... and thus by completeness of M it converges toapoint m ∗. Sinced (m ∗,f (m ∗ )) ≤ d (m ∗ ,m i )+ d (m i,f (m i )) + d(f (m i ), f (m ∗ )) ≤ (1 + k) d (m ∗ ,m i )+ kid (m 0 ,m 1 ) is arbitrarily small, ... cl(A)=int(S\A),S\int(A)=cl(S\A), and cl(A)=A ∪ der(A);(iii) cl( )= int( )= ∅, cl(S)=int(S)=S, cl(cl(A )) = cl(A), and int(int(A )) = int(A);(iv) cl(A ∪ B)=cl(A) ∪ cl(B), der(A ∪ B)=der(A) ∪ der(B), and int(A ∪ B) ⊃ int(A) ∪ ... d (m 1 ,m 2 )= d (m 2 ,m 1 ) (symmetry); and M3 . d (m 1 ,m 3 ) ≤ d (m 1 ,m 2 )+ d (m 2 ,m 3 ) (triangle inequality).A metric space is the pair (M, d); if there is no danger of confusion, just write M for (M, d).Taking...