Download Cauchy sequences. Definition: A sequence (xn) is said to be a

Cauchy sequences.
Definition: A sequence (xn ) is said to be a Cauchy sequence if given any ε > 0, there
exists K ∈ N such that
|xn − xm | < ε for all n, m ≥ K.
Thus, a sequence is not a Cauchy sequence if there exists ε > 0 and a subsequence
(xnk : k ∈ N) with
|xnk − xnk+1 | ≥ ε for all k ∈ N.
3.5.5 A sequence converges if and only if it is a Cauchy sequence.
Proof. If a sequence converges, it is easy to see that it must be a Cauchy sequence
(exercise). Next, if it is a Cauchy sequence, then it is bounded (why) and hence it
has a convergent subsequence say (xnk ) by the Bolzano-Weierstrass theorem. Suppose
lim(xnk : k ∈ N) = x, it is easy to check that lim(xn ) = x (exercise).
Example If (sn ) is a sequence such that |sn − sn+1 | ≤ rn for all n ∈ N, 0 ≤ r < 1,
then the sequence converges. In particular, if sn = x1 + x2 + · · · + xn where |xn | ≤ rn ,
then (sn ) is a Cauchy sequence and hence converges.
3.5.6 (a) x1 = 1, x2 = 2 and xn+2 = (xn+1 + xn )/2 for all n.
1
(−1)n
(b) sn = −1 + + · · · +
.
2!
n!
(c) |xn+1 − xn | ≤ r|xn − xn−1 | for all n ≥ 2.
Example 3.5.10. Solution of x3 − 7x + 2 = 0.
Define xn+1 = 17 (x3n + 2) for suitable choice of x1 .
Indeed, one can also define u = sup{x : x3 − 7x + 2 < 0}.
1
2
Construction of Real numbers by Cauchy Sequence.
Recall that we have constructed the set of rational numbers. However, we have yet
shown that there exists a set of real numbers, i.e., a complete ordered field. We will
now construct this set using rational numbers.
Since we have yet to define real numbers. We will say (xn ) is a Cauchy sequence of
rational numbers if given any positive rational number r, there exists K ∈ N such
that |xn − xm | < r for all n, m ≥ K.
Let S be the collection of Cauchy sequences of rational numbers. We say (xn ) ∼ (yn )
if given any positive rational number ε, there exists K ∈ N such that |xn − yn | < ε
for all n ∈ K. We then show that this defines an equivalence relation on S.
Exercise: (xn ) ∼ (xn+K : n ∈ N).
We will now define R as the set of equivalence classes in S and let us identify any
rational number r with the equivalence class that contains the sequence (r). Note
that if (a) ∼ (b) where a, b ∈ Q, then a = b; see Homework/Exercise.
We will now show that R is a field.
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The only more difficult part is to show that each nonzero equivalence class has a
multiplicative inverse.
First, we will need a simple observation.
Fact: If (xn ) is not a Cauchy sequence that is equivalence to 0, then there exists
M, K ∈ N such that |xn | > 1/M for all n ≥ K. It is then clear that (xn ) is
equivalence to a sequence (yn ) of rational numbers with absolute value more than
1/M .
Using the above sequence, we now claim that the sequence (1/yn ) is Cauchy and
clearly it is the multiplicative inverse of (xn ).
We next show that R is an ordered field.
Similar to the previous argument, we will take the positive real numbers to be the set
of equivalence classes (yn ) with yn > r for all n for some positive rational number r.
Finally, we show that it is complete.
In general, let ((y(k)n : n ∈ N) : k ∈ N) be a Cauchy sequence of Cauchy sequences
of rational numbers. Then note that lim(y(k) : k ∈ N) = (zn : n ∈ N) for some
Cauchy sequence of rational numbers (zn ). Hence every Cauchy sequence converges.
For more details, see the Appendix at the end of the note.
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Now suppose S is a nonempty set of equivalence classes of Cauchy sequences that is
bounded above. For each fixed n ∈ N, consider the set ( recall that we have identified
constant sequences (r) with rational numbers r)
Tn = {k ∈ Z : k/2n is an upper bound of S}.
First, let us show that it is nonempty. Let X be an upper bound of S. We may assume
that X > 0. Now, note that 1/X > 0 and there exists rational number r = m/K,
m, K ∈ N such that 0 < m/K < 1/X. Clearly 1/X > 1/K and hence K2n ∈ Tn .
We next show that Tn has a least number. For simplicity, let us first assume that
the set has a nonnegative real number (an equivalence class of Cauchy sequences
of rational numbers). Then it has a least element say n0 by a consequence of well
ordering property of natural number. Note that in − 1 6∈ Tn and hence (in − 1)/2n
is not an upper bound while in /2n is an upper bound. (Note that we have identified
the number in /2n to the constant sequence (r) with r = in /2n . Define an = in /2n
and check that (an ) is a Cauchy sequence of rational numbers since |an − am | ≤ 1/2n
for any m ≥ n. Finally, we need to show that [(in /2n )] is the least upper bound of S.
