Prove that if [imath]A \subseteq B \subseteq \mathbb{R}[/imath] then [imath]\sup A \le \sup B[/imath]

[Ozma]

Prove that if [imath]A \subseteq B \subseteq \mathbb{R}[/imath] with [imath]A \ne \emptyset[/imath] and [imath]B \ne \emptyset[/imath], then [imath]\sup A \le \sup B[/imath].

I tried this: by definition of supremum, for each [imath]\epsilon>0[/imath] there exists [imath]a_\epsilon \in A[/imath] such that [imath]\sup A-\epsilon<a_\epsilon[/imath]. But [imath]A \subseteq B[/imath], so since [imath]a_\epsilon \in A[/imath] it is [imath]a_\epsilon \in B[/imath]. But, by definition of supremum, from [imath]a_\epsilon \in B[/imath] it follows that [imath]a_\epsilon \le \sup B[/imath]. So, for each [imath]\epsilon>0[/imath] it is [imath]\sup A-\epsilon<\sup B[/imath]. This implies [imath]\sup A \le \sup B[/imath]. Is this correct?

 

[pka]

Prove that if [imath]A \subseteq B \subseteq \mathbb{R}[/imath] with [imath]A \ne \emptyset[/imath] and [imath]B \ne \emptyset[/imath], then [imath]\sup A \le \sup B[/imath].

How do you know that [imath]A\text{ or }B [/imath] is bounded above?
Is that a part that you failed include?

[imath][/imath][imath][/imath][imath][/imath]

 

[Steven G]

Suppose A=(2,3,4,5,6,7,....} and B={1,2,3,4,5,....}
Then A⊆B⊆R. Does it follow that sup A ≤ sup B?

 

[MaxMath]

I think the question presupposes the existence of sup A and sup B, which is ok.

The question is intuitively straightforward (of course not saying this is sufficiently a proof). Proof by contradiction comes up in my mind.

 

[Ozma]

I am sorry for the late answer. As pka said, there is the further hypothesis that the sets are bounded above. Thanks to everyone for checking this, and sorry for being imprecise.

 

[pka]

I am sorry for the late answer. As pka said, there is the further hypothesis that the sets are bounded above. Thanks to everyone for checking this, and sorry for being imprecise.

If [imath]\alpha=\sup(A)\;\&\;\beta=\sup(B)[/imath] and [imath]\beta<\alpha[/imath]
Now [imath]\alpha-\beta >0.[/imath] so that [imath](\exists a\in A)\left[a>(\alpha-\beta)=\beta\right][/imath][imath][/imath]
Knowing that [imath]a\in B[/imath] means the we have a contradiction. YOU EXPLAIN WHY.
If not [imath]\beta<\alpha[/imath] then it must be [imath]\alpha\le\beta.[/imath][imath][/imath]

 

[MaxMath]

If [imath]\alpha=\sup(A)\;\&\;\beta=\sup(B)[/imath] and [imath]\beta<\alpha[/imath]
Now [imath]\alpha-\beta >0.[/imath] so that [imath](\exists a\in A)\left[a>(\alpha-\beta)=\beta\right][/imath][imath][/imath]
Knowing that [imath]a\in B[/imath] means the we have a contradiction. YOU EXPLAIN WHY.
If not [imath]\beta<\alpha[/imath] then it must be [imath]\alpha\le\beta.[/imath][imath][/imath]

I think that's it. But probably you had a typo in it (or otherwise it's just I don't understand math language). My take would be---

Let [imath]\alpha=\sup(A)\;\&\;\beta=\sup(B)[/imath]; suppose [imath]\beta<\alpha[/imath].

Since both [imath]\alpha[/imath] and [imath]\beta[/imath] are real numbers, there must exist a real number [imath]\gamma[/imath] that meets [imath]\beta<\gamma<\alpha[/imath] (say [imath]\gamma=(\alpha+\beta)/2[/imath]).

It follows that [imath]\gamma \in A[/imath] (because [imath]\alpha=\sup(A)[/imath] and [imath]\gamma < \alpha[/imath], otherwise [imath]\alpha[/imath] is not [imath]\sup(A)[/imath]).

It then follows that [imath]\gamma \in B[/imath] (because [imath]A \subseteq B[/imath]).

This means [imath]\gamma \leq \beta[/imath] (because [imath]\beta=\sup(B)[/imath] and [imath]\gamma \in B[/imath].)

This contradicts with [imath]\gamma > \beta[/imath]. So it must be true that [imath]\alpha \leq \beta[/imath].
Q.E.D

I think the question presupposes the existence of sup A and sup B, which is ok.

By the way, this is not precise. The question only presupposes the existence of [imath]\sup(B)[/imath]. The existence of [imath]\sup(A)[/imath] follows naturally from this.

 

[MaxMath]

No I think my answer is wrong.

 

[MaxMath]

Let me try again.

Let [imath]\alpha=\sup(A)\;\&\;\beta=\sup(B)[/imath]; suppose [imath]\beta<\alpha[/imath].

Then [imath]\exist \gamma \in A[/imath]: [imath]\beta<\gamma \leq \alpha[/imath] (otherwise [imath]\beta = \sup (A)[/imath] i.e. [imath]\alpha \neq \sup (A)[/imath]).

Since [imath]A \subseteq B[/imath], [imath]\gamma \in B[/imath].

It follows that [imath]\gamma \leq \beta[/imath] (because [imath]\beta=\sup(B)[/imath]).

