accumulation points and the bolzano weierstrass theorem

Why Bolzano-Weierstrass?The Bolzano-Weierstrass Theorem

Accumulation Points and the BolzanoWeierstrass Theorem

Dr. Lance Nielsen

January 26, 2007

Dr. Lance Nielsen Accumulation Points

Outline

1 Why Bolzano-Weierstrass?CompletenessHeine-Borel TheoremWhy Bolzano?

2 The Bolzano-Weierstrass Theorem

Outline

CompletenessHeine-Borel TheoremWhy Bolzano?

Outline

Completeness of R

DefinitionAn ordered field F is complete if every nonempty subset A of Fwhich is bounded above has a least upper bound.

ExampleR is complete, as was proved using Dedekind cuts.It is worth noting that completeness for other spaces, forexample Rn, does not depend on order properties.

Completeness of R

Outline

Statement of the Theorem

TheoremLet A ⊂ R. A is compact in R if and only if A is closed andbounded.

Compactness

A subset A of a toplogical space X is compact if everyopen cover of A admits a finite subcover.Closed and bounded intervals in R are compact as arefinite unions of closed and bounded intervals. (ByHeine-Borel.)Arbitrary intersections of compact sets are compact. (Theproof is an easy exercise.)

Compactness

Outline

Comments

1 The completeness of the real number system has beendiscussed in terms of the least upper bound property.

2 Our first consequence of the completeness of the real linewas the Heine-Borel Theorem, which dealt withcompactness.

3 A second consequence of completeness of the real line isthe Bolzano-Weierstrass Theorem.

The Bolzano-Weierstrass Theorem involves infinitebounded sets of real numbers and shows that in such a setthere is a at least one number around which points in theset "accumulate" or "cluster".This is also related to compactness, as we will see later onin our discussion of sequences.We want a theorem like Bolzano-Weierstrass sincebounded sets don’t have to be closed and our question isthen "how far from compact are they?".

Comments

Definition of Accumulation Point

1 Let S be a subset of a topological space X . A point x ∈ Xis an accumulation point or cluster point of S if every openset U, x ∈ U, is such that U ∩ S 6= ∅.

2 In R, this definition can be recast as follows: x ∈ R is anaccumulation point for S if for every ε > 0 there is a points ∈ S such that 0 < |s − x | < ε.

Definition of Accumulation Point

1 Let S be a subset of a topological space X . A point x ∈ Xis an accumulation point or cluster point of S if every openset U, x ∈ U, is such that U ∩ S 6= ∅.

2 In R, this definition can be recast as follows: x ∈ R is anaccumulation point for S if for every ε > 0 there is a points ∈ S such that 0 < |s − x | < ε.

Examples

1 For S = {1n : n ∈ N}, there is only one accumulation point,

namely 0. It is obvious that 0 /∈ S.2 The set Z has no accumulation points.3 Every real number is an accumulation point of Q. (Why?)

Examples

Bolzano-Weierstrass Theorem

TheoremLet S ⊂ R be a nonempty infinite set. The S has at least oneaccumulation point.

Proof of Bolzano-Weierstrass

Proof:Let S ⊂ R be bounded above by b and below by a. Definethe set P as follows: a real number p is in P if and only ifthere are at most a finite number of elements of S that areless than p.

Note that if p ∈ P and q ≤ p, then q ∈ P because anyelement of S less than q is less than p.a ∈ P and so P 6= ∅.

P has a least upper bound: This follows at once since P isbounded above and since R is complete. Let λ = sup P.

Proof Continued

λ is an accumulation point of S: Given any ε > 0, there aretwo cases to consider:

λ ∈ P: Then λ + ε is not in P, since λ is a least upper boundfor P. This tells us that there are an infinite number ofelements of S that are less than λ + ε. But only a finitenumber of these same elements are less than or equal to λ,since λ ∈ P. So there must be at least one element of S in(λ, λ + ε). It is clear that this element is an accumulationpoint for S.λ /∈ P: Since λ is the l.u.b. for P, there is some elementp ∈ P such that λ− ε ≤ p < λ. Since p ∈ P, there are only afinite number of elements of S that are less than p. On theother hand, λ /∈ P; therefore λ is greater than or equal to aninfinite number of elements of S. So there must be at leastone element s ∈ S betwen p and λ. This finishes the proof.

Proof Continued

accumulation points and the bolzano weierstrass theorem

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