Note that ((in − 1)/2n ) ∈ [(in /2n )].
In general if X is a set with a ”distance function”, we say it is complete if and only
if every Cauchy sequence in the set converges.
Rn is complete. It is not an ordered field for n > 1. But it is complete, i.e., every
Cauchy sequences converges.
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Homework/Exercise:
A. Show that multiplication of ”real numbers” is well defined.
B. Let a, b ∈ Q. Prove that the constant sequences (a) ∼ (b) if and only if a = b.
C. Show that if we define the positive real numbers as above, then the set of ”real
numbers” satisfies properties 2.1.5.
Sec 3.5: all questions except 1,3 and 7.
(4th ed is the same as 3rd ed)
Additional questions for HW4: Sec 3.5: 6,8,12.
Sec 3.6: all questions except 1,6,10.
(4th ed is the same as 3rd ed)
Homework5: 5,7,8 in the homework on limsup and liminf below, Questions B and
Sec 3.6: 1, 6, 10.
Properly divergent sequences
We say lim(xn ) = ∞ if given any real number α ∈ R, there exists K ∈ N such that
xn > α for all n ≥ K.
Note that then the sequence diverges. Similarly, we could define the meaning of
lim(xn ) = −∞.
Example : lim(n) = ∞ and hence lim(nk ), lim(n!), lim(rn ) = ∞ (if r > 1.)
Moreover, if lim(xn+1 /xn ) = L > 1 where (xn ) is a sequence of positive real numbers,
then lim(xn ) = ∞.
A simple fact (3.6.4)
Suppose xn ≤ yn for all n. Then
lim(xn ) = ∞ =⇒ lim(yn ) = ∞.
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Operation with infinite limits:
Suppose lim(xn ) = x and lim(yn ) = l, l could be infinite while x ∈ R. Then
(i) lim(xn + yn ) = x + l (x + ∞ = ∞ and x − ∞ = −∞.)
(ii) lim(xn yn ) = lx if x 6= 0 (a × ∞ = ∞ if a > 0.)
(iii) lim(xn /yn ) = 0 if l is infinite and yn 6= 0 for all n.
Theorem 3.6.5 Let (xn ), (yn ) be sequences of positive real numbers and lim(xn /yn ) =
L > 0 (L 6= ∞). Then lim(xn ) = ∞ if and only if lim(yn ) = ∞.
Limit superior and limit inferior
Definition Let {an )∞
n=1 be a sequence of real numbers. We define its set of cluster
points as follows:
S(an ) = {c ∈ R : there exists a subsequence (ank : k ∈ N) of (an ) that converges to c}.
Examples
(i) S((−1)n ) = {1, −1}.
(ii) S( 12 , 1, 13 , 23 , 1, 14 , 42 , 34 , 1, 15 , · · · ) = [0, 1].
(iii) S(sin n) =?.
Definition Let (an ) be a bounded sequence. We define
lim(an ) = lim an = limsup an = sup S(an )
n→∞
n→∞
lim(an ) = lim an = liminf an = inf S(an )
n→∞
n→∞
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Useful Fact lim(an ) = l if and only if
(i) there exists a subsequence of (an ) that converges to l and
(ii) lim(ank ) ≤ l for any convergent subsequence (ank ) of (an ).
Proof. It is clear that l = limn→∞ an if (i) and (ii) hold. Conversely, let limn→∞ an = l.
Then (ii) follows immediately from the definition. To prove (i), by the definition of
the supremum, for each k ∈ N there exists an uk ∈ S(an ) such that l ≥ uk > l −
1
.
2k
But there exists a subsequence (akm ) of (an ) that converges to uk . By choosing
a subsequence of the subsequence, we may assume that |akm − uk | < 1/2m for each
m ∈ N. Moreover, we may assume that km < (k+1)m for k, m ∈ N (again by choosing
subsequences). We now note that (akk ) is a subsequence of (an ) that converges to l.
Remark By modifying the above argument, we can also show that the set of cluster
points is closed, i.e., if there exists (uk ), uk ∈ S(an ) for all k such that uk → u, then
u ∈ S(an ).
Examples
(1) limn→∞ sin(nπ/4) = 1.
Proof. It is clear that sin(nπ/4) ≤ 1 for all n ∈ N and lim(sin(8n + 2)π/4) = 1.
(2) Let an =


 1+ 1
n
if n is odd
Then lim(an ) = 1 and lim(an ) = −2.
if n is even.
