Math 311 W08
Day 6
Section 2.4
1. Read example 2.31 it gives another example of finding the limit of a
recursively defined sequence like the example we did last time:
#% 2
if n = 1
$
%& 2 + a n"1 if n > 1
!
2. Definition. Consider a sequence {an}. Let {nk}be a sequence of
natural numbers that is strictly increasing; that is n1 < n2 < n3 < •••.
Then the sequence {bk} defined by bk = an k for every index k is called a
subsequence of the sequence {an}.
3. Intuitively you can think of building a subsequence by choosing terms
! moving farther along the original
in the original sequence always
sequence. (Picture)
4. Examples:
a. Consider the sequence defined by an = 1 + (-1)n and the
sequence of natural numbers defined by nk = 2k.
The subsequence bk = an k = 1 + (-1)2k = 2 for all values of k. This is
the subsequence consisting of all the 2’s in the original sequence.
(If we used the sequence defined by nk = 2k - 1 we would get the
subsequence!consisting of all the 0’s in the original sequence.)
b. If we take any sequence {an} and use the sequence of natural
numbers given by nk = k + 7, we get the subsequence given by
an k = ak+7. This is the subsequence obtained by simply starting
with the 8th term of the original sequence rather than the 1st.
!
5. Theorem (Prop 2.30) If a sequence converges then all of its
subsequences converge to the limit of the original sequence.
See proof in book (make sure it makes sense to you).
6. Note that in Example (a) above each of the subsequences mentioned
were convergent (they were constant) even though the original
sequence did not converge.
a. Some non-convergent sequences do not have any subsequences
that converge. For example, the sequence defined by an = n for
all indices n does not have any convergent subsequences.
b. However, bounded sequences always have a convergent
subsequence. To prove this we need to prove another theorem
first.
7. Theorem (2.32). Every sequence has a monotone subsequence.
The Idea of the Proof: Suppose that you tried to make an increasing
sequence. You would pick a first term and then move along the sequence
always picking larger (or at least equal) terms. Suppose you got stuck.
This would mean that you found a term of the sequence that was bigger
than all the rest of the terms. Let’s call a term like this a peak term.
There are two cases.
Case 1: There are only finitely many peak terms. Then we can make an
increasing sequence easily. Go to the last peak term and start from the
next term, you can then go happily along making your subsequence
because you will not run into any more peak terms (so there will always
be a larger or equal term to choose).
Case 2: There are infinitely many peak terms. Then these peak terms
form a decreasing subsequence. Why? Because each peak term is bigger
than all the remaining terms (and so is bigger than the next peak term).
Theorem (2.33). Every bounded sequence has a convergent
subsequence.
Proof. By the previous theorem, we can find a monotonic
subsequence. This subsequence is clearly bounded (since the original
sequence is bounded). So by the Monotone Convergence Theorem,
this sequence converges.
8. Definition. A set of real numbers S is said to be sequentially compact
provided that every sequence {an} in S has a subsequence that
converges to a point that belongs to S.
9. Examples:
a. (0, 2) is not sequentially compact. {1/n} is a sequence in this set
that converges to 0. Thus every subsequence also converges to
0, which is not an element of the set (0, 2).
b. [0, 2] is sequentially compact. Since any sequence in [0, 2] is
clearly bounded (by 2), we know by the previous theorem that it
has a convergent subsequence. And by Theorem 2.22, we also
know that limit of this subsequence must be in [0, 2].
10. Theorem (2.36). The Sequential Compactness Theorem. Let a and
b be numbers such that a < b. Then the interval [a, b] is sequentially
compact.
Proof: (Really done in the previous example). Any sequence in [a, b]
will be bounded and hence will have a convergent subsequence by the
previous theorem. The limit of this subsequence will be in [a, b] by
Theorem 2.22.