The Completeness Axiom .... and Sohrab Exercise 2.1.29 ....

[Math Amateur]

I am reading Houshang H. Sohrab's book: "Basic Real Analysis" (Second Edition).

I am focused on Chapter 2: Sequences and Series of Real Numbers ... ...

I need help with Exercise 2.1.29 ...

Exercise 2.1.29 (including the Completeness Axiom) reads as follows:https://www.physicsforums.com/attachments/7088I am unable to make a meaningful start on the above exercise ...

... my questions are as follows:

What exactly does "equivalence" mean in the above context?

Does "equivalence" mean: Supremum Property $$\Longleftrightarrow$$ Infimum Property ... .. ? If not ... what does it mean ...?

How should I start and proceed with this exercise?Peter

 

[Opalg]

What exactly does "equivalence" mean in the above context?

Does "equivalence" mean: Supremum Property $$\Longleftrightarrow$$ Infimum Property ... .. ?

Yes. To prove this, you could start by assuming that the Supremum Property holds. You then want to show that the Infimum Property holds. So assume that $S$ is a nonempty set bounded below by some number $t\in\Bbb{R}$. Following Sohrab's hint, show that the set $-S$ is bounded above by $-t$. Apply the Supremum Property to show that $\sup(-S)$ exists, and deduce that $S$ has an infimum, namely $-\sup(-S).$

You will then have proved that Supremum Property $\Longrightarrow$ Infimum Property. The reverse implication is essentially the same proof with signs changed as appropriate.

 

[Math Amateur]

Yes. To prove this, you could start by assuming that the Supremum Property holds. You then want to show that the Infimum Property holds. So assume that $S$ is a nonempty set bounded below by some number $t\in\Bbb{R}$. Following Sohrab's hint, show that the set $-S$ is bounded above by $-t$. Apply the Supremum Property to show that $\sup(-S)$ exists, and deduce that $S$ has an infimum, namely $-\sup(-S).$

You will then have proved that Supremum Property $\Longrightarrow$ Infimum Property. The reverse implication is essentially the same proof with signs changed as appropriate.

Thanks Opalg ... was just working through what you suggested and struck a problem ... to illustrate problem consider the following ... basically following your advice ...

Assume $$S \subset \mathbb{R}$$ where $$s \ne \emptyset$$ ...

... and further ... assume Supremum Property (AoC) holds ...

We want to demonstrate that, given the above, the Infimum Property holds ...

... that is ... assume also that $$S$$ is bounded below by some real number $$t$$ ... that is $$t \le s \ \forall s \in S$$ ...

... and then we have to show that S has an infimum, $$ \text{inf} (S) \in \mathbb{R}$$ ... Now ... we have $$t \le s$$ ... so $$-t \ge -s$$ ...

... so $$-S = \{ -s \in \mathbb{R} \ | \ s \in S \}$$ is bounded above by $$-t$$ ...

... therefore ... by the Supremum Property, $$-S$$ has a least upper bound, $$\text{Sup} (-S)$$ ...
NOW ... I want to conclude that therefore ... $$S$$ has a greatest lower bound $$- \text{Sup} (-S)$$ ...

... intuitively, it seems a very reasonable step ...

BUT ... how do I prove it ... ?Can you help ...?

Peter
NOTE/EDIT

I am becoming suspicious that to finish the above proof we may need Proposition 2.1.30 ... which is after! the Completeness Axiom and Exercise 2.1.29 ... Proposition 2.1.30 reads as follows:View attachment 7128
View attachment 7129If it is needed to finish my proof then surely Sohrab should have presented Proposition 2.1.30 before Exercise 2.1.29 ...

So ... it is of interest to me whether the above proof I presented (based on Opalg's guidance) can be finished without resort to Proposition 2.1.30 ...

Peter

 

[Opalg]

Let $t = \sup(-S)$. You want to show that $-t = \inf(S)$. To do that, you could certainly use Prop. 2.1.30. But you should also be able to do it from the definitions of sup and inf. I'm not sure what Sohrab uses for these definitions. I assume that the definition of $-t = \inf(S)$ would be that (1) $-t$ is a lower bound for $S$, and (2) if $r$ is also a lower bound for $S$ then $r\leqslant -t$. In that case, (1) follows from the fact that $t$ is an upper bound for $-S$. To prove (2), you should show that
$-r$ is an upper bound for $-S$. The definition of sup then implies that $-r \geqslant t$, so that $r\leqslant -t$, as required.

 

[Math Amateur]

Let $t = \sup(-S)$. You want to show that $-t = \inf(S)$. To do that, you could certainly use Prop. 2.1.30. But you should also be able to do it from the definitions of sup and inf. I'm not sure what Sohrab uses for these definitions. I assume that the definition of $-t = \inf(S)$ would be that (1) $-t$ is a lower bound for $S$, and (2) if $r$ is also a lower bound for $S$ then $r\leqslant -t$. In that case, (1) follows from the fact that $t$ is an upper bound for $-S$. To prove (2), you should show that
$-r$ is an upper bound for $-S$. The definition of sup then implies that $-r \geqslant t$, so that $r\leqslant -t$, as required.

Hi Opalg ... thanks again for the help ..

You wrote:

" ... ... the definitions of sup and inf. I'm not sure what Sohrab uses for these definitions. ... ... "Sohrab defines Sup and Inf on partially ordered sets ... here are Sohrab's definitions of partial order through to Sup and Inf ...
View attachment 7148
https://www.physicsforums.com/attachments/7149Hope that helps ...

Peter

 

[Math Amateur]

Hi Opalg ... thanks again for the help ..

You wrote:

" ... ... the definitions of sup and inf. I'm not sure what Sohrab uses for these definitions. ... ... "Sohrab defines Sup and Inf on partially ordered sets ... here are Sohrab's definitions of partial order through to Sup and Inf ...

Hope that helps ...

Peter

We have shown that $$-S = \{ -s \in \mathbb{R} \ | \ x \in S \}$$ is bounded above by $$-t$$ ...

But ... since $$-S$$ is bounded above, by the AoC $$-S$$ has a supremum ...

Let $$k = \text{Sup}(-S)$$

$$\Longrightarrow -s \le k \text{ for all } s \in S$$ ... ... ... (1)

and

... if $$r$$ is another upper bound for $$-S$$ ... then we have $$k \le r$$ ... ... ... (2)Now ... (1) $$\Longrightarrow s \ge -k$$ for all $$s \in S$$

... that is .. $$-k \le s \text{ for all } s \in S$$ ... in other words, $$-k$$ is an lower bound for S ... ... ... (3)

... and ...

... (2) $$\Longrightarrow$$ if $$-r$$ is another lower bound on $$S$$ then we have $$-k \ge -r$$ ... ... ... (4)(3) (4) $$\Longrightarrow$$ that $$-k = \text{inf} (S)$$ ...Is that correct?

Is it the best way to prove it ...?NOTE: I'm a bit concerned that (4) may not be correct ... of if it is correct, then it is not well expressed ... any comments ...

Peter

 

[Opalg]

I'm a bit concerned that (4) may not be correct ... of if it is correct, then it is not well expressed ... any comments ...

To explain (4) I would say that if $r$ is a lower bound for $S$ then $r\leqslant s$ for all $s$ in $S$. Therefore $-r\geqslant -s$ for all $-s$ in $-S$, so that $-r$ is an upper bound for $-S$. It follows that $-r\geqslant k=\sup(-S)$ and therefore $r\leqslant -k$. That shows that $-k$ is the greatest lower bound for $S$.