Distance functions allow us to talk concretely about limits and convergence
of sequences.
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=1∞converges 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∞.
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!
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, xi→0xi→0, but 0∉M0∉M, i.e., the sequence converges to something that lives outside of our space.
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.
A metric space (M,d(x,y))(M,d(x,y)) is complete if every Cauchy sequence in MM is convergent in MM.
- 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 d∞d∞, the above example is no longer Cauchy.)
- QQ is not complete, but RR is.