Review

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Review

• Open Cover If a family of open sets whose union
  contains a set S, the family is called an open
  cover of S.
• Subcover If a subcollection of an open cover of
  S is also an open cover of S, then the
  subcollection is called a subcover.

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Review

• Compact Sets A set S is said to be compact if
  every open cover of S contains a finite subcover.
• Examples
• Any finite set S is compact.
• Sa, b is compact.
• A closed subset of a, b is compact.

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Review

• Examples
• 4. S(0, 1) is not compact
• 5. SR is not compact.
• Lemma (Nested Interval Theorem)
• Let be a collection of
  closed intervals such that
• Then empty set.

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• Theorem ( Heine-Borel) A subset S of R is compact
  if and only if S is closed and bounded.
• More Theorem. If a bounded subset S of R contains
  infinitely many points, then there exists at
  least one point in R that is an accumulation
  point of S.
• Theorem Let FKa a is in A be a family of
  compact sets of R. Suppose that the intersection
  of any finite subfamily of F is nonempty. Then
• nKa a is in A is not empty.

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Chapter 4 Sequences

• Convergence
• Limit Therorems
• Monotone Sequences and Cauchy Sequences
• Subsquences

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4.16 Convergence

• Sequences
• Definition
• A sequence is a function whose domain is the set
  N of natural numbers.
• A sequence is usually denoted by s.
• The value of s at n is denoted by sn instead of
  s(n) and is called the n-th term of the sequence.
• 2) Example

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Examples

• a) sn 2n-1
• (1, 3, 5, 7, ..)
• b) sn (-1)n
• (-1, 1, -1, 1, .)
• c) sn 1-1/n
• (0, 1/2, 2/3, 3/4, 4/5, )

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Convergence

• 2. Convergence
• 1) Definition
• A sequence sn is said to converge to the
  real number s provided that
• for each egt0 there exists a real number K
  such that for all n in N, ngtK implies that
• sn slt e.

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Convergence

• If sn converges to s, then s is called the limit
  of the sequence sn and we write
• limn?8 sn s, lim sn s, or sn ?s.
• If a sequence does not converge to a real number,
  it is said to diverge.
• 2) Example
• a) sn 1-1/n, lim sn 1

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Examples

• Given egt0, let K1/ e. Then for any ngtK we have
  1-1/n-1
• -1/n
• 1/nlt1/K e
• So lim sn 1

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Practice

• sn 21/n2, prove that lim sn 2.

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Practice

• Proof. Given egt0, let K . Then for any ngtK
  we have
• 21/n2-2
• 1/n2
• 1/n2lt1/K2 e
• So lim sn 2

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Homework

• 16.1, 16.2, 16.3 on page 164
• (optional) 14.3 (c), (d), 14.11 on page 144-145.

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