Re: Real Numbers and Supremum and Infimum

On Aug 6, 1:48 pm, hagman <goo...@xxxxxxxxxxxxx> wrote:
On 6 Aug., 09:08, kusbo <kubsot...@xxxxxxxxx> wrote:



On Aug 2, 11:24 pm, Angus Rodgers <twir...@xxxxxxxxxxx> wrote:

On Sat, 2 Aug 2008 11:02:01 -0700 (PDT), kusbo

<kubsot...@xxxxxxxxx> wrote:
Thanks , but i don't understand how (x+x)/2 < (x+y)/2
and (x+y)/2 < (y+y)/2 . I will be grateful , if you can give me
steps by which you came to above conclusion.
What am i missing here? These theorems looks obvious. It is now more
than 24 hrs , i am not even able to solve exercise from this
introductory chapter. Can you tell me what parts i need to study again
to solve this problem or similar problems.
Thanks

Working with familiar elementary properties of numbers can be
quite mindbending, can't it?

The answer to your question is not unique, and in particular it
depends on what postulates and definitions you're working with.

(I don't have Apostol's book to hand.)

But one possibility is that you have the following postulates
and definitions (in a different order, with different wording,
and with unpredictable status as postulates or definitions or
theorems - but all at any rate true!):

(i) A real number x is negative if and only if -x is positive.

(ii) Every non-zero real number is either positive or negative.

(iii) x < y means that y - x is positive. (So x > 0 is equivalent
to x being positive, and x < 0 is equivalent to x being negative.)

(iv) The sum of two positive real numbers is positive.

From (iv), we have at once that if x > 0, then 2x > 0. Also, if
x < 0, then -x > 0, therefore -(2x) = 2(-x) > 0, i.e. 2x < 0.

So we have 2x > 0, or = 0, or < 0 according as x > 0, = 0, or < 0.
Call this result (v).

(Don't forget to make sure that you understand why -(2x) = 2(-x)!)

From (v), we infer that if x > 0, then x/2 > 0 (because x = 2(x/2),
by the definition of x/2). Call this result (vi).

Now, we are supposing that x < y. By (iii), this means y - x > 0.

Therefore, by (vi), (y - x)/2 > 0.

Therefore, (x + y)/2 - (x + x)/2 = (y - x)/2 > 0.

But, by (iii) again, this means (x + x)/2 < (x + y)/2.

Q.E.D.

(Don't forget to make sure you understand why (x + y)/2 - (x + x)/2
= (y - x)/2!)

The proof of (x + y)/2 < (y + y)/2 is similar. Try it! Or else do
it differently. (It's largely a matter of taste.) Can you improve
on the above proof of (x + x)/2 < (x + y)/2? Or just vary it a bit?

--
Angus Rodgers
(twirlip@ eats spam; reply to angusrod@)
Contains mild peril

Sorry to bother you too soon :) . I read the intro chapter again ,
this time slowly . If possible can verify the proof of above ,which
i attempted using fundamental properties of supremum and infimum.
Thanks in advance .

; If x and y are arbitrary real numbers with x < y , prove that
; there is atleast one real z satisfying x < z < y .

;Proof :
; Let S be a set of real numbers such that {a | x <= a , a < y }
; Properties for this set S is as follows :
; 1. S is non empty , since x is in the S
; 2. S is bounded below and x=inf S and x=min S
; 3. S is bounded above and y=sup S

How exactly do you know that y = sup S?
Since S is bounded by y, we clearly have sup S <= y.
But couldn't it be the case - as of now - that sup S < y?
Clearly sup S >= x. So just to prove that sup S > x, you need
that there exists a number z in S with x < z < y, but
this is the theorem you want to prove!



; We know that if S has a infimum , then for some z in S we have
; z < inf S + h
; where h is a positive integer

; therefore , z < x + h
; Hence we have x < z ,

? How this? You only have x <= z and z < x+h right here.



also z is smaller than y , since y is
; least upper bound
; and y is not in set S , therfore we have
;
; x < z < y
;Hence there exists atleast one real number say z , which satisfy the
above equality x < z < y.
;

; Theorem which have been proved in the book , which i made use
; of in above proof :

; Let h be a given positive integer and let S be a set of
; real numbers .
; (a) If S has supremum , then for some x in S we have
; x > sup S - h
;
; (b) If S has a infimum , then for some x in S we have
; x < inf S + h

I assume that the statement reads "positive real" instead of "positive
integer", doesn't it?

hagman

You seems right when you say sup S <= y . Since y is upper bound ,
and if there is a least upper bound , then it must <= y.

so, we just say that y is a upper bound , makes no other assumption
as of now.

But isn't inf S = x , since there x=min S .

For some z in S this is true :
z < inf S + h
where h is positive integer

Proof of this is something like this ,
;Consider a non-empty set S of real number bounded below and let
;h be a positive integer.

;Since set is bounded below , and non empty , then it must have
;greatest lower bound. Then following is true for some z in S
; z < inf S + h
;because if z >= inf S + h for all z in S then
;inf S + h will be greatest lower bound for S
;hence there for we must have a z < inf S + h for at least
;one z in S .

So if above is correct then atleast for some z in S
z < x + h , and z cannot be z > x + h , since x > x + h is not
possible and we know x < y , therefore x < z < y

Right? or am i missing something here
.

Relevant Pages

  • Re: analysis exercise,
    ... I see this is true from a picture but I could not come up with a direct ... I tried supposing sup S> inf T. Then pictorially, ... so pick any t1 in T. Is it an upper bound for the set S? ...
    (sci.math)
  • Re: analysis exercise,
    ... On Wed, 30 Aug 2006, vsgdp wrote: ... t is upper bound of S. ... Thus sup S <= t, ... Thus sup S <= inf T ...
    (sci.math)
  • Re: Idiosyncratic (or idiotic?) notation for lim inf, etc.
    ... >All the results concern a sequence of integrable functions. ... >use `Sup s' and `Inf s' to denote the upper and lower sequences ... >lim always exist, ...
    (sci.math)
  • Idiosyncratic (or idiotic?) notation for lim inf, etc.
    ... All the results concern a sequence of integrable functions. ... Then Sup s is a decreasing sequence, and Inf s an increasing ... lim always exist, ...
    (sci.math)
  • Re: Non-bounce bleed defence
    ... If we take a look at the most widely spread disciplines among vampires ... Obfuscate: 18 (5 inf, 13 sup) ...
    (rec.games.trading-cards.jyhad)