2Convergence of sequences
IA Analysis I
2.4 Cauchy sequences
The third characterization of real numbers is in terms of Cauchy sequences.
Cauchy convergence is an alternative way of defining convergent sequences
without needing to mention the actual limit of the sequence. This allows us to
say
{
3
,
3
.
1
,
3
.
14
,
3
.
141
,
3
.
1415
, ···}
is Cauchy convergent in
Q
even though the
limit π is not in Q.
Definition (Cauchy sequence). A sequence (
a
n
) is Cauchy if for all
ε
, there is
some
N ∈ N
such that whenever
p, q ≥ N
, we have
|a
p
− a
q
| < ε
. In symbols,
we have
(∀ε > 0)(∃N)(∀p, q ≥ N) |a
p
− a
q
| < ε.
Roughly, a sequence is Cauchy if all terms are eventually close to each other
(as opposed to close to a limit).
Lemma. Every convergent sequence is Cauchy.
Proof.
Let
a
n
→ a
. Let
ε >
0. Then
∃N
such that
∀n ≥ N
,
|a
n
− a| < ε/
2.
Then ∀p, q ≥ N, |a
p
− a
q
| ≤ |a
p
− a|+ |a − a
q
| < ε/2 + ε/2 = ε.
Lemma. Let (
a
n
) be a Cauchy sequence with a subsequence (
a
n
k
) that converges
to a. Then a
n
→ a.
Proof.
Let
ε >
0. Pick
N
such that
∀p, q ≥ N
,
|a
p
− a
q
| < ε/
2. Then pick
K
such that n
K
≥ N and |a
n
K
− a| < ε/2.
Then ∀n ≥ N, we have
|a
n
− a| ≤ |a
n
− a
n
K
| + |a
n
K
− a| <
ε
2
+
ε
2
= ε.
An important result we have is that in
R
, Cauchy convergence and regular
convergence are equivalent.
Theorem (The general principle of convergence). Let
F
be an ordered field with
the monotone-sequence property. Then every Cauchy sequence of F converges.
Proof.
Let (
a
n
) be a Cauchy sequence. Then it is eventually bounded, since
∃N
,
∀n ≥ N
,
|a
n
− a
N
| ≤
1 by the Cauchy condition. So it is bounded. Hence by
Bolzano-Weierstrass, it has a convergent subsequence. Then (
a
n
) converges to
the same limit.
Definition (Complete ordered field). An ordered field in which every Cauchy
sequence converges is called complete.
Hence we say that R is a complete ordered field.
However, not every complete ordered field is (isomorphic to)
R
. For example,
we can take the rational functions as before, then take the Cauchy completion
of it (i.e. add all the limits we need). Then it is already too large to be the reals
(it still doesn’t have the Archimedean property) but is a complete ordered field.
To show that completeness implies the monotone-sequences property, we
need an additional condition: the Archimedean property.
Lemma. Let
F
be an ordered field with the Archimedean property such that
every Cauchy sequence converges. The
F
satisfies the monotone-sequences
property.
Proof.
Instead of showing that every bounded monotone sequence converges, and
is hence Cauchy, We will show the equivalent statement that every increasing
non-Cauchy sequence is not bounded above.
Let (a
n
) be an increasing sequence. If (a
n
) is not Cauchy, then
(∃ε > 0)(∀N)(∃p, q > N) |a
p
− a
q
| ≥ ε.
wlog let p > q. Then
a
p
≥ a
q
+ ε ≥ a
N
+ ε.
So for any N , we can find a p > N such that
a
p
≥ a
N
+ ε.
Then we can construct a subsequence a
n
1
, a
n
2
, ··· such that
a
n
k+1
≥ a
n
k
+ ε.
Therefore
a
n
k
≥ a
n
1
+ (k −1)ε.
So by the Archimedean property, (a
n
k
), and hence (a
n
), is unbounded.
Note that the definition of a convergent sequence is
(∃l)(∀ε > 0)(∃N)(∀n ≥ N) |a
n
− l| < ε,
while that of Cauchy convergence is
(∀ε > 0)(∃N)(∀p, q ≥ N) |a
p
− a
q
| < ε.
In the first definition,
l
quantifies over all real numbers, which is uncountable.
However, in the second definition, we only have to quantify over natural numbers,
which is countable (by the Archimedean property, we only have to consider the
cases ε = 1/n).
Since they are equivalent in
R
, the second definition is sometimes preferred
when we care about logical simplicity.