A uncountable set of +reals.Show for any real R,there finitly distinct a,b,c,.. in A s.t a+b+c+..>=R?

For the sake of future attempters, the QUESTION is:


A is an infinite set of positive real numbers. Show that, for any real R, there exist finite and distinct real numbers a,b,c,... in A such that a + b + c + ... %26gt;= R





Answer:


By the condition of the problem,


R lies in A.





Say r is the positive integer closest to R, such that r%26gt;R.





Now, since A is the set of positive reals, one subset of the


set is the set of positive integers.





Now, by the principle of finite induction


if S is the set of positive integers {1,2,3,...,k} then whenever k lies in S, k+1 also lies in S.





Hence, we state r+1 lies in the set A.


Now since R%26lt;r, from the above statement, it follows that


R+1 lies in A. ... (I)





Let us now assume that, for a real R, and for finite and distinct a,b,c,.....such that a,b,c,.... includes the entire set A except R,





a + b + c + ... %26lt; R


This implies


a%26lt;R


b%26lt;R


c%26lt;R .......... (II)


..


..


But as stated in (I), the real number R+1 also lies in set A


Statements (II), however, inevitably yield





R+1 %26lt; R , a contradiction.


Hence, our initial assumption


a+b+c +... %26lt; R is incorrect.





Thus, there exists atleast one set of finite and distinct reals


a,b,c, such that


a+b+c+... %26gt;=R


--------------------------------------...


Correction:


Yes, I do agree with Ron, my proof is valid only for the special case wherein :


1. R lies in A.


2. A is the entire set of real numbers.


Kindly excuse me for the errors, as I attempted the proof in a state of stupor and torpor... :)





The complete proof requires additional consideration as follows:





Firstly consider that the set A bound by its largest and smallest element, occupies a band on the number line.





1. If R lies outside of A and lies to the left of the A-band, then the proof is trivial. (this accounts for R%26lt;0 too)


2. If R%26gt;0 and lies to the right of the A-band,


then, it is possible to find


n such that n = R/X where X is the smallest element of the set A. let N be the positive integer closest to and larger than n.


Now, since R is a finite real number, n and thus N are also finite.


Hence the addition of the first N numbers of the set A arranged in ascending order should give u the required result.





3. If R lies within the A-band (either within or without A), the proof easily follows from the reasoning used in 2 above.





Hope this goes towards clarifying.

A uncountable set of +reals.Show for any real R,there finitly distinct a,b,c,.. in A s.t a+b+c+..%26gt;=R?
Homework.





Let A_n={x in A: x%26gt;1/n}.





Some A_n must be uncountable.
Reply:For future voters, read the question carefully.


The long answer above is incorrect: R can be negative, but A is restricted to positive reals, so R is not necessarily in A as was assumed. The answerer also erroneously equated an uncountably infinite set of positive reals with the entire set of positive reals. The open real interval (0,1) is an uncountably infinite set of positive reals and 3 is a positive real not in that set.


The short answer given by mathematician above is concise and a good lead for homework.
Reply:Sorry, I can't answer this without knowing what the set A consists of.