Topology. Metric Space.

Quote:

Originally Posted by vercammen

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Do you expect someone to do the problem(s) for you?

Each of these follows from this theorem:
$(\forall x,~y)\left[|D(A,x)-D(A,y)|\le d(x,y)\right]$

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

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

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