11.1.2 Cauchy Sequences, Completeness
Of course it does not go the other way. For example, you could let xn =
and it has a
convergent subsequence but fails to converge. Here d
and the metric space is just
However, there is a kind of sequence for which it does go the other way. This is called a Cauchy
is called a Cauchy sequence if for every ε >
0 there exists N such that if
m,n ≥ N, then
Now the major theorem about this is the following.
Theorem 11.1.16 Let
be a Cauchy sequence. Then it converges if and only if any subsequence
This was just done above.
⇐= Suppose now that
is a Cauchy sequence and lim
. Then there exists N1
if k > N1,
From the definition of what it means to be Cauchy, there exists N2
that if m,n ≥ N2,
2. Let N ≥
. Then if
k ≥ N,
then nk ≥ N
It follows from the definition that limk→∞xk = x. ■
Definition 11.1.17 A metric space is said to be complete if every Cauchy sequence converges.
Another nice thing to note is this.
Proposition 11.1.18 If
is a sequence and if p is a limit point of the set S
then there is a subsequence
Proof: By Theorem 11.1.7, there exists a sequence of distinct points of S denoted as
none of them equal
. Thus B
contains infinitely many different points of the set
, this for every r.
Let xn1 ∈ B
is the first index such that xn1 ∈ B
have been chosen, the ni
increasing and let 1 > δ1 > δ2 >
where xni ∈ B
Let xnk+1 ∈ B
is the first index such that xnk+1
is contained B
Another useful result is the following.
Lemma 11.1.19 Suppose xn → x and yn → y. Then d
Proof: Consider the following.
and the right side converges to 0 as n →∞. ■
First are some simple lemmas featuring one dimensional considerations. In these, the metric space is ℝ
and the distance is given by
First recall the nested interval lemma. You should have seen something like it in calculus, but
this is often not the case because there is much more interest in trivialities like integration
Lemma 11.1.20 Let
for all k
. Then there exists a point p in
Proof: We note that for any k,l,ak ≤ bl. Here is why. If k ≤ l, then
If k > l, then
It follows that for each l,
Hence supkak is a lower bound to the set of all bl and so it is no larger than the greatest lower bound. It
Pick x ∈
. Then for every
k,ak ≤ x ≤ bk
. Hence x ∈∩k=1∞
Lemma 11.1.21 The closed interval
is compact. This means that if there is a collection of
open intervals of the form
whose union includes all of
, then in fact
is contained in
the union of finitely many of these open intervals.
Proof: Let C be a set of open intervals the union of which includes all of
to admit a finite subcover. That is, no finite subset of
has union which contains
. Then this must be
the case for one of the two intervals
be the one for which this is so. Then split
it into two equal pieces like what was just done and let I2
be a half for which there is no finite subcover of
sets of C
. Continue this way. This yields a nested sequence of closed intervals I1 ⊇ I2 ⊇
and by the
above lemma, there exists a point x
in all of these intervals. There exists U ∈C
such that x ∈ U.
However, for all n large enough, the length of In is less than min
actually contained in
contrary to the construction. Hence
is compact after all.
As a useful corollary, this shows that ℝ is complete.
Corollary 11.1.22 The real line ℝ is complete.
is a Cauchy sequence in
. Then there exists M
Why? If there is no convergent subsequence, then for each x ∈
there is an open set
for only finitely many values of k
is compact, there are
finitely many of these open sets whose union includes
. This is a contradiction because
for all k ∈ ℕ
so at least one of the open sets must contain xk
for infinitely many k.
is a convergent subsequence. By Theorem 11.1.16
the original Cauchy sequence converges to some
Example 11.1.23 Let n ∈ ℕ. ℂn with distance given by
is a complete space. Recall that
. Then ℂn is complete. Similarly ℝn is
To see that this is complete, let
be a Cauchy sequence. Observe that for each j,
That is, each component is a Cauchy sequence in ℂ
is a Cauchy sequence. Similarly
is a Cauchy sequence. It
follows from completeness of ℝ
shown above, that these converge. Thus there exists aj,bj
and so xk → x showing that ℂn is complete. The same argument shows that ℝn is complete. It is easier
because you don’t need to fuss with real and imaginary parts.