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 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 , 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 . Its length is (b-a)/2. Bisect this interval to get another subinterval of length which contains infinitely many points of S and denoted by []. Inductively we construct a sequence of intervals each containing infinitely many points of S and of length which tends to zero as yielding as a limit point of S. The method is extended to an infinite bounded set . Let S [a , b] × [a , b] × . . . × [a , b] = T. Repeating the steps in the previous case (i.e for ) for each interval in this product we will get limit points y_{1}, y_{2} ),…,y_{n } from the first, second , . . . , n^{th } intervals of T which finally yields a limit point (y_{1}, y_{2} ,…,y_{n) } 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** **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 (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 [**a _{1} ,b_{1}**] =

**I**. Let

_{1}**S**=

_{1}**I**

_{1}**S**. Now

**S**is an infinite bounded subset of

_{1}**I**. Then there exists a subinterval of

_{1}**I**of length 1/2 which contains infinitely many points of

_{1}**S**. Denote this interval by[

_{1}**a**] =

_{2},b2**I2**. Let

**S2**=

**I2**

**S**. In the next step we will get a subinterval of length 1/3 containing infinitely many points of

_{1}**S2**.Inductively we construct such intervals [

**a**] to get a sequence of subintervals of [

_{k},b_{k}**a,b**] namely, [

**a**] , [

_{1},b_{1}**a2 ,b2**], . . . , [

**a**] , . . . such that for each we have

_{k},b_{k}(i) [**a _{k+1} ,b_{k+1}**] [

**a**] (ii)

_{k},b_{k}**b**=

_{k}-a_{k}**1/k**(iii)

**a**

_{k}-b_{k}**contains infinitely many points of S**which implies that {

**a**} is an increasing bounded sequence and {

_{k}**b**} is a decreasing bounded sequence and hence both are convergent. Since

_{k}**b**=

_{k}-a_{k}**1/k**

**0**, it follows that

**, say**. It will be proved x is a limit point of

**S**.

Let > 0 and consider an open interval . Then, since , there exists a positive integer **m** such that both **a _{k}** and

**b**are in for all

_{k}**k**

**m**, i.e.,

**I**

_{k}= [

**a**]

_{k,}b_{k}**k**

**m**. Since each

**I**

_{k}contains infinitely many points of

**S**, therefore also contains infinitely many points of

**S**. Thus,

**x**is a limit point of

**S**.

Next, we prove the theorem for where .

Let **S** be an infinite bounded subset of . Then there exists a rectangle **T** = [**c _{1},d_{1}**] × [

**c**]× . . . × [

_{2},d_{2}**C**] in containing

_{n},d_{n}**S**in which the length of each interval is

**f >1**, i.e.,

**d**=

_{i – }c_{i}**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 and then we consider the intervals **[c _{i},c_{i}+r], [c_{i}+r, c_{i}+2r],…, [c_{i}+(m-1)r, c_{i}+mr]**, I = 1,2, . . . ,n each of length

**r**. Then the number of all possible rectangles formed by these intervals (for instance,

**[c**

_{1}+r,c_{1}+2r] x**[c**is one such rectangle) is m

_{2}+r,c_{2}+2r] x … x [c_{n}+r,c_{n}+2r]^{n}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

**[c**, then we change it to

_{i}+(m-1)r,c_{i}+mr]**[d**so that

_{i}-r,d_{i}]**J**is now a subrectangle of

**T**.

Therefore, for each positive integer **k**there is a subrectangle **J _{k}** 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

**J**of

_{k }= [a_{k1},b_{k1}] x [a_{k2},b_{k2}] x … x [a_{kn},b_{kn}]**T**which contains infinitely many points of

**S**and the length of each interval in

**J**is

_{k}**1/k.**

Since , **for each i= 1,2, . . . ,n,** the sequence {**a _{ki}**} is increasing and bounded above, the sequence {

**b**} is decreasing and bounded below and hence they are convergent.

_{ki}Also, as for each **I = 1,2, . . . , n** implies that .Call this limit **x _{i}** . Then the point x = (

**x**) is the limit of the sequence {(

_{1},x_{1},…, x_{i},…, x_{n}**a**)}. To prove that this point is also a limit point of

_{k1},a_{k2},…, a_{kn}**S**.

Let and consider an open rectangle V = . Since , there exists a positive integer **m _{1}** such that , that is, . Similarly, there exist positive integers

**m**, … etc such that . Let m = maximum{

_{2},m_{3}**m**}. Then, for each we have . Since each

_{1},m_{2},…,m_{n }**J**contains infinitely many points of

_{k }**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,