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pengkuan

Continuity and uncountability

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48 minutes ago, studiot said:

that was no reason to pour out vitriol.

 

I did no such thing. I characterized your error, not you personally. I added detail regarding unrestricted comprehension so readers could Google the relevant facts. What you SAID was garbage. That is an objective fact. Not vitriol. And "I think in pictures, not algebra," is unconvincing coming from someone whom I've seen lay out brilliant technical responses to questions of engineering math. You're not lacking in algebra by any means. you're simply lacking some basics in abstract math. Basics easily studied on Wikipedia. 

Edited by wtf

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3 hours ago, studiot said:

However since the OP is indifferent to my help I see no purpose being served in my further presence in this thread.

I'm very grateful to your help. Sorry for not replying you more.

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2 minutes ago, pengkuan said:

I'm very grateful to your help. Sorry for not replying you more.

So did you catch my point about the different terminology used by different mathematicians?

This would be important to you since English is not your first language.

I have basically been restricting my input to try to help you make sense of what others are telling you.

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2 hours ago, uncool said:

No, that definition of finitude is not the definition used to prove that a set is countable, because countability and infinitude are different concepts.

Yes, you are right. 

 

3 hours ago, uncool said:

However, I want to draw your attention to something. Note that every fraction appears after a finite number of steps. That is, if you gave me a fraction, I could tell you exactly at what step you reach it - and I could represent the number by writing it as "1 + 1 + 1 + 1" and eventually stop. For example, 2/4 is reached in the 12th step, that is, the 1+1+1+1+1+1+1+1+1+1+1+1th step. 

So, the set of the rationals is countable because every i/j corresponds to a finite number n in the counting order. At nth step, we stop the count. Although the set itself is infinite, all counting numbers are finite.

But for the power set of N, the set of all even numbers corresponds to the infinite binary sequence 1010101010...... To reach it we cannot stop counting because this sequence is not finite. Can we say that  the power set of N is not countable because the counting numbers of infinite subsets are infinite?

27 minutes ago, studiot said:

So did you catch my point about the different terminology used by different mathematicians?

This would be important to you since English is not your first language.

I have basically been restricting my input to try to help you make sense of what others are telling you.

Yes. I agree that I have difficulty in seizing the exact sense of the discussion. 

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1 hour ago, pengkuan said:

So, the set of the rationals is countable because every i/j corresponds to a finite number n in the counting order. At nth step, we stop the count. Although the set itself is infinite, all counting numbers are finite.

I'm not sure what you mean by "at the nth step, we stop the count"; if you mean "for each i/j, we can stop the count at some n and be at i/j", yes. The other two parts are both correct. 

1 hour ago, pengkuan said:

But for the power set of N, the set of all even numbers corresponds to the infinite binary sequence 1010101010...... To reach it we cannot stop counting because this sequence is not finite.

That is a way to look at it; that your attempt to match N to its powerset doesn't work for the set of all even numbers because there is no finite number that matches it.

1 hour ago, pengkuan said:

Can we say that  the power set of N is not countable because the counting numbers of infinite subsets are infinite?

Not really. Countability is about the existence of some matching - some bijection; the fact that some map doesn't work as a bijection doesn't mean there can't be another. For example: the set {1, 1/2, 1/3, 1/4, 1/5, ...} with an added 0 (i.e. union with {0}) is countable, even though the map n -> 1/n clearly "misses" 0. It is countable because of the bijection 1 -> 0, n -> 1/(n - 1) for n > 2.

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10 hours ago, studiot said:

that was no reason to pour out vitriol.

 

I apologize. I get carried away sometimes. No personal malice intended. 

 

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21 hours ago, pengkuan said:

However, I want to draw your attention to something. Note that every fraction appears after a finite number of steps. That is, if you gave me a fraction, I could tell you exactly at what step you reach it - and I could represent the number by writing it as "1 + 1 + 1 + 1" and eventually stop. For example, 2/4 is reached in the 12th step, that is, the 1+1+1+1+1+1+1+1+1+1+1+1th step.

 

19 hours ago, uncool said:

I'm not sure what you mean by "at the nth step, we stop the count"; if you mean "for each i/j, we can stop the count at some n and be at i/j", yes. The other two parts are both correct. 

I was restating your explanation above with my words.

