Another proof of Bolzano-Weierstrass Theorem

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Authors

S. KHONGSIT

Mathematics Department, Lady Keane College, Shillong -793001, Meghalaya, India
shailanstar@gmail.com

Abstract

Bolzano-Weierstrass theorem for infinite sets states that every infinite bounded subset of R^n has a limit point. In this article a proof of this theorem is given which is similar but different from that given in ‘Mathematical analysis’ by Tom Apostol.

Keywords Bolzano-Weierstrass Theorem, Proof.

Introduction

There are two versions of Bolzano-Weierstrass theorem, the first says that every bounded sequence of real numbers has a convergent subsequence (Apostol, 1997) and the second states that every infinite bounded set in R^{n} has a limit point (Bartle and Sherbert, 2012). Karl Weierstrass developed the theorem independently and published years after Bolzano’s first proof and which was initially called the Weierstrass theorem until Bolzano’s earlier work was rediscovered (Merzbach and Boyer, 1991).

Following is a summary of the method used in (Bartle and Sherbert, 2012) for the proof of the theorem:

If S be an infinite bounded subset of \mathbb{R}, then there is an interval [a,b] containing S. Bisect the interval [a,b] to get one sub-interval containing infinitely many points of S. Let this be denoted by [a_{1},b_{1}]. Its length is (b-a)/2. Bisect this interval to get another subinterval of length (b-a)/2^{2}) which contains infinitely many points of S and denoted by [a_{2},b_{2}]. Inductively we construct a sequence of intervals \left \{ [a_{k},b_{k}] \right \}each containing infinitely many points of S and of length \left ( b-a \right )/2^{n} which tends to zero as n \to \infty yielding lim_{k\rightarrow \infty }a_{k} = lim_{k\rightarrow \infty }b_{k}as a limit point of S. The method is extended to an infinite bounded set S\subseteq R^{n}. Let S \subseteq [a , b] × [a , b] × . . . × [a , b] = T. Repeating the steps in the previous case (i.e for S\subseteq R^{1} ) for each interval in this product we will get limit points y1, y2 ),…,y from the first, second , . . . , nth  intervals of T which finally yields a limit point (y1, y2 ,…,yn)  of S.

We prove the theorem as follows:

Proof: First we will prove the theorem for the case n = 1. Let S be an infinite bounded subset of R.Then there exists a closed interval [a,b]=I which contains S. We choose this interval so that b-a > 1

We will show that for any positive number r < b-a, however small it may be, there is a sub-interval of  I of length r which contains infinitely many points of S.

Let r >0 and r< b-a. Choose a positive integer m such that a + (m-1)r \leq b <a+mr. Then each of the intervals [a,a+r], [a+r,a+2r], . . . , [a+(m-1)r,a+mr], is of length r and their union contains the interval [a,b]. At least one of these finite number of intervals contains infinitely many points of S. Call one such interval as J. But since we want such an interval to be contained in I,if J is the last interval [a+(m-1)r ,a+mr] , then we choose J=[b-r,b]. Thus we have proved our claim.

As a result of this, for each n\in \mathbb{N} (the set of all natural numbers), there is a subinterval of [a,b] of length 1/n which contains infinitely many points of S.

Therefore, there is a subinterval of [a,b] of length 1 which contains infinitely many points of S. Call this [a1 ,b1] =I1. Let S1 = I1 \cap S. Now S1 is an infinite bounded subset of I1. Then there exists a subinterval of I1 of length 1/2 which contains infinitely many points of S1. Denote this interval by[a2 ,b2] =I2. Let S2 = I2 \cap S1. In the next step we will get a subinterval of length 1/3 containing infinitely many points of S2 .Inductively we construct such intervals [ak ,bk] to get a sequence of subintervals of [a,b] namely, [a1 ,b1] , [a2 ,b2], . . . , [ak ,bk] , . . . such that for each k\in \mathbb{N} we have

(i) [ak+1 ,bk+1\nsubseteq [ak ,bk] (ii) bk-ak = 1/k (iii) ak-bk contains infinitely many points of S which implies that {ak} is an increasing bounded sequence and {bk} is a decreasing bounded sequence and hence both are convergent. Since bk-ak = 1/k \rightarrow 0, it follows that lim_{k\rightarrow \infty }a_{k}=lim_{k\rightarrow \infty }b_{k}=x, say. It will be proved x is a limit point of S.

