By Mohamed A. Khamsi

ISBN-10: 0471418250

ISBN-13: 9780471418252

Provides updated Banach area results.

* positive factors an in depth bibliography for out of doors reading.

* offers targeted routines that elucidate extra introductory fabric.

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Now let e > 0 and choose n so that 1/n < e. It follows that if x 6 M there exists p 6 P such that d(x,p) < 1/n < e, and this proves that ~P = M. m Finally, we mention another interesting fact about separability. If (M, d) is separable then the space of all nonempty bounded closed subsets of M endowed with the Hausdorff metric need not be separable. 23). Another basic topological concept which arises often is that of connectedness. 9 A metric space (M, d) is said to be connected if M is not the union of two disjoint nonempty open subsets of M.

Then f has a unique fixed point x, and lim d ( / " (x), x) = 0 for every x G M. Proof. Fix x € M and let xn — Tn(x), the argument into two steps. Step 1. lim d(xn,xn+\) n—»oo n—»oo n = 1,2, · · ■. As before, we break = 0. Proof. 0 < limsupd(a; n ,a; n + 1 ) < lim sup V'" (d{x,X\)) n—»oo n—·οο = 0. Step 2. {xn} is a Cauchy sequence. Proof. Since rpn (t) —► 0 for t > 0, V (e) < ε f° r a ny ε > 0. In view of Step 1, given any ε > 0 it is possible to choose n so that d(xn+ï,xn) <ε-ψ(ε). Now let K (χη,ε) = {x e M : d (x, xn) < ε).

It is universal with respect to separability in the following profound sense! 13 Every separable metric space is isometric with a subset ofC[0,1] It is possible to give an elementary (but quite detailed) proof of this fact; see, for example, [145], p. 192. :e totally bounded. 14 Every compact metric space (M,d) is separable. Proof. Let (M,d) be a compact metric space and for each n € N, let Pn be a subset of M which is maximal with respect to the property: if x,y € Pn then d(x,y) > 1/n (a simple induction argument suffices for this).

### An Introduction to Metric Spaces and Fixed Point Theory by Mohamed A. Khamsi

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