Hi,
Your proof looks fine to me. I think maybe Tom was thinking along these lines:
Let and . Then sup S=x:
As you point out, for any and so x is an upper bound of S. Let . Choose n with -- n can be such that . Then and . Done.
The problem appears in Tom Apostol's Calculus, Volume 1, pages 31,32. This is introductory material, related to foundations of real number system.
Quote:
No proof is given and as often happens to me, contrary to the claim, I could not do the thing easily.Since we shall do very little with decimals in this book, we shall not develop their properties in any further detail except to mention how decimal expansions may be defined analytically with the help of the Least-Upper-Bound axiom.
If x is a given positive real number, let denote the largest integer Having chosen , we let denote the largest integer such that
More generally, having chosen we let denote the largest integer such that
Let S denote the set of all numbers
obtained this way for n=0,1,2... Then S is nonempty and bounded above, and it is easy to verify that x is actually the least upper bound of S.
So I need a proof that sup(S)=x.
For this purpose I should use Least-Upper-Bound axiom which roughly states that any set of real numbers that has upper bound has supremum too (which is real number).
I did try to prove it, resulting somewhat ugly and cumbersome proof, which most surely has weaknesses. I'm giving it here in hope someone could
comment on it. My attempt at the proof uses two intermediate facts.
a/ If a,x,y are real numbers which satisfy equalities for any then . It is proven in the textbook as consequence of the Least-Upper-Bound axiom
b/ for any .
The proof is by induction on n. Case n=1 is obviously true, so assuming
is true we need to prove
Since
if we manage to prove that
we shall have
.
This is easy
which is true for all .
Now to the proof of the main statement.
By construction of x,
for any and x is an upper bound of S.
By Least-Upper-Bound axiom, S has supremum and it's a real number. Denote . Then
for any .
Combining the two last inequalities with b/ we can conclude that
for any .
And because y is least upper bound we have
for any positive integer n. Applying a/ gives us the equality x=y.
I'd appreciate any comments as well as one-liner proof by contradiction for example.
Live long and prosper ;-).
Hi,
Your proof looks fine to me. I think maybe Tom was thinking along these lines:
Let and . Then sup S=x:
As you point out, for any and so x is an upper bound of S. Let . Choose n with -- n can be such that . Then and . Done.