Proof. Let (ank ) be a convergent subsequence of (an ). It is clear that min{1 +
1
, 7
nk nk

 7 −2
n
− 2} ≤ ank ≤ max{1 + n1k , n7k − 2} and hence −2 ≤ lim(ank ) ≤ 1. Finally, since
lim(a2k ) = −2 and lim(a2k+1 ) = 1, the answer follows.
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Theorem Let (an ) be a sequence of real numbers. If it is bounded above we let
Uk = sup{an : n ≥ k}.
And if (an ) is bounded below we let
Lk = inf{an : n ≥ k}.
If (an ) is bounded, then lim(an ) = lim(Uk ) and lim(an ) = lim(Lk ).
Useful properties of limit superior and limit inferior
(i) lim(an ) ≤ u if and only if given any ε > 0, there exists an N > 0 such that
an < u + ε for n > N
(∗)
Proof: ( =⇒ ) done in class using Uk = sup{an : n ≥ k}.
(⇐) Suppose on the contrary that lim(an ) = a > u. Then (an ) has a subsequence
(ank ) that converges to a. Take ε = (a − u)/2. Then there exists N 0 ∈ N such that
|ank − a| < ε for all k ≥ N 0 . In particular, we have
u + ε > ank > a − ε if k ≥ max{N, N 0 } where N is given in (∗).
which is impossible as u + ε = (a + u)/2 = a − ε.
(ii) (Comparison of limit superior) Let an ≤ bn for n ∈ N. Then lim(an ) ≤ lim(bn ).
(iii)
lim(an )+lim(bn ) ≤ lim(an +bn ) ≤ lim(an )+lim(bn ) ≤ lim(an +bn ) ≤ lim(an )+lim(bn ).
(iv) lim(−an ) = − lim(an ).
(v) If limn→∞ an = limn→∞ an = l, then (an ) converges to l.
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Remark
(i) If a set S is not bounded above then we write sup S = ∞, similarly we write
inf S = −∞ if S is not bounded below. Clearly, if ∅ 6= A ⊂ B, then
inf B ≤ inf A ≤ sup A ≤ sup B.
We will usually write inf ∅ = ∞ and sup ∅ = −∞.
(ii) We can extend the definition of limit superior to unbounded sequences as follows:
if (an ) is not bounded, it is still possible to define limit superior and limit inferior.
However, we will have to define it by sup S(an ) and inf S(an ) and include ∞ and −∞
as possible limits.
Homework on limit superior and inferior
(1) Given lim(an ) ≥ 0 and (bn ) is bounded, show that
lim an bn = lim an lim bn .
n→∞
n→∞
n→∞
(2) Show that a convergent sequence has at least a maximum or minimum element.
(3) Let an > 0 for n ∈ N. If lim an+1 /an = r, show that {an }∞
n=1 converges if
n→∞
r < 1. What if r ≥ 1?
(4) Show that
lim
n→∞
ÃZ
π/6
0
!1/n
n
sin xdx
1
= .
2
(5) Given limn→∞ an ≥ 5 with an > −3 for n ∈ N, show that
4an
≥ 5/2.
n→∞ 3 + an
lim
(6) Show that the set of cluster points of the sequence ((−1)n ) is {−1, 1}.
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(7) Given limn→∞ an ≥ a, show by using the definition that
a1 + a2 + a3 + · · · + an
≥ a.
n→∞
n
lim
(8) Show that
lim(an )+lim(bn ) ≤ lim(an +bn ) ≤ lim(an )+lim(bn ) ≤ lim(an +bn ) ≤ lim(an )+lim(bn ).
Appendix We will give a sketch to show that any Cauchy sequence of equivalence
classes of Cauchy sequence of rational numbers converges to a (an equivalence class
of) Cauchy sequence of rational numbers.
Indeed, first we will need to choose a better representative from each equivalence
class, we will choose them (by taking subsequences) such that
|y(k)n − y(k)m | <
1
2k+n
if m ≥ n.
Moreover, we can choose a subsequence (y(mk )) of the Cauchy sequence such that
|y(mk ) − y(ml )| < 1/2k+1 if l ≥ k.
First note we have for all l ≥ k,
|y(mk )n − y(ml )n | < 1/2k for all n ≥ 2.
Indeed, for each l ≥ k, there exists N ∈ N such that |y(mk )N − y(ml )N | < 1/2k+1 .
Hence
|y(mk )n −y(ml )n | ≤ |y(mk )n −y(mk )N |+|y(mk )N −y(ml )N |+|y(ml )N −y(ml )n | < 1/2k
for all n ≥ 2.
Claim: we may take zn = y(mn )n .
It is easy to see that it is a Cauchy sequence and lim([y(mk )]) = [(zn )] and hence
lim([y(k)]) = [(zn )].