Bolzano Weierstrass Theorems I
James K. Peterson
Department of Biological Sciences and Department of Mathematical Sciences
Clemson University
September 8, 2017
Outline
The Bolzano Weierstrass Theorem
Extensions to <2
Bounded Infinite Sets
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Bolzano-Weierstrass Theorem: Convergent Subsequences of Bounded Sequences, Study notes of Biological Sciences
An outline and proof of the Bolzano-Weierstrass Theorem, which states that every bounded sequence with an infinite range has at least one convergent subsequence. The proof is given for sequences in the real numbers, and extensions to higher dimensions are discussed.
What you will learn
- What is the Bolzano-Weierstrass Theorem?
- Why does every bounded sequence with an infinite range have at least one convergent subsequence?
- How is the Bolzano-Weierstrass Theorem proven?
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Bolzano Weierstrass Theorems I
James K. Peterson
Department of Biological Sciences and Department of Mathematical Sciences Clemson University
September 8, 2017
Outline
The Bolzano Weierstrass Theorem
Extensions to <^2
Bounded Infinite Sets
Theorem
Bolzano Weierstrass Theorem Every bounded sequence with an infinite range has at least one convergent subsequence.
Proof
As discussed, we have already shown a sequence with a bounded finite range always has convergent subsequences. Now we prove the case where the range of the sequence of values{a 1 , a 2... , } has infinitely many distinct values. We assume the sequences start at n = k and by assumption, there is a positive number B so that −B ≤ an ≤ B for all n ≥ k. Define the interval J 0 = [α 0 , β 0 ] where α 0 = −B and β 0 = B. Thus at this starting step, J 0 = [−B, B]. Note the length of J 0 , denoted by ` 0 is 2 B. Let S be the range of the sequence which has infinitely many points and for convenience, we will let the phrase infinitely many points be abbreviated to IMPs.
Proof
Step 1: Bisect [α 0 , β 0 ] into two pieces u 0 and u 1. That is the interval J 0 is the union of the two sets u 0 and u 1 and J 0 = u 0 ∪ u 1. Now at least one of the intervals u 0 and u 1 contains IMPs of S as otherwise each piece has only finitely many points and that contradicts our assumption that S has IMPS. Now both may contain IMPS so select one such interval containing IMPS and call it J 1. Label the endpoints of J 1 as α 1 and β 1 ; hence, J 1 = [α 1 , β 1 ]. Note 1 = β 1 − α 1 = 12 0 = B We see J 1 ⊆ J 0 and
−B = α 0 ≤ α 1 ≤ β 1 ≤ β 0 = B
Since J 1 contains IMPS, we can select a sequence value an 1 from J 1. Step 2: Now bisect J 1 into subintervals u 0 and u 1 just as before so that J 1 = u 0 ∪ u 1. At least one of u 0 and u 1 contain IMPS of S.
Proof
I (^) From our proposition, we have proven the existence of three sequences, (αp )p≥ 0 , (βp )p≥ 0 and (p )p≥ 0 which have various properties. I The sequencep satisfies p = (1/2)p− 1 for all p ≥ 1. Since 0 = 2B, this means 1 = B, 2 = (1/2)B, 3 = (1/ 22 )B leading to `p = (1/ 2 p−^1 )B for p ≥ 1. I
−B = α 0 ≤ α 1 ≤ α 2 ≤... ≤ αp ≤... ≤ βp ≤... ≤ β 2 ≤... ≤ β 0 = B
I (^) Note (αp )p≥ 0 is bounded above by B and (βp )p≥ 0 is bounded below by −B. Hence, by the completeness axiom, inf (βp )p≥ 0 exists and equals the finite number β; also sup (αp )p≥ 0 exists and is the finite number α.
