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NortonH

End Points of Cantor Set - Countable?

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28 minutes ago, studiot said:

Did you read the (impressive) list of proofs that I linked to disputing this?

I used a notification link and missed your post - sorry.

I looked at https://math.stackexchange.com/questions/421580/is-there-a-rational-number-between-any-two-irrationals  from your links.

First two proofs wrong. I'm not going to refute unlimited incorrect proofs. Less impressive than one link to a wrong proof.

 

 

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57 minutes ago, taeto said:

A well-ordering is pretty much precisely the example of an ordering < such that if x is any given element, there is one particular element y such that x < y, and there is no z for which x < z < y. The usual ordering < of the real numbers does not have this property, since if x and y are any real numbers with x < y, then z = (x+y)/2 does lie between x and y and is different from both.

A well-ordering is pretty much precisely the example of an ordering < such that if x is any given element, there is one particular element y such that x < y,

Oh my. What is the immediate predecessor of the ordinal \(\omega\)? Or did that fall under the escape clause of "pretty much precisely," which means in this case "not at all?"

 

Edited by wtf

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10 minutes ago, taeto said:
22 minutes ago, Carrock said:

I used a notification link and missed your post - sorry.

I looked at https://math.stackexchange.com/questions/421580/is-there-a-rational-number-between-any-two-irrationals  from your links.

First two proofs wrong. I'm not going to refute unlimited incorrect proofs. Less impressive than one link to a wrong proof.

 

 

So all the proofs are just wrong, but you are not going to tell us where they go wrong?

How many disproofs would you and Studiot require to be (impressive)?

The first two mapped irrational numbers to rational numbers and then generated a new rational number proving you can always generate new rational numbers between irrationals.

I'm not sure if I can prove to your satisfaction that there are more irrational numbers than rationals so I'm done here.

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15 minutes ago, wtf said:

A well-ordering is pretty much precisely the example of an ordering < such that if x is any given element, there is one particular element y such that x < y,

Oh my. What is the immediate predecessor of the ordinal ω ? Or did that fall under the escape clause of "pretty much precisely," which means in this case "not at all?"

For \(x = \omega\) you have \(y = \omega +1\), so no escape clause needed.

18 minutes ago, Carrock said:

How many disproofs would you and Studiot require to be (impressive)?

One is generally enough.

19 minutes ago, Carrock said:

The first two mapped irrational numbers to rational numbers and then generated a new rational number proving you can always generate new rational numbers between irrationals..

Excellent. So you have now seen how to prove that between two real numbers there is a rational number. Now that you have seen how that works, why do you insist that there is anything wrong with it?

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7 minutes ago, taeto said:

For x=ω you have y=ω+1 , so no escape clause needed.

 

For x=ω you have y=ω+1, so no escape clause needed.

Sorry about that, I read that wrong. But your definition is not the standard one, so I'm wondering if it's equivalent or imprecise. Will get back to you on that.

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2 minutes ago, wtf said:

Sorry about that, I read that wrong. But your definition is not the standard one, so I'm wondering if it's equivalent or imprecise. Will get back to you on that.

No problem at all. 

I am being vague on purpose in this thread.  If caught, it is fine too. Somehow maybe I will escape... 

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1 minute ago, taeto said:

No problem at all. 

I am being vague on purpose in this thread.  If caught, it is fine too. Somehow maybe I will escape... 

The counterexample is the  integers. That is not a well ordering but each element has a unique successor.

IMO this thread has veered off course in the sense that discussing the definition of a well-order, or the mysterious properties of the rationals inside the reals, doesn't really help us understand the Cantor set. But clearly everyone else is happy so I'll stay out of it.

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1 minute ago, wtf said:

The counterexample is the  integers. That is not a well ordering but each element has a unique successor.

Nice catch +1. In the context of the thread it is not relevant though, because we are dealing with just positive distances.

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19 minutes ago, taeto said:
37 minutes ago, Carrock said:

The first two mapped irrational numbers to rational numbers and then generated a new rational number [from those rationals] 'proving' you can always generate new rational numbers between irrationals. [edited to reduce sarcasm]

 

Excellent. So you have now seen how to prove that between two real numbers there is a rational number. Now that you have seen how that works, why do you insist that there is anything wrong with it?

 

So you do think irrational numbers are the same as rational numbers.

 

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21 minutes ago, Carrock said:

So you do think irrational numbers are the same as rational numbers.

No, I am sure that a number such as \(\sqrt{2}\) is an irrational number which is not a rational number. Come to think of it, I am even convinced that no irrational number is at all a rational number, purely by definition of those concepts. What prompts you to ask such a question?

