Prove Sup S = c

  • #1
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0

Homework Statement


S≡{x|x∈ℝ,x≥0,x2 < c} Prove Sup S = c

Homework Equations




The Attempt at a Solution


Since x in the set real numbers, there are two cases for x: x < 1 or 0 <= x <=1

if 0 <= x <=1, then x < c + 1 since c is positive.
if x < 1, then x^2 < c < x*c + x = x(c+1)
thus x < c+1, by the completeness axiom, S has a least upper bound, let's called it b.

I am stuck at this step, can anyone give me a hit or a suggestion to keep going. Thanks
 

Answers and Replies

  • #2
35,062
6,796

Homework Statement


S≡{x|x∈ℝ,x≥0,x2 < c} Prove Sup S = c

Homework Equations




The Attempt at a Solution


Since x in the set real numbers, there are two cases for x: x < 1 or 0 <= x <=1
It would seem to me that the two cases are ##0 \le x \le 1## and ##x > 1##. Is there some reason you're not considering ##x > 1##?
HaLAA said:
if 0 <= x <=1, then x < c + 1 since c is positive.
if x < 1, then x^2 < c < x*c + x = x(c+1)
thus x < c+1, by the completeness axiom, S has a least upper bound, let's called it b.

I am stuck at this step, can anyone give me a hit or a suggestion to keep going. Thanks
 
  • #3
621
419
Set S consists of all real numbers ##x## such that ##x\geq 0## and ##x^2 < C##.
We know ##C>0##, for if ##C<0##, there exists no real number that satisfies ##x^2 < C## and if ##C=0##, then ##x^2 < 0##, which is also impossible.
Provided that ##C>0## then we can say set S is non-empty and then apply the axiom:
If a non-empty subset has an upper bound, then it has a least upper bound.

If we establish ##b## as the candidate for the least upper bound, then ##b## satisfies certain critera:
1) ##x\leq b## for every ##x\in S## (in other words, ##b## has to be an upper bound)
2)For every ##\varepsilon >0##, there exists an ##x_0## such that ##b-\varepsilon < x_0## (meaning ##b## is the least of upper bounds)
EDIT:
The second criterion could also be written as:
2a)For every ##y\in\mathbb{R}## such that ##y < b##, there exists an ##x_0\in S## such that ##y < x_0##
2 and 2a mean the same thing.

We can agree that ##b## is an upper bound, now fix an arbitrary ##\varepsilon >0## and show that ##b## is indeed the least of upper bounds and in the finale it follows immediately after that ##\sqrt{C} = b##.
 
Last edited:
  • #4
Surely sup S = sqrt c not c ? Unless I'm missing something ?
 
  • #6
ok then so the proof of the OP would proceed as follows:

suppose Sup S = sqrt c + epsilon where epsilon > 0. Then (Sup S - epsilon /2)^2 > c (contradicting the definition of Sup S)
suppose Sup S = sqrt c - epsilon. Then (sup S + epsilon/2)^2 <c (again contradicting the definition of Sup S)
so Sup S = sqrt c
 
  • #7
621
419
ok then so the proof of the OP would proceed as follows:

suppose Sup S = sqrt c + epsilon where epsilon > 0. Then (Sup S - epsilon /2)^2 > c (contradicting the definition of Sup S)
suppose Sup S = sqrt c - epsilon. Then (sup S + epsilon/2)^2 <c (again contradicting the definition of Sup S)
so Sup S = sqrt c
Well, hurts my eyes without TeX, for one, but more importantly, do not post complete solutions. I merely referred to the definition in an attempt to hint at how to proceed.
 
  • #9
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It would seem to me that the two cases are ##0 \le x \le 1## and ##x > 1##. Is there some reason you're not considering ##x > 1##?
That is a typo
 

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