COMPLETION OF A METRIC SPACE

Transcription

1 COMPLETION OF A METRIC SPACE HOW ANY INCOMPLETE METRIC SPACE CAN BE COMPLETED REBECCA AND TRACE Given any incomplete metric space (X,d), ( X, d X) a completion, with (X,d) ( X, d X) where X complete, and X is not closed in X. Specifically, the closure of X will, with respect to d X, be X Given any Cauchy sequence {x n } 1. Show equivalence relation n=1 in X we introduce the formal limit LIM x n. we say LIM x n = LIM y n if lim d(x n, y n ) = 0. This forms an equivalence relation on X 1.1. Reflexivity. It can be easily seen shown that LIM x n = LIM x n since x X lim d(x n, x n ) = 0. So the relation is reflexive Symmetry. Since d is a metric, then x,y d(x,y) = d(y,x) if lim d(x n, y n ) = 0, then lim d(y n, x n ) = 0. So the relation is symmetric Transitive. If LIM x n = LIM y n, and LIM y n = LIM z n then that means that lim d(x n, y n ) = 0, and that lim d(y n, z n ) = 0. Then by the triangle equality 0 d(x n, z n ) d(x n, y n ) + d(y n, z n ) 0 lim d(x n, z n ) lim d(x n, y n ) + lim d(y n, z n ) 0 lim d(x n, z n ) By the squeeze theorem lim d(x n, z n ) = 0 Since x,y d(x,y) 0 then we can say that lim d(x n, z n ) = 0. Showing that the relation is transitive. 1

2 2 REBECCA AND TRACE 2. Show( X, d) is a metric space Let X be the space of all formal limits of Cauchy sequences in X, with the above relation. We define a metric d X : X X R + by setting d X(LIM x n, LIM y n ).= lim d(x n, y n ) Well defined. Let x=x, y= y, we would like to show that d X(x, y) = d X(x, y ). It suffices to show that d X(x, y) = d X(x, y ) d X(x n, y n ) d X(x n, y n) + d X(y n, y n ) d X(x n, y n) d X(x n, y n ) + d X(y n, y n) note : d X(y n, y n ) = d X(y n, y n) d X(y n, y n) d X(x n, y n ) d X(x n, y n) d X(y n, y n) lim d X (y n, y n) lim d X (x n, y n ) lim d X (x n, y n) lim d X (y n, y n) 0 lim d X (x n, y n ) lim d X (x n, y n) 0 By the squeeze theorem lim d X (x n, y n ) = lim d X (x n, y n) 2.2. Gives the structure of a metric space. Since d X(LIM x n, LIM y n ).= lim d(x n, y n ) then LIM x n, LIM y n. (1) d X is non negative: since d is a metric, d(x n, y n ) 0 for all n > 0, hence (2) Identity holds: d X(LIM x n, LIM y n ) = lim d(x n, y n ) 0 By definition of LIM: d X(LIM x n, LIM y n ) = lim d(x n, y n ) = 0 LIM x n = LIM y n (3) d X is symmetric: since d is a metric: d(x n, y n ) = d(y n, x n ) for all n > 0, hence d X(LIM x n, LIM y n ) = lim d(x n, y n ) d X(LIM y n, LIM x n ) = lim d(y n, x n ) lim d(y n, x n ) = lim d(x n, y n ) d X(LIM x n, LIM y n ) = d X(LIM y n, LIM x n )

3 COMPLETION OF A METRIC SPACE HOW ANY INCOMPLETE METRIC SPACE CAN BE C (4) The triangle inequality holds: since d is a metric for all n > 0 d(x n, z n ) d(x n, y n ) + d(y n, z n ), hence lim d(x n, z n ) lim d(x n, y n ) + lim d(y n, z n ) but since d X(LIM x n, LIM y n ) = lim d(x n, y n ) d X(LIM y n, LIM z n ) = lim d(y n, z n ) d X(LIM x n, LIM z n ) = lim d(x n, z n ) we conclude that d X(LIM x n, LIM z n ) d X(LIM x n, LIM y n )+d X(LIM y n, LIM z n ). 3. Show (X,d) is a subspace of ( X, d) Let x X, with LIM x X we say that x = y LIM x = LIM y. Let s assume LIM x = LIM y then d X(LIM x, LIM y).= lim d(x, y)= 0 by definition. Similarly for the other direction. This shows that d X = d X which means we that d x can be represented by d X x which shows that X is a subspace of X 4. Show ( X, d) is complete Let ϕ :X X be defined by ϕ(x) := LIM{x}. Then ϕ is an isometry since for x and y in X we have, ϕ(x) := LIM{x} and ϕ(y) := LIM{y} and d X(ϕ(x), ϕ(y)) = lim d(x, y) = d(x, y). Definition of Density of ϕ(x) X: ϕ(x) is dense in X if given any x X and ɛ > 0 we can find a point x X so that ϕ(x) is ɛ-close to x. Now we show how to find x in X given x and ɛ. Given any x = LIM{x n } in X with {x n } Cauchy in X, ε > 0 N > 0, n, m N d(x n, x m ) < ε, m = N n N, d(x n, x N ) ε lim d(x n, x N ) ε Let x = x N X then ϕ(x) := LIM{x N } and d X( x, ϕ(x)) = lim d(x n, x N ) ε.

