MHB Sequence of Intervals - Rudin, Theorem 2.38

[Math Amateur]

I am reading Walter Rudin's book, Principles of Mathematical Analysis.

Currently I am studying Chapter 2:"Basic Topology".

Although I can basically follow it, I am concerned that I do not fully understand the proof of Theorem 2.38.

Rudin, Theorem 2.38 reads as follows:https://www.physicsforums.com/attachments/3789In the above proof we read the following:

" ... ... If $$m$$ and $$n$$ are positive integers then

$$a_n \le a_{m+n} \le b_{m+n} \le b_m$$

so that $$x \le b_m$$ for each $$m$$. ... ..."This appears to me to be true ... ... BUT ... ...

Why doesn't Rudin simply say the following:

"Let $$m$$ be a positive integer.

Then $$a_m \le b_m$$ ...

so that $$x \le b_m$$ for each $$m$$. "Since my statement is simpler than what Rudin says, I feel that I must be missing something and my analysis above must be wrong ...

Can someone point out why my proof is defective and thus clarify this issue?

Hope someone can help ... ...

Peter

 

[Euge]

Why doesn't Rudin simply say the following:

"Let $$m$$ be a positive integer.

Then $$a_m \le b_m$$ ...

so that $$x \le b_m$$ for each $$m$$. "

Hi Peter,

The implication you have here is false. It would've been true if $x$ was the infimum of $E$ (since then $x \le a_m \le b_m$ for all $m$) but in fact $x$ is the supremum of $E$. He uses the necessary fact that for all $n$, $a_n \le b_m$ for all $m$ (notice here how $n$ is independent of $m$). With this, we know that for each $m$, $b_m$ is an upper bound for $E$. Hence, by definition of $x$, $x \le b_m$ for all $m$. Makes sense?