Let \epsilon > 0 and consider an open interval ]x-\epsilon ,x+\epsilon [. Then, since lim_{k\rightarrow \infty }a_{k}=lim_{k\rightarrow \infty }b_{k}=x, there exists a positive integer m such that both ak and bk are in  ]x-\epsilon ,x+\epsilon [ for all k \geq m , i.e., Ik = [ak,bk] \subseteq  ]x-\epsilon ,x+\epsilon [ \forall k \geq m. Since each Ik contains infinitely many points of S, therefore ]x-\epsilon ,x+\epsilon [ also contains infinitely many points of S. Thus, x is a limit point of S.

Next, we prove the theorem for \mathbb{R}^{n} where n\geq 1.

Let S be an infinite bounded subset of \mathbb{R}^{n}. Then there exists a rectangle T = [c1,d1] × [c2,d2]× . . . × [Cn,dn] in \mathbb{R}^{n} containing S in which the length of each interval is f >1, i.e.,di – ci = f > 1 for each I = 1,2, . . . , n.

As in the case n=1, if 0 < r < f we choose a positive integer m such that (m-1)r\leq f< mr and then we consider the intervals [ci,ci+r], [ci+r, ci+2r],…, [ci+(m-1)r, ci+mr], \forall I = 1,2, . . . ,n each of length r. Then the number of all possible rectangles formed by these intervals (for instance, [c1+r,c1+2r] x [c2+r,c2+2r] x … x [cn+r,cn+2r] is one such rectangle) is mn and their union contains T. At least one of these rectangles contains infinitely many points of S. Denote one such rectangle by J. Again, as in the case n=1, if there is an interval in J which is the the last interval [ci+(m-1)r,ci+mr], then we change it to [di-r,di] so that J is now a subrectangle of T.

Therefore, for each positive integer kthere is a subrectangle Jk of T containing infinitely many points of S in which the length of each interval in it is 1/k. Inductively, we construct a sequence of nested closed subrectangles Jk = [ak1,bk1] x [ak2,bk2] x … x [akn,bkn] of T which contains infinitely many points of S and the length of each interval in Jk is 1/k.

Since J_{1}\supseteq J_{2}\supseteq ...\supseteq J_{k}\supseteq ... , for each i= 1,2, . . . ,n, the sequence {aki} is increasing and bounded above, the sequence {bki} is decreasing and bounded below and hence they are convergent.

Also, b_{ki}-a = 1/k \rightarrow 0 as k\rightarrow \infty for each I = 1,2, . . . , n implies that lim \rightarrow _{k\rightarrow \infty }b_{ki} = lim_{k\rightarrow \infty }a_{ki} .Call this limit xi . Then the point x = (x1,x1,…, xi,…, xn) is the limit of the sequence {(ak1,ak2,…, akn)}. To prove that this point is also a limit point of S.

Let \delta > 0 and consider an open rectangle V = ]x_{1}-\delta ,x_{1}+\delta [X]x_{2}-\delta ,x_{2}+\delta [X...X]x_{n}-\delta ,x_{n}+\delta. Since lim_{k\rightarrow \infty }a_{k1} = lim_{k\rightarrow \infty }b_{k1}=x_{1}, there exists a positive integer m1 such that b_{k1},a_{k1}\in ]x_{1}-\delta,x_{1}+\delta [\forall k\geq m_{1} , that is, \square [a_{k1},b_{k1}]\subseteq ]x_{1}-\delta ,x_{1}+\delta [\forall k\geq m_{1}. Similarly, there exist positive integers m2,m3, … etc such that [a_{k2},b_{k2}]\subseteq]x_{2}-\delta ,x_{2}+\delta [\forall k\geq m_{2},...,[a_{kn},b_{kn}]\subseteq ]x_{n}+\delta [\forall k\geq m_{n} . Let m = maximum{ m1,m2,…,m}. Then, for each k\geq m we have J_{k}=[a_{k1},b_{k1}]X[a_{k2},b_{k2}]X...X[a_{kn},b_{kn}]\subseteq V. Since each Jcontains infinitely many points of S, therefore, V will also contain infinitely many points of S, showing that x is a limit point of S. This completes the proof.

References

Apostol, T.M. 1997. Mathematical Analysis, Narosa Publishing House, New Delhi, .ISBN-81-85015-66-X.

Bartle, R.G. and Sherbert, D.R. 2012, Introduction to Real Analysis, Wiley India (P.) Ltd.,New Delhi, ISBN-13:978-81-265-1109-9.

Merzbach, U.C. and Boyer, C.B. 1991. A History of Mathematics, New York: John Wiley & Sons,

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