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Now, using the lemma, we can give the proof of the Bolzano-Weierstrass Theorem.

If { a n } is a sequence of real numbers, this theorem is an immediate consequence of part (4) of the preceding lemma.

If a n = b n + c n i is a sequence of complex numbers, and if { a n } is bounded, then { b n } and { c n } are both bounded sequences of real numbers.See [link] . So, by the preceding paragraph, there exists a subsequence { b n k } of { b n } that converges to a real number b . Now, the subsequence { c n k } is itself a bounded sequence of real numbers, so there is a subsequence { c n k j } that converges toa real number c . By part (2) of [link] , we also have that the subsequence { b n k j } converges to b . So the subsequence { a n k j } = { b n k j + c n k j i } of { a n } converges to the complex number b + c i ; i.e., { a n } has a cluster point. This completes the proof.

There is an important result that is analogous to the Lemma above, and its proof is easily adapted from the proof of that lemma.

Let { a n } be a bounded sequence of real numbers. Define a sequence { y n } by y n = i n f k n a k . Prove that:

  1. { y n } is nondecreasing and bounded above.
  2. y = lim y n is a cluster point of { a n } .
  3. If z is any cluster point of { a n } , then y z . That is, y is the minimum of all the cluster points of the sequence { a n } . HINT: Let { α n } = { - a n } , and apply the preceding lemma to { α n } . This exercise will then follow from that.

The Bolzano-Wierstrass Theorem shows that the cluster set of a bounded sequence { a n } is nonempty. It is also a bounded set itself.

The following definition is only for sequences of real numbers. However, like the Bolzano-Weierstrass Theorem, it is of very basicimportance and will be used several times in the sequel.

Let { a n } be a sequence of real numbers and let S denote its cluster set.

If S is nonempty and bounded above, we define lim sup a n to be the supremum sup S of S .

If S is nonempty and bounded below, we define lim inf a n to be the infimum i n f S of S .

If the sequence { a n } of real numbers is not bounded above, we define lim sup a n to be , and if { a n } is not bounded below, we define lim inf a n to be - .

If { a n } diverges to , then we define lim sup a n and lim inf a n both to be . And, if { a n } diverges to - , we define lim sup a n and lim inf a n both to be - .

We call lim sup a n the limit superior of the sequence { a n } , and lim inf a n the limit inferior of { a n } .

  1. Suppose { a n } is a bounded sequence of real numbers. Prove that the sequence { x n } of the lemma following [link] converges to lim sup a n . Show also that the sequence { y n } of [link] converges to lim inf a n .
  2. Let { a n } be a not necessarily bounded sequence of real numbers. Prove that
    lim sup a n = inf n sup k n a k = lim n sup k n a k .
    and
    lim inf a n = sup n inf k n a k = lim n inf k n a k .
    HINT: Check all cases, and use [link] and [link] .
  3. Let { a n } be a sequence of real numbers. Prove that
    lim sup a n = - lim inf ( - a n ) .
  4. Give examples to show that all four of the following possibilities can happen.
    1.   lim sup a n is finite, and lim inf a n = - .
    2.   lim sup a n = and lim inf a n is finite.
    3.   lim sup a n = and lim inf a n = - .
    4. both lim sup a n and lim inf a n are finite.

The notions of limsup and liminf are perhaps mysterious, and they are in fact difficult to grasp.The previous exercise describes them as the resultof a kind of two-level process, and there are occasions when this description is a great help. However, thelimsup and liminf can also be characterized in other ways that are more reminiscent of the definition of a limit. These other ways are indicated in the next exercise.

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Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
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