Also you are interfering with an existing thread, seemingly trying to support some kind of private agenda. If you insist on not answering the questions that are asked to you, the moderators may ban you. 

Edited by taeto

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29 minutes ago, taeto said:

Nice catch +1. In the context of the thread it is not relevant though, because we are dealing with just positive distances.

LOL The "it is not relevant" defense. As if inaccuracy may be forgiven, as long as it is accompanied by irrelevance! That the story you're sticking with? 

[The tone at my end is playful joshing, sometimes that can be misconstrued but I'm just having fun here].

But yes I'll stipulate that the Peano axioms imply a well-ordered set. That's because of Axiom 8, which says that 0 is not the successor of any number. So Well-ordering is baked in to the definition of the natural numbers. https://en.wikipedia.org/wiki/Peano_axioms#Formulation

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1 minute ago, wtf said:

LOL The "it is not relevant" defense. As if inaccuracy may be forgiven, as long as it is accompanied by irrelevance! That the story you're sticking with? 

Just keep rubbing it in, please. I did admit you were right, so what more do you want? And anyway, this is the internet, and I can stick with whichever story I want. 

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59 minutes ago, Carrock said:

So you do think irrational numbers are the same as rational numbers.

 

Again in the spirit of playful pickiness, I will take a shot at a rational defense of a YES position on that issue. Rationals are pretty much the same as reals, in the scheme of things.

It it surely the case in a literal sense that the rationals and reals are different sets. One is a proper subset of the other. It's also true that the rationals differ from the reals in many important respects. The rationals are countable, that's important. They serve different roles in math. The rationals let you solve all the linear equations like ax + b = 0. That was a big deal back in the day but we need a lot more from our numbers these days! Modern science requires a mathematical model of continuity. And that's what the reals do. Algebraically and topologically complete. The very idea of flow. 

I hope you will grant that I am not ignorant in the many profoundly important ways in which rationals are different than reals. Yes I argue that in the scheme of things, the rationals and the reals are more alike than different.

They are very different mathematically. Sure. But if you're not into math, one mathematical abstraction is like another. In school some screechy old math teacher said the fractions were like 5/7 and the "improper fractions" are like 33 & 1/3, and that the real numbers were some "infinite decimals" and don't ask what that means because we won't tell you. 

I say this: The average human being walking around the planet; or even around a modern college campus; does not understand or remember or care about the difference between a rational and a real number. They don't even know what the words mean.

So yes, among those of us who care, the rationals are countable and they're all algebraic and computable, and almost all reals aren't any of those things. The rationals are essentially the result of a finite process; and the reals are the result of an infinitary process. So yes, we regard the rationals and reals as being very different.

But a cat is more like a giraffe than a giraffe is like the number 3. You see my point I hope. Cats and giraffes are different, if you are a biologist. Or in the process acquiring a pet. But they are much more alike, as animals, as either of them is to the number 3, or the Eiffel tower, or the concept of civilization. 

So I say that the rationals and the reals are MUC more alike than different. Only specialists think there's a difference worth mentioning. That is true of everything. A biologist will insist a cat is not a giraffe. Everyone else can plainly see they're both animals, and not bridges and television sets.

If you are a research team from Mars sent to study earth, you will conclude that cats and giraffes are more like each other; and the rational and real numbers are more like each other; than any combination of cats and real numbers or giraffes and rationals could ever be. 

Well thanks for asking. And for reading.

Edited by wtf

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It's too late at night for me to engage brain before making a proper reply. You've reminded me that in school some screechy old maths teacher said "There is no such thing as infinity." "Shut up and calculate" was all he ever taught us. Shame no one stole his answer book.

More later.

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6 hours ago, taeto said:

That is a very reasonable comment. It brings us to a more familiar scenario. Suppose we started with the set of all real numbers, and then we took away all the rational numbers from it. Then we have RQ . Removing a countable set Q from the uncountable set R produces another uncountable set, which has all its pairs of distinct points separated. Are you comfortable with that?

Yes. That is a good example. A countable number of removals (of rationals) leaves behind an uncountable number of irrationals. I guess we can remove the algebraics as well and get the same result.

OK. So now i am still stuck on the conundrum of the 2^n and 2^aleph0.

Why does one not approach the other as n approaches aleph0?

The only thing I can think is that the set of end points is well ordered and the set of rationals is not. But then the set of integers is...

So I am still not sure how to resolve that conundrum. Any ideas?