4 4 REBECCA AND TRACE This is the first step to completeness: Cauchy sequences in the dense subset ϕ(x) are convergent in X. If {x k } is Cauchy in X and {x k } ϕ(x) = A with A dense in X, x k = ϕ(x k ), with x k X Then given ε > 0, N > 0, n, m > N and since ϕ is an isometry d X(x n, x m ) = d X(ϕ(x n ), ϕ(x m )) = d(x n, x m ) < ε {x n }in X is Cauchy for {x k } ϕ(x) The reverse implication also holds. We just showed that if x k = ϕ(x k ) = LIM{x k } then {x k } is Cauchy in X {x k } X is Cauchy in X. Let x = LIM{x n } X. We need to show that the Cauchy sequence x k x in X. So we need to have that for ε > 0 N > 0 with k > N d x (x k, x) = d x (LIM{x k }, LIM{x n }) = lim d(x k, x n ) < ε. The last inequality holds for k, n N because the sequence {x k } is Cauchy in X. Given ( X, d x ) metric space, we see that A=ϕ(X) is a subset of X such that every Cauchy sequence in A converges in X. And since A is dense in X we can show that X is complete, by Lemma below, finishing the proof. Lemma: Let (Z, ρ) be a metric space, A a dense subset of Z. Suppose that given any Cauchy sequence {a n } A then the sequence is convergent in Z. Then Z is complete. Proof: Take a Cauchy sequence {z k } Z, since A is dense then for each z k we can find a Cauchy sequence {a k n} A such that lim ρ(ak n, z k ) = 0. This means that given ɛ > 0 there is N k > 0 such that for all n N k ρ(a k n, z k ) ɛ/3.

5 COMPLETION OF A METRIC SPACE HOW ANY INCOMPLETE METRIC SPACE CAN BE C Increasing if necessary the values of N k we can ensure that N 1 < N 2 < N 3 <.... particular, if we choose n = N k get for all k 1 ρ(a k N k, z k ) ɛ/3. Claim 1: {a n := a n N n } n 1 is Cauchy in A, hence convergent to some z 0 Z by hypothesis: ρ(a n, a m ) ρ(a n, z n ) + ρ(z n, z m ) + ρ(z m, a m ) ρ(a n, a m ) ρ(a n N n, z n ) + ρ(z n, z m ) + ρ(a m N m, z m ) ρ(a n, a m ) ɛ/3 + ɛ/3 + ɛ/3, for the middle term we used the hypothesis that the sequence {z k } is Cauchy. Hence there is z 0 Z such that lim a n = z 0 so then for n N ρ(a n, z 0 ) ɛ/2 Finally we show that the Cauchy sequence {z n } converges to that z 0. Claim 2: d(z n, z 0 ) < ɛ for n N d(z n, z 0 ) d(z n, a n ) + d(a n, z 0 ) d(z n, z 0 ) ɛ/2 + ɛ/2 ɛ. In Lastly, we let X = Z, d x = ρ and A = A Then we have by the Lemma that X is in fact complete. *** A final note to the reader: You will recall that it took us half a semester to show the completion of the reals. Because we did all of that work, and we already knew that the reals were complete, we were able to show the completion of a given metric space in much less time. We were able to define our metric d X : X X R + with d X(LIM x n, LIM y n ).= lim d(x n, y n ). because we know that the right-hand-side limit exists. Since we have that, {x n } n 1 and {y n } n 1 are Cauchy sequences in X then the sequence of real numbers {d(x n, y n )} n 1 is a Cauchy sequence in R and since R is complete the sequence must be convergent to a real number L = lim d(x n, y n ).***