So, in the counting of \(\mathbb{Q}\), the ratio i/j corresponds to the number n which is finite. For the ratio i/1, the number of count is: \[n=\frac{i(i+1)}{2}\] So, if n is a finite number, i is also finite. Does this mean that the number i is not allowed to go to infinity? In this case, \(\mathbb{Q}\) does not completely cover the plane \(\mathbb{N}\times \mathbb{N}\) . Can we say that \(\mathbb{Q}\) is countable only because i and j are not allowed to have infinite value?

Edited by pengkuan
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On 11/21/2018 at 1:17 PM, pengkuan said:

So, if n is a finite number, i is also finite. Does this mean that the number i is not allowed to go to infinity?

I'm not sure what you mean by that. i can be an arbitrarily large finite number. 

On 11/21/2018 at 1:17 PM, pengkuan said:

Can we say that Q is countable only because i and j are not allowed to have infinite value?

We can say that the rationals are countable because we can construct a bijection between the natural numbers and the rational numbers. That's all there is to it.

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On 2018/11/23 at 7:59 AM, uncool said:

I'm not sure what you mean by that. i can be an arbitrarily large finite number. 

I mean that because you say “i can be an arbitrarily large finite number.”, i and j cannot be infinitely big, for example, a number with infinitely many digits like 9517452…… or \(10^{\infty }\). 

In this case, can we write the set of natural numbers as {1,2,3,…,n-1,n}, with n being a finite number with arbitrarily large value?
 

On 2018/11/23 at 7:59 AM, uncool said:

We can say that the rationals are countable because we can construct a bijection between the natural numbers and the rational numbers. That's all there is to it.

If i had value 9517452…… , then the ratio i/1 would  not have corresponding natural number.

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2 hours ago, pengkuan said:

If i had value 9517452……

How often does it happen that you actually have a value like that? 

I am not a violent person. But I suggest that if you come across a

landlord who tells you that the rent per month in US$ amounts to

9517452……, then you are absolutely in the right to punch them on

the nose, using up to medium strength.

 

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7 minutes ago, taeto said:

How often does it happen that you actually have a value like that? 

I am not a violent person. But I suggest that if you come across a

landlord who tells you that the rent per month in US$ amounts to

9517452……, then you are absolutely in the right to punch them on

the nose, using up to medium strength.

 

Hey, last time I rented the White House that was the rent.

But I hear Big T has doubled it since.

 

:)

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I wonder if it might be helpful to point out the difference between the ordinal numbers and the cardinal numbers, since we seem to restricting the discussion to the natural numbers \(\mathbb{N}\)

So an ordinal number roughly speaking describes the position of an element in an ordered set. In contrast a cardinal number describes the size of a set, ordered or otherwise. Notice these are quite different concepts.

Now, by construction the natural numbers are ordered (there is a theorem that any set can be ordered - the proof is hellacious and not relevant here).

So it is fairly easy to see that, for any subset of \(\mathbb{N}\) (it's ordered recall) if there exists a largest ordinal \(n\) then this corresponds to the cardinality of our subset and it must be finite.

Otherwise the gloves are off. The largest non-finite ordinal, by an arbitrary convention is denoted as \(\omega\). This is still an ordinal., and can in no way denote the cardinality of a non finite subset of \(\mathbb{N}\) e.g \(\mathbb{N}\) itself. For this we use the arbitrary symbol \(\aleph_0\).

Any help?

Edited by Xerxes

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23 minutes ago, Xerxes said:

The largest non-finite ordinal, by an arbitrary convention is denoted as \(\omega\)

Oh my. You don't believe in \(\omega + 1\)? Perhaps you meant the smallest non-finite ordinal. And by order, perhaps you meant well-order. I'll leave it here as to not appear to be piling on.

ps -- Ok I'll pile on just a little bit more.

> So it is fairly easy to see that, for any subset of N (it's ordered recall) if there exists a largest ordinal n then this corresponds to the cardinality of our subset

Really? That's fairly easy to see? It's not even true as you expressed it, and I'm not even sure what you're trying to say. The smallest ordinal larger than any of the elements of {2, 4, 6} is 7, but the cardinality of that set is 3. The largest ordinal in the set is 6.I couldn't understand what you're getting at. "If there exists a largest ordinal? There is no largest ordinal. Can you clarify your thoughts please?

Edited by wtf

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ps -- The larger point is that OP seems to believe that there are natural numbers that are infinite; and can't distinguish between the fact that there are infinitely many natural numbers, but each one is finite. On that basis, I don't think the ordinals are going to reduce the confusion in this thread. 

If as @studiot says I sounded "vitriolic" my apologies once again. I am staying out of this thread from now on. 