This contradicts with [imath]\gamma > \beta[/imath]. So it must be true that [imath]\alpha \leq \beta[/imath].
Q.E.D

I hope this is a valid proof (if it is, I don't believe this is the best one!). Phew!

 

[Steven G]

If [imath]\alpha=\sup(A)\;\&\;\beta=\sup(B)[/imath] and [imath]\beta<\alpha[/imath]
Now [imath]\alpha-\beta >0.[/imath] so that [imath](\exists a\in A)\left[a>(\alpha-\beta)=\beta\right][/imath][imath][/imath]
Knowing that [imath]a\in B[/imath] means the we have a contradiction. YOU EXPLAIN WHY.
If not [imath]\beta<\alpha[/imath] then it must be [imath]\alpha\le\beta.[/imath][imath][/imath]

pka, can you please explain why (α−β)=β? Or am I not reading this correctly.

 

[Steven G]

Let me try again.

Let [imath]\alpha=\sup(A)\;\&\;\beta=\sup(B)[/imath]; suppose [imath]\beta<\alpha[/imath].

Then [imath]\exist \gamma \in A[/imath]: [imath]\beta<\gamma \leq \alpha[/imath] (otherwise [imath]\beta = \sup (A)[/imath] i.e. [imath]\alpha \neq \sup (A)[/imath]).

Since [imath]A \subseteq B[/imath], [imath]\gamma \in B[/imath].

It follows that [imath]\gamma \leq \beta[/imath] (because [imath]\beta=\sup(B)[/imath]).

This contradicts with [imath]\gamma > \beta[/imath]. So it must be true that [imath]\alpha \leq \beta[/imath].
Q.E.D

I hope this is a valid proof (if it is, I don't believe this is the best one!). Phew!

Nicely done. It is crystal clear to me and right on target!

 

[pka]

pka, can you please explain why (α−β)=β? Or am I not reading this correctly.

It should read [imath]\alpha{\Large-}(\alpha -\beta)=\beta[/imath]

 

[Ozma]

Ok, another attempt: let [imath]x \in A[/imath] be arbitrary. Since [imath]A\subseteq B[/imath], it is [imath]x \in B[/imath]. So, by definition of supremum, [imath]x \le \sup B[/imath]. This means that [imath]\sup B[/imath] is an upper bound for the set [imath]A[/imath]. But the supremum is the least upper bound, hence [imath]\sup A \le \sup B[/imath].

But I still not sure about my proof in #1 (with the added hypotheses that [imath]A,B[/imath] are bounded above). Can someone confirm it or tell me if it is wrong?

 

[MaxMath]

Ok, another attempt: let [imath]x \in A[/imath] be arbitrary. Since [imath]A\subseteq B[/imath], it is [imath]x \in B[/imath]. So, by definition of supremum, [imath]x \le \sup B[/imath]. This means that [imath]\sup B[/imath] is an upper bound for the set [imath]A[/imath]. But the supremum is the least upper bound, hence [imath]\sup A \le \sup B[/imath].

This is a nice one!

 

[MaxMath]

To answer your question, I don’t think your first attempt is correct, or to be more precise, I don’t think it’s sufficient. Because you start with [math]\epsilon>0[/math] rather than [math]\epsilon\geq0[/math], due to the lack of the “=“ here, we cannot get “=“ in [math]\sup(A)\leq\sup(B)[/math] in what was to be proved.

What we can reach is only [math]\sup(A)<\sup(B)[/math], which we now know is incorrect. So there must be other logical problems.

 

[MaxMath]

This leads me to think that a change from [math]\epsilon>0[/math] to [math]\epsilon\geq0[/math] probably suffices. But I need to think about this more carefully.

 

[Ozma]

I don't agree with this reasoning, because if [imath]x<y+\epsilon[/imath] for each [imath]\epsilon>0[/imath] then [imath]x \le y[/imath].

 

[MaxMath]

I don't agree with this reasoning, because if [imath]x<y+\epsilon[/imath] for each [imath]\epsilon>0[/imath] then [imath]x \le y[/imath].

This is really nuanced, so much so that I almost have to agree with your point. Because logically this statement (above) is absolutely true.

But what I mean, I guess, is that your proof is not a 'tight' fit of the problem. I'll try to find a better way of describing this. I might be wrong here though!

By the way, finding problems in an argument is a very interesting business!!

I admit that what I said above is logically incorrect —

What we can reach is only [math]\sup(A)<\sup(B)[/math]

Maybe I should have said—

What we can tightly reach is only [math]\sup(A)<\sup(B)[/math]
But anyway this is not a good argument, to say the least, I know!

 

[MaxMath]

Ah, I had another look, and now I think your proof might actually be valid, though it's a bit difficult to follow. It's based on this 'lemma'—

If for any [imath]\epsilon>0[/imath], there is [imath]A-\epsilon<B[/imath], then [imath]A\leq B[/imath].
This is because [imath]A-\epsilon<B[/imath] means [imath]A-B<\epsilon[/imath]. So if [imath]0<\epsilon[/imath] and [imath]\epsilon[/imath] may be as small as we please, then [imath]A\leq B[/imath]!

(To be honest, I think this may only be taken as a 'sketch' of proof of this 'lemma', not strictly one. The more shared is this view, the less sufficient is your proof.)

Thanks for your disagreement by the way!

 

[Ozma]

Thank you for the ideas exchange! The formal proof of the lemma is by contradiction.