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# Completeness

Module by: Mark A. Davenport. E-mail the author

Distance functions allow us to talk concretely about limits and convergence of sequences.

## Definition 1

Let (M,d(x,y))(M,d(x,y)) be a metric space and {xi}i=1{xi}i=1 be a sequence of elements in MM. We say that {xi}i=1{xi}i=1converges to x*x* if and only if for every ε>0ε>0 there is an NN such that d(xi,x*)<εd(xi,x*)<ε for all i>Ni>N. In this case we say that x*x* is the limit of {xi}i=1{xi}i=1.

## Definition 2

A sequence {xi}i=1{xi}i=1 is said to be a Cauchy sequence if for any ε>0ε>0 there is an NN such that d(xi,xj)<εd(xi,xj)<ε for every i,j>Ni,j>N.

It can be shown that any convergent sequence is a Cauchy sequence. However, it is possible for a Cauchy sequence to not be convergent!

## Example 1

Suppose that M=(0,2)M=(0,2), i.e., the open interval from 0 to 2 on the real line, and let d(x,y)=|x-y|d(x,y)=|x-y|. Consider the sequence defined by xi=1ixi=1i. {xi}{xi} is Cauchy since for any εε we can set NN such that 1N<ε21N<ε2, so that |xi-xj||xi|+|xj|<ε2+ε2=ε|xi-xj||xi|+|xj|<ε2+ε2=ε. However, xi0xi0, but 0M0M, i.e., the sequence converges to something that lives outside of our space.

## Example 2

Suppose that M=C[-1,1]M=C[-1,1] (the set of continuous functions defined on [-1,1][-1,1]) and let d2d2 denote the L2L2 metric. Consider the sequence of functions defined by

f i ( t ) = 0 if t - 1 i i t 2 + 1 2 if - 1 i < t < 1 i 1 if t 1 i . f i ( t ) = 0 if t - 1 i i t 2 + 1 2 if - 1 i < t < 1 i 1 if t 1 i .
(1)

For j>ij>i we have that

d 2 ( f i , f j ) = ( j - i ) 2 6 j 3 i . d 2 ( f i , f j ) = ( j - i ) 2 6 j 3 i .
(2)

This goes to 0 for j,ij,i sufficiently large. Thus, the sequence {fi}i=1{fi}i=1 is Cauchy, but it converges to a discontinuous function, and thus it is not convergent in MM.

## Definition 3

A metric space (M,d(x,y))(M,d(x,y)) is complete if every Cauchy sequence in MM is convergent in MM.

## Example 3

• M=[0,1],d(x,y)=|x-y|M=[0,1],d(x,y)=|x-y| is complete.
• (C[-1,1],d2)(C[-1,1],d2) is not complete, but one can check that (C[-1,1],d)(C[-1,1],d) is complete. (This space works because using dd, the above example is no longer Cauchy.)
• QQ is not complete, but RR is.

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