Remarks about Two Theorems in Principles of Mathematical Analysis

The first chapter of the classic book Principles of Mathematical Analysis (WALTER RUDIN, Third Edition) is the real and complex number systems. Theorem 1.20 of the chapter is extracted from the construction of real number R, and it provides a good illustration of what one can do with the least-upper-bound property. Besides, theorem 1.21 proves the existence of nth roots of positive real numbers. Remarks were given because of the importance of the two theorems. Particularly, the proof of “Hence there is an integer m (with−m2 ≤ nx ≤ m1 ) such that m − 1 ≤ nx < m . ”which is not mentioned in theorem 1.20 was given in 2.2. A loophole “For every realx > 0 and every integern > 0 there is one and only one realy such that yn = x.” and a flaw of “If t = x 1+x then 0 < t < 1. Hence t n < t < x.” of theorem 1.21 were indicated in 3.1 and 3.2.

Theorem 1.20 (a) Ifx ∈ R, y ∈ R, and x > 0, then there is a positive integer n such that nx > y .
(b) If x ∈ R, y ∈ R , and x < y , then there exists a p ∈ Q such that x < p < y.
Theorem 1.21For every real x > 0 and every integer n > 0 there is one and only one real y such that y n = x.These two theorems have been proofed previously.(Rudin, 1976, p. 9-11).
1.1 Put A = {nx|x > 0, n ∈ Z + } .If (a)were false, that is to say, "If x ∈ R, y ∈ R, and x > 0, then there is not a positive integer n such that nx > y ."Namely, for ∀n ∈ Z + , nx ≤ y.Then y would be an upper bound of A, A ⊂ R, A φ (For example 1 ∈ A ),and A is bounded above, then, according to the definition of least-upper-bound property(see α is the least upper bound of A, α − x is smaller than α, and α − x is not an upper bound of A . Hence α − x < mx for some positive integer m .(Assumethis is not correct, then ∀m ∈ Z + , such that α − x ≥ mx, which is contradictory to α − x is not an upper bound of A . ) Which is impossible, since α is an upper bound of A .This is a contradiction, the assumption which (a) were false is not correct.
The proof is complete.
1.2 Sometimes, the proposition below is satisfied.
If x ∈ R, y ∈ R, and x > 0, then there is a negative integer n such that nx > y.
But it is not correct for all the situations, for example, if x ∈ R >0 , y ∈ R >0 , for all negative integer n , the multiplication of n and x is smaller than the positive numbery in R.
¢ www.ccsenet.org/jmrISSN: 1916-9795 2. Remarks about theorem 1.20 (b) 2.1 Since x < y ,we have y − x > 0, 1 ∈ R, and (a) furnishes a positive integer n such that n(y − x) > 1. Since 2.2 Here we will give the proof of "Hence there is an integer m (with We now know that −M φ and −α is an upper bound for −M , hence, sup(−M) exist, according to the least-upper-bound property.Denote β = sup(−M) .And −β is a lower bound for M is immediate from the fact that β is an upper bound for −M .Next we show that −β is the greatest lower bound, we let γ > −β and prove that γ is not a lower bound for M. Now −γ < β , hence, −γ is not an upper bound for −M , so there exists m ∈ −M such that m > −γ.Then −m ∈ M and −m < γ.
Hence,γ is not a lower bound for M .Hence, −β = in f (M).
To complete the proof, another proposition( * * ) is needed: ( * * ) For any x ∈ R , if 0 ≤ x < m then there is a positive integer k, such that k − 1 ≤ x < k.
(2) Assume 0 ≤ x < m, then there is a positive integer k, such that k − 1 ≤ x < k.
We consider0 ≤ x < m + 1, there are two cases: (i)If 0 ≤ x < m, according to the assumption, there is a positive integer k, such that k − 1 ≤ x < k.
On the other hand, for any y ∈ R, in the case of y ≥ 0, 1 > 0, according to the theorem 1.20 (a), there is a positive integer n, such that n • 1 > y.According to ( * * ) , there is a positive integer k, such that k − 1 ≤ y < k.
In the case of y < 0, −y > 0, according to( * * ), there is a positive integer k, such that k − 1 ≤ y < k.
According to the hints at the beginning of 2.2, nx ∈ R, there exists an integer m (with−m 2 ≤ nx ≤ m 1 ) such that m − 1 ≤ nx < m.
Then the proposition above is satisfied.)Step1 Let M be a nonempty set of real numbers which is bounded below, let −M be the set of all numbers −m , where m ∈ M. We will prove that in f (M) exists and in f (M) = −sup(−M).(* ) Solution: First note that −M is nonempty and bounded above.Indeed, M contains some element m , and then −m ∈ −M; moreover, M has a lower bound α, α ∈ R, ∀m ∈ M, m ≥ α .Then −α ∈ R, ∀ − m ∈ −M, −m ≤ −α , and −α is an upper bound for −M .