3 hours ago, Carrock said:

I used a notification link and missed your post - sorry.

I looked at https://math.stackexchange.com/questions/421580/is-there-a-rational-number-between-any-two-irrationals  from your links.

First two proofs wrong. I'm not going to refute unlimited incorrect proofs. Less impressive than one link to a wrong proof.

 

 

So can you give us a counterexample? Show us a couple of irrationals with no rational between them?

If not then it seems clear to me that the original proof stands.

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9 hours ago, NortonH said:

So can you give us a counterexample? Show us a couple of irrationals with no rational between them?

If not then it seems clear to me that the original proof stands.

What is the original proof you're referencing?

If that proof is flawed it doesn't stand.

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Studiot or someone listed a few but it is pretty easy to demonstrate that between p,q irrational there is z rational.

YOU are the one claiming that the hypothesis is false so I asked you for a counterexample.

Do you have one? If not, then what is your reason for claiming that it is not always possible to find a rational between two irrationals?

If you are here just to troll and waste everyone's time i will ignore you. No doubt I will be suspended for 'failing to answer relevant questions' or some such crap but that doesn't matter. I am not really interested in indulging clowns.

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11 hours ago, NortonH said:

OK. So now i am still stuck on the conundrum of the 2^n and 2^aleph0.

Why does one not approach the other as n approaches aleph0?

Maybe you could say that it is a kind of discontinuity. I.e. \(\lim_{n\to \infty} 2^n \neq 2^{\lim_{n\to \aleph_0} n}\). 

Expressed like that, it seems suspect that the LHS means the limit of a series of numbers, whereas the RHS appears to be a limit of sets, where the \(n\) has to be understood as something like \(n = \{0,1,2,\ldots,n-1\}.\) 

For the LHS we would usually write \(\lim_{n\to \infty} 2^n = \infty\) to mean that the series is divergent, i.e. it does not have a proper limit, but it increases without bound in some regular manner. But that is only a symbolic notation, there is no inherent meaning, no mathematical object to attach, to the \(\infty\) symbol that appears on the RHS of that expression.

For the RHS, if we force the sequence \(n\) to be considered as a sequence of sets, then we have no choice other than \(\lim_{n\to \aleph_0} n = \aleph_0,\) and the RHS becomes \( 2^{\aleph_0} = c\), i.e. the continuum.

You end up being forced to interpret the LHS as a limit of sets: \(\lim_{n\to \aleph_0} 2^n\) which is a limit of sets all elements of which are finite sets. Even in the limit, there is no way that you somehow get to include infinite sets as elements.

Maybe more philosophically, we can also argue that even though we write \(\lim_{n\to \infty}\) and we think "\(n\) approaches infinity", this is entirely different from writing \(\lim_{x\to 0}\) and think "\(x\) approaches 0." In the latter case \(x\) actually "gets infinitely closer" to 0 in the limit, closer than distance \(\varepsilon\) for every \(\varepsilon > 0\). Whereas in the expression  \(\lim_{n\to \infty}\) every \(n\) is still infinitely far away from "\(\infty\)". Perhaps this is what causes the discontinuity; that there always remains an infinite gap between \(n\) and its limit?  

Edited by taeto

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1 hour ago, NortonH said:

Studiot or someone listed a few but it is pretty easy to demonstrate that between p,q irrational there is z rational.

YOU are the one claiming that the hypothesis is false so I asked you for a counterexample.

Do you have one? If not, then what is your reason for claiming that it is not always possible to find a rational between two irrationals?

If you are here just to troll and waste everyone's time i will ignore you. No doubt I will be suspended for 'failing to answer relevant questions' or some such crap but that doesn't matter. I am not really interested in indulging clowns.

"YOU are the one claiming that the hypothesis is false"

No. I'm claiming that the proofs are not valid. Significant difference. You're moving the goalposts. A false proof that 'irrationals can always be surrounded by rationals' is consistent with my saying 'irrationals cannot always be surrounded by rationals.'

15 hours ago, Carrock said:

....First two proofs wrong. I'm not going to refute unlimited incorrect proofs.

So you won't can't provide a single example of a valid proof or provide your own 'easy' proof.

You're effectively claiming that a hypothesis without proof or a flawed proof is true unless a counterexample is provided. duh.

I'll put you on the ignore list so we can both stop wasting time.:)

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You have been asked by others to show where the proof is flawed and have refused.

Thanks for playing.

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Posted (edited)

I have quite a different take on counting the Cantor set.

LINK DELETED

Edited by Strange

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