y$, and $D$ is the Hausdorff metric defined by $D(A,B)=\inf\lbrace r>0: N_r(A)\supset B, N_r(B)\supset A\rbrace,$ $N_r(S) =\lbrace x\in C: d(x,S)\lt r\rbrace$ being the $r$-neighborhood of $S$.

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It's wrong: Take $x=0$ and $y=e_1=(1,0,0,\dots)$. Then $f(x)=e_1$ and $f(y)=e_2=(0,1,0,\dots)$, so $r(x) = 1/2$ and $r(y) = \sqrt2/2$. Now it is not hard to see that $p = (-\sqrt7/4,3/4,0,\dots)$ is the point in $Ty = B(f(y),r(y))\cap B$ farthest away from $f(x)$. From this one deduces that $ D(Tx,Ty) = \sqrt{(1+\sqrt7/4)^2+(3/4)^2} - 1/2 = \sqrt{2+\sqrt7/2} - 1/2 > 1 = \|x-y\|. $ (As $Ty$ is the larger chunk, geometric considerations show that $Ty \subseteq N_r(Tx)$ implies $Tx \subseteq N_r(Ty)$, for all $r>0$.)

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    @TCL: You're quite right, sorry; thanks for pointing that out. As you see from my corrected answer, now I do understand what you're doing with the $p$ `:-)`.2010-12-18