Proof
I (^) So if we fix p, it should be clear the number βp is an upper bound for all the αp values ( look at our inequality chain again and think about this ). Thus βp is an upper bound for (αp )p≥ 0 and so by definition of a supremum, α ≤ βp for all p. Of course, we also know since α is a supremum, that αp ≤ α. Thus, αp ≤ α ≤ βp for all p. I (^) A similar argument shows if we fix p, the number αp is an lower bound for all the βp values and so by definition of an infimum, αp ≤ β ≤ βp for all the αp values I (^) This tells us α and β are in [αp , βp ] = Jp for all p. Next we show α = β.
Proof
I (^) Let > 0 be arbitrary. Since α and β are in Jp whose length is `p = (1/ 2 p−^1 )B, we have |α − β| ≤ (1/ 2 p−^1 )B. Pick P so that 1 /(2P−^1 ) < . Then |α − β| < . But > 0 is arbitrary. Hence, by a previous propostion, α − β = 0 implying α = β. I We now must show ank → α = β. This shows we have found a subsequence which converges to α = β. We know αp ≤ anp ≤ βp and αp ≤ α ≤ βp for all p. Pick > 0 arbitrarily. Given any p, we have
|anp − α| = |anp − αp + αp − α|, add and subtract trick ≤ |anp − αp | + |αp − α| triangle inequality ≤ |βp − αp | + |αp − βp | definition of length = 2 |βp − αp | = 2 (1/ 2 p−^1 )B.
Choose P so that (1/ 2 P−^1 )B < / 2. Then, p > P implies |anp − α| < 2 /2 = . Thus, ank → α.
Theorem
Bolzano Weierstrass Theorem in <^2 Every bounded sequence of vectors with an infinite range has at least one convergent subsequence.
Proof
We will just sketch the argument. The sequence of vectors looks like
xn =
[
x 1 n x 2 n
]
where each element in the sequence is a two dimensional vector. Since this sequence is bounded, there are positive numbers B 1 and B 2 so that
−B 1 ≤ x 1 n ≤ B 1 and −B 2 ≤ x 2 n ≤ B 2
A little thought also shows
Theorem
Bolzano Weierstrass Theorem in <^4 Every bounded sequence of vectors with an infinite range has at least one convergent subsequence.
Proof
We now bisect each edge of what is called a 4 dimensional hypercube and there are now 16 pieces at each step, at least one of which has IMPs. The vectors are now 4 dimensional but the argument is quite similar.
POMI allows us to extend the result to
Theorem
Bolzano Weierstrass Theorem in <n Every bounded sequence of vectors with an infinite range has at least one convergent subsequence.
Proof
We have done the basis step and in the induction step we assume it is true for n − 1 and show it is true for n. We now bisect each of the n edges of what is called a n dimensional hypercube and there are now 2 n^ pieces at each step, at least one of which has IMPs. The vectors are now n dimensional but the argument is again quite similar.
A more general type of result can also be shown which deals with sets which are bounded and contain infinitely many elements.
Definition
Let S be a nonempty set. We say the real number a is an accumulation points of S if given any r > 0, the set
Br (a) = {x : |x − a| < r }
contains at least one point of S different from a. The set Br (a) is called the ball or circle centered at a with radius r.
Example
S = (0, 1). Then 0 is an accumulation point of S as the circle Br (0) always contains points greater than 0 which are in S, Note Br (0) also contains points less than 0. Note 1 is an accumulation point of S also. Note 0 and 1 are not in S so accumulation points don’t have to be in the set. Also note all points in S are accumulation points too. Note the set of all accumulation points of S is the interval [0, 1].
Example
S = ((1/n)n≥ 1. Note 0 is an accumulation point of S because every circle Br (0) contains points of S different from 0. Also, if you pick a particular 1/n in S, the distance from 1/n to its neighbors is either 1 /n − 1 /(n + 1) or 1/n − 1 /(n − 1). If you let r be half the minimum of these two distances, the circle Br (1/n) does not contain any other points of S. So no point of S is an accumulation point. So the set of accumulation points of S is just one point, { 0 }.