Rudin Theorem 1.20 doubt

  • #1
https://www.physicsforums.com/showthread.php?t=253563

There is this post on this theorem. But the issue I wanted to discuss isn't discussed there fully(so far as I could follow the thread). So I am creating this thread for that issue, so as to avoid necro-post in that thread.

Pg 9

If x in R, y in R, and x<y, then there exists a p in Q such that x<p<y.

Proof:
Since x<y, we have y-x>0, and the archimedean property furnishes a positive integer n such that
n(y-x)>1.

Apply the archimedean property again, to obtain positive integers m1 and m2 such that m1>nx, m2>-nx. Then

-m2<nx<m1.

Hence there is an integer m ( with -m2<= m <= m1) such that

m-1 <= nx < m

If we combine these inequalities, we obtain

nx<m<= 1 + nx < ny.

Since n>0, it follows that

x < m/n < y. This proves (b) with p = m/n.

My doubt is regarding m. Why do we have to find an m which lies between -m2 and m1.

Why can't we directly say that there exists an m1 which is greater than nx (by Archimedian property).
And then go on to say that there would be an m less than or equal to m1 which is such that nx lies between m-1 and m.Why do we have to show that -m2 is less than nx. Do we have to establish that nx lies in an interval which has both a lower and upper bound(-m2 and m1 respectively). Can't we simply establish that nx lies in an interval which has an upper bound ( i.e m1) and then go on to show existence of m ? Why do we need m2

*I am sorry if I am assuming anything which is not obvious and hence needed to be proven. I am new to analysis , hence I am not too sure of what is obvious and what is not. *


He might have needed the m1 and m2 because we cannot just pick any number. We therefore used archimedian property to find m1 and m2, and then again to find a m in between them but a little to the "right" of nx.
.

is the above , the correct reason ? Still I am having difficulty understanding the need for -m2.
 

Answers and Replies

  • #2
22,129
3,297
The thing is that we have to construct m such that

[tex]m-1\leq nx<m[/tex]

For this, we choose

[tex]m=\min\{a\in \mathbb{N}~\vert~nx<a\}[/tex]

But in order for that minimum to exist, we need the set [itex]\{a\in \mathbb{N}~\vert~nx<a\}[/itex] to have a lower bound. This is exactly what [itex]-m_2[/itex] provides...
 
  • #3
ok got it . Thanks.

But 1 doubt still remains though .
in this set you pointed out -

[tex]m=\min\{a\in \mathbb{N}~\vert~nx<a\}[/tex]

shouldn't 'a' belong to the set of integers instead of the set of naturals.
As any set of naturals will have a minimum. But any set of integers may not have a minimum.
 
  • #4
22,129
3,297
ok got it . Thanks.

But 1 doubt still remains though .
in this set you pointed out - [tex]{a\in \mathbb{N}~\vert~nx<a\}[/tex] ,

shouldn't 'a' belong to the set of integers instead of the set of naturals.
As any set of naturals will have a minimum. But any set of integers may not have a minimum.

Yes, of course. It needs to be the integers :blushing:
 
  • #5
Thanks. Sorry didn't get the latex code correct.:cry:
DOUBT resolved!
 

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