Confusion with finite sets vs an infinite set of natural numbers

[pzkpfw]

17 hours ago, Boltzmannbrain said:

...

I wanted the list to have all n of the set of all natural numbers N.

Then you simply need to make your mind up on whether n is infinite or not.

If it's not, then your list is finite and has an end:

8888 { 1, 2, 3, ..., 8887, 8888 }

If it is, then your list is also infinite and has no end:

... { 1, 2, 3, ... }

The only "issue" comes in limiting n to some finite number, then wondering why the list doesn't contain all of N. Plainly, if you limit n to 1,000, then 1,001 won't be in your list.

[Boltzmannbrain]

On 4/28/2023 at 5:42 AM, Genady said:

I wouldn't say in this case, "as n goes to infinity", because n doesn't "go" at all here, but rather is one arbitrary member of the set / sequence.

Also, the expression, "for all n element of N", doesn't make sense to me in this case.

One could write that there is mapping from the set N to set of rows such that every n in N is mapped to a row {1,2,3, ..., n}. See, e.g., Mapping | mathematics | Britannica.

 

Okay, that makes sense. +1

 

I have been thinking about your argument earlier regarding whether or not all the sets contain the set N.  What if I just put the last natural number in each set, so it would look like this.

1 {1}

2 {2}

3 {3}

.

.

.

n {n}

.

(Every n in N is mapped to a row)

Then I ask the same sort of the same question.  Does every natural number exist in the list of each set?  

This would seem like I should run into the same kind of problem as the argument you made.  But it would seem contradictory or at least counterintuitive that every n is not in each set.

 

I am quite confused. 

On 4/28/2023 at 11:06 AM, studiot said:

I am sorry you felt like that since nothing could be further from the truth.

You are most definitely not an idiot since greater minds that yours or mine have been baffled by this question.

I have noticed since that I mistakenly assumed your set S to be a set of number when in fact I see now that you stated clearly a set of sets.

 

This brings us to the crux of greater minds since this is exactly the situation brought about by Russel's Paradox.

That is when you try to apply Cantor - ZFC naive set theory to infinite sets that cannot be members of themselves.

 

This is why Russell and Whitehead introduced the theory of types or classes, which is basically a reclassification of sets introducing a hierarchy of set types.

This also paved the way for 'orders of logic'  so ZFC is first order, infinite sets of sets is second order which is needed to correctly analyse  infinite first order sets of numbers.

In general you need a higher order of logic than the one you want to analyse and there are an unknown, perhaps indefinite or infinