Analysis question

[kaelbu]

The problem : Let \(\displaystyle a,b \epsilon \mathbb{R}\) show that if \(\displaystyle a \le b_1\) for every \(\displaystyle b_1 > b\) then \(\displaystyle a \le b\).
My solution: I have no idea where to even begin, because this statement does not seem to be true to me. For example, if a = 5 and b[sub:3348a1nh]1[/sub:3348a1nh] = 6 and b = 2. But the exercise doesn't seem to allow "This statement is untrue" as an answer.
Thanks for any help in advance

 

[mmm4444bot]

the exercise doesn't seem to allow "This statement is untrue" as an answer.

I'm sure that's not what they expect. An instruction of the form given tells us that a ? b (whenever those two conditions are met). They are asking for some sort of proof for this "fact".

Your counterexample makes me ponder, "Is there some typographical error in the materials" ?

Perhaps somebody intended to write b > b[sub:2val1cf6]1[/sub:2val1cf6], instead ? :?

 

[pka]

The problem : Let \(\displaystyle a,b \epsilon \mathbb{R}\) show that if \(\displaystyle a \le b_1\) for every \(\displaystyle b_1 > b\) then \(\displaystyle a \le b\).

Suppose \(\displaystyle b<a\). Then \(\displaystyle b<\frac{a+b}{2}<a\).

But from the given we have \(\displaystyle b<\frac{a+b}{2}\) which implies \(\displaystyle a\le\frac{a+b}{2}<a\).

That is a contradiction. Thus \(\displaystyle a\le b\).

 

[mmm4444bot]

Well, this thread interests me now because I must be misinterpreting something in the problem statement. I had thought the scenario fairly straightforward, but I'll have to wait until I'm not hung over to focus on pka's post. :?

 

[kaelbu]

I am having a bit of trouble understanding this statement:

But from the given we have \(\displaystyle b<\frac{a+b}{2}\) which implies \(\displaystyle a\le\frac{a+b}{2}<a\).

Are we assuming that \(\displaystyle b_1 = \frac{a+b}{2}\)? If so, why is this?
Secondly, I don't understand how the first equation implies the second...except that it also indicates that \(\displaystyle b_1 = \frac{a+b}{2}\).
I think my problem is that I don't understand how b[sub:sj5x3m03]1[/sub:sj5x3m03] is defined.

 

[pka]

I am having a bit of trouble understanding this statement:

But from the given we have \(\displaystyle b<\frac{a+b}{2}\) which implies \(\displaystyle a\le\frac{a+b}{2}<a\).

Are we assuming that \(\displaystyle b_1 = \frac{a+b}{2}\)? If so, why is this?
Secondly, I don't understand how the first equation implies the second...except that it also indicates that \(\displaystyle b_1 = \frac{a+b}{2}\).
I think my problem is that I don't understand how b[sub:1k23qe8i]1[/sub:1k23qe8i] is defined.

The question has this meaning.
For any number \(\displaystyle \beta\) if \(\displaystyle b<\beta\) then we know that \(\displaystyle a\le\beta\).

That is the given. From which you are asked to prove that \(\displaystyle a\le b.\)

Perhaps it is unfortunate that the author used \(\displaystyle b_1\) instead of a general term such as \(\displaystyle \beta\).
Nonetheless, this is a well known theorem.

 

[kaelbu]

I feel like I'm being a bit dense, but I still don't understand how the following is evident from the given:

But from the given we have \(\displaystyle b<\frac{a+b}{2}\) which implies \(\displaystyle a\le\frac{a+b}{2}<a\).

 

[mmm4444bot]

Suppose \(\displaystyle b < a\).

Then \(\displaystyle b < \frac{a+b}{2} < a\). Okay. This relationship is a direct result of the supposition in the line above it.


But from the given we have \(\displaystyle b < \frac{a+b}{2}\) Meaning, we get this "from the supposition" versus "from the given [statement]" ?

Or, perhaps pka intended to type "from the given we have a ? (a + b)/2" ?

That would seem to imply the following.

which implies \(\displaystyle a \le \frac{a+b}{2} < a\).

If we somehow know that a ? (a + b)/2 from the given problem statement, then I have not perceived how, yet. :?


Here is how I'm interpreting the question.

Since b[sub:lnsusajp]1[/sub:lnsusajp] is greater than b, we have b[sub:lnsusajp]1[/sub:lnsusajp] ? (b, ?)

In other words, b[sub:lnsusajp]1[/sub:lnsusajp] has no least value.

Now, we are told that for any possible b[sub:lnsusajp]1[/sub:lnsusajp] within the interval (b, ?) the number a must either be the same as b[sub:lnsusajp]1[/sub:lnsusajp] or less than it.

We are trying to prove -- under this given scenario -- that number a cannot possibly lie between b and b[sub:lnsusajp]1[/sub:lnsusajp]; so there are only two possibilities: a must either be the same as b or less than it.

I hope that this view of the exercise is accurate.

I cannot seem to avoid thinking that number a could lie between b and b[sub:lnsusajp]1[/sub:lnsusajp] because no matter how close b[sub:lnsusajp]1[/sub:lnsusajp] lies to the right of b, there will always be room for some number a. Disproving this notion seems to be the point of the exercise, maybe.

(This thread has put a new crease on my cerebellum.)

MY EDITS: I meant the other "left" and "a gap" was a poor choice of words :roll: and changed "if" to "since"

 

[pka]

Let’s try to fix this mess.
Suppose that each of \(\displaystyle a~\&~b\) is a number and \(\displaystyle a\) has the property that for any \(\displaystyle \beta\) if \(\displaystyle b<\beta\) then \(\displaystyle a\le \beta\). Now prove that \(\displaystyle a\le b\)

PROOF by contradiction.
Suppose that \(\displaystyle b<a\). We know that \(\displaystyle b<\frac{b+a} {2}< a\).
But means that \(\displaystyle b<\beta=\frac{b+a}{2}\) which implies that \(\displaystyle a\le\frac{b+a}{2}\). That is a contradiction because \(\displaystyle \frac{b+a}{2} <a.\)
Therefore, \(\displaystyle a\le b.\)

 

[daon]

Here's a direct proof using the "epsilon argument" (which i believe was proved prior to a similar question in bartle's baby analysis). In the proof an element of contradiction is hidden, so this is essentially the same as the above.

Assume \(\displaystyle a \neq b\). Let \(\displaystyle \epsilon > 0\).

Then by assumption we have \(\displaystyle a \le b+\epsilon\) or \(\displaystyle a-b \le \epsilon\).

Since this must hold for all positive \(\displaystyle \epsilon\), \(\displaystyle a-b \le 0\) or (since \(\displaystyle a\neq b\)) \(\displaystyle a < b\).

 

[mmm4444bot]

Suppose that each of \(\displaystyle a~\&~b\) is a number and \(\displaystyle a\) has the property that for any \(\displaystyle \beta\) if \(\displaystyle b<\beta\) then \(\displaystyle a\le \beta\). Now prove that \(\displaystyle a\le b\)

PROOF by contradiction.

Suppose that \(\displaystyle b<a\). We know that \(\displaystyle b<\frac{b+a} {2}< a\).

But [this] means that \(\displaystyle b<\beta=\frac{b+a}{2}\) which implies that \(\displaystyle a\le\frac{b+a}{2}\). That is a contradiction because \(\displaystyle \frac{b+a}{2} <a.\)

Therefore, \(\displaystyle a\le b.\)

I'm thinking that I understand everything in this post, except for how to determine that ? is the average of a and b (under the supposition).

I'm adding daon's post to my to-do list.

Can anybody post the name of this well-known theorem?

Cheers

 

[pka]

Re:

I'm adding daon's post to my to-do list.
Can anybody post the name of this well-known theorem?

This trival result does not rise to an important enough level to deserve a name.

As for the version daon posted here is the problem from Bartle's honors calculus.
If \(\displaystyle a~\&~b\) are real numbers such that \(\displaystyle \left( {\forall \varepsilon > 0} \right)\left[ {a \leqslant b + \varepsilon } \right]\) then prove that \(\displaystyle a\le b\).
The proof is so simple. If \(\displaystyle b<a\) then let \(\displaystyle \varepsilon = a - b > 0\).
The contradiction follows at once.

 

[daon]

Yep, first thing to prove is the easier result if \(\displaystyle a\) is a non-negative real number such that \(\displaystyle a \le \epsilon\) for every positive number \(\displaystyle \epsilon\), then \(\displaystyle a=0\) - read: "a non-negative number smaller than every positive number must be zero". Make sense? Note this is a special case of the kaelbu's question, which is what I referred to as the "epsilon argument", I don't know what else to call it

As trivial as it looks, this result is used (in my learning travels anyway) countless times to build up many theorems in analysis and measure theory. Those "gaps" between real numbers you speak of mmm, don't exist in the real numbers, but do so in the surreal number system (which I don't know much about).

 

[mmm4444bot]

This trival result does not rise to an important enough level to deserve a name.

A trivial well-known theorem, eh? Sounds like an imposter, to me. :wink:

 

[mmm4444bot]

Those "gaps" between real numbers you speak of mmm, don't exist

Yes, you are correct; the Real number line is infinitely dense. "A gap" was a poor choice of words. I edited my post.

I often need visual references in my mind, when dealing with the Real numbers. This is particularily true while sorting through these types of Real-analysis proofs.

I learned that between any two Real numbers there lie an infinite number of additional Real numbers.

So, for instance, we can say that there is no smallest positive number. No matter how close to the right of zero you "look" at, there is always room for more positive numbers between that point and zero.

Before I try harder to understand this thread, I would like to confirm that my interpretation in green is not a derailment.

 

[daon]

Yes, in fact the proof would have been the same even if restricted to rational numbers or any dense set. If it helps, let \(\displaystyle a_n\) be a sequence which tends to \(\displaystyle b\) from the right (monotonically). Then it is the same result to show: if \(\displaystyle a \le a_n\) for all n, then \(\displaystyle a \le b\).

Showing these two statements, above and in the original post, are equivalent might help your intuition.

edit: It doesn't matter which sequence you pick, so assume what you'd like. For instance \(\displaystyle b+2^{-n}\).

 

[mmm4444bot]

let \(\displaystyle a_n\) be a sequence which tends to \(\displaystyle b\) from the right

if \(\displaystyle a \le a_n\) for all n, …

Regarding sequences where a[sub:1gd4v4jp]n[/sub:1gd4v4jp] denotes the nth term, I learned that symbol a is an abbreviation for a[sub:1gd4v4jp]1[/sub:1gd4v4jp].

This is why I read your inequality above to say that the nth term in your example sequence can be greater than the first term.

I do not understand how this is possible with a sequence of decreasing terms. :?

On the plus side, thank you for confirming for me that there is room for an infinite number of Real values between any pair of Real numbers. At least I got one thing right.

 

[pka]

Re:

On the plus side, thank you for confirming for me that there is room for an infinite number of Real values between any pair of Real numbers. At least I got one thing right.

I find most of this posting incomprehensible.
BUT here is a comment.
For any open interval \(\displaystyle (a,b) \sim \mathbb{R}\).
That is there is a one-to-one and onto mapping from \(\displaystyle (a,b)\leftrightarrow \mathbb{R}\).

So no matter how ‘small’ you think \(\displaystyle (a,b)\) is there is still as many real numbers in \(\displaystyle (a,b)\) as there are numbers in \(\displaystyle \mathbb{R}\).

If that drives you “mad”, do not worry it most probability drove Cantor mad.

 

[daon]

Re:

Regarding sequences where a[sub:33lq0wz4]n[/sub:33lq0wz4] denotes the nth term, I learned that symbol a is an abbreviation for a[sub:33lq0wz4]1[/sub:33lq0wz4].

This is why I read your inequality above to say that the nth term in your example sequence can be greater than the first term.

I do not understand how this is possible with a sequence of decreasing terms. :?

On the plus side, thank you for confirming for me that there is room for an infinite number of Real values between any pair of Real numbers. At least I got one thing right.

Jumping from what today's "Calculus" is and abstract analysis can be frustrating, so let us try a more concrete example.

Suppose that \(\displaystyle x_n = b+2^{-n}\) (\(\displaystyle n\) may vary through all the real numbers if you wish as \(\displaystyle x_n > b\) for any real number \(\displaystyle n\)).

Then it is clear that \(\displaystyle \lim_{n \to \infty}x_n = b\), right?

Say we want to prove this: "If \(\displaystyle 2\) is less than every number in the set \(\displaystyle \{x_n\}\), then \(\displaystyle 2\) can be at most the number \(\displaystyle b\)."

Said another way, if \(\displaystyle 2 < x_n\) for every \(\displaystyle n\), then \(\displaystyle b\) must be at least \(\displaystyle 2\).

It is not anything fancy, and when you see it it should become obvious. The assumption is that 2 is less than everything in the sequence. You cannot find a counter-example unless the hypothesis is satisfied and the conclusion is contradicted. That is where the original poster's counter example failed.

So the assumption is that 2 is less than \(\displaystyle b+\frac{1}{2}\). It is less than \(\displaystyle b+\frac{1}{4}\), \(\displaystyle b+\frac{1}{128}\), \(\displaystyle b+\frac{1}{4096}\), ... etc.

The indirect proof goes like this: if what we are trying to show is not true, i.e., if \(\displaystyle 2>b\), then we can find some real number \(\displaystyle m\) (in fact infinitely many) such that \(\displaystyle 0 < (\frac{1}{2})^m < (2-b)\). But then \(\displaystyle b+2^{-m} < 2\). That is, an element of our set (\(\displaystyle x_m\)) is less than \(\displaystyle 2\)! This goes against our assumption, so the claim that \(\displaystyle 2>b\) is false.

For example, if \(\displaystyle b = 1.999999923245673463.....\) (take it to be some irrational) then you can verify \(\displaystyle b+2^{-30}\) is less than 2... which does not satisfy out assumption.

So the "best" \(\displaystyle b\) can be is exactly the number \(\displaystyle 2\). If its anything smaller, no matter what valid sequence we pick, a contradiction like this will happen.

I hope that helped.

But yeah, on your last point, it is quite mind-blowing. Its easy to say "its common knowledge" and do the proofs and think you "know" something... but really there is much more going on in \(\displaystyle \mathbb{R}\)than most realize. Just for fun: there are-tremendously-more real numbers between \(\displaystyle 0.00000000001\) and \(\displaystyle 0.00000000002\) than their are rational coordinates in any \(\displaystyle n-\)dimensional space.