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On 2018/11/21 at 12:58 AM, uncool said:

That is a way to look at it; that your attempt to match N to its powerset doesn't work for the set of all even numbers because there is no finite number that matches it.

 

On 2018/11/24 at 11:22 PM, taeto said:

But I suggest that if you come across a

landlord who tells you that the rent per month in US$ amounts to

9517452……, then you are absolutely in the right to punch them on

the nose, using up to medium strength.

 

On 2018/11/26 at 3:10 AM, wtf said:

OP seems to believe that there are natural numbers that are infinite; and can't distinguish between the fact that there are infinitely many natural numbers, but each one is finite.

It seems that everyone thinks that natural numbers have finite values while the entire set in infinite. I'm OK with that.

But in this case, the length of the set of all even numbers is finite, because it's a natural number. Actually, one cannot pass from a finite number, the length of a finite set, to infinite number, the length of a infinite set, which is the finite set when its length is stretched to infinity.

On 2018/11/24 at 11:30 PM, studiot said:

Hey, last time I rented the White House that was the rent.

But I hear Big T has doubled it since.

Actually, if the price is 1111....., you can double it, 2*1111...=2222...

On 2018/11/25 at 8:08 PM, Xerxes said:

I wonder if it might be helpful to point out the difference between the ordinal numbers and the cardinal numbers, since we seem to restricting the discussion to the natural numbers.

Thanks for your help. I think within the set of natural number, the ordinal numbers are natural numbers.

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25 minutes ago, pengkuan said:

But in this case, the length of the set of all even numbers is finite, because it's a natural number.

By the "length" of the set, do you mean the cardinality?

If so, why do you think it is a natural number?

Do you think the cardinality of all sets is a natural number? 

26 minutes ago, pengkuan said:

Thanks for your help. I think within the set of natural number, the ordinal numbers are natural numbers.

The ordinal numbers are. But what about the cardinality?

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2 hours ago, Strange said:

By the "length" of the set, do you mean the cardinality?

No. Length is the number of member of a series. So, it is a natural number.

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1 hour ago, pengkuan said:

No. Length is the number of member of a series. So, it is a natural number.

That's what cardinality means. And no, that doesn't explain why it should be a natural number.

There are infinitely many even numbers. There is no natural number n such that the set of even numbers is in bijection with {0, 1, ..., n - 1}.

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10 hours ago, pengkuan said:

So, it is a natural number.

Why?

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20 hours ago, uncool said:

That's what cardinality means. And no, that doesn't explain why it should be a natural number.

There are infinitely many even numbers. There is no natural number n such that the set of even numbers is in bijection with {0, 1, ..., n - 1}.

 

11 hours ago, Strange said:

Why?

I will change my theory to handle infinity.

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Perhaps it would solve the philosophical confusion if a line is defined to be a set of ordered constructable  points, where a point is a  computable total function by at least one terminating algorithm.    That way there is explicit clarity that there are no "holes" in our field of entities  - except perhaps for the partial functions corresponding to non-terminating algorithms  that cannot be ordered by their outputs -  and that there is only a countable number of entities describable by mathematics. 

Edited by TheSim

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2 hours ago, TheSim said:

Perhaps it would solve the philosophical confusion if a line is defined to be a set of ordered constructable  points, where a point is a  computable total function by at least one terminating algorithm.    That way there is explicit clarity that there are no "holes" in our field of entities  - except perhaps for the partial functions corresponding to non-terminating algorithms  that cannot be ordered by their outputs -  and that there is only a countable number of entities describable by mathematics. 

The constructible real line doesn't satisfy the Intermediate value theorem. Hell of a poor model of the continuum, don't you agree? Contrary to your claim that there are no holes, the constructible real line is full of holes, one hole where each noncomputable real used to be. There are many Cauchy sequences that do not converge. Worst model of the continuum ever. 

Edited by wtf

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It also doesn't really get rid of the philosophical confusion - I'd even argue it adds to it. Someone should have a firm grasp of the basics of countability and set theory before attempting to get into computability theory.

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7 hours ago, TheSim said:

Perhaps it would solve the philosophical confusion if a line is defined to be a set of ordered constructable  points, where a point is a  computable total function by at least one terminating algorithm.  

I do not think that a line is a set of points just because points cannot be fill all holes. But this is another story.

 

7 hours ago, TheSim said:

there is only a countable number of entities describable by mathematics. 

I agree that real numbers are countable. 

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And that's a nice demonstration of what I said.

The set of real numbers is not countable, pengkuan.

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