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- Oct 20th 2012, 12:38 AMvercammenTopology. Metric Space.
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- Oct 20th 2012, 05:42 AMPlatoRe: Topology. Metric Space.
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Do you expect someone to do the problem(s) for you?

Each of these follows from this theorem:

$\displaystyle (\forall x,~y)\left[|D(A,x)-D(A,y)|\le d(x,y)\right]$

The proof follows directly from the definition: $\displaystyle D(A,x)=\inf\{d(x,a):~a\in A\}~.$

Note that if $\displaystyle \delta>0$ then $\displaystyle \exists b\in A$ such that $\displaystyle D(A,x)\le d(x,b)<D(A,x)+\delta$.

Also note that $\displaystyle \forall a\in A$ we know $\displaystyle D(A,x)\le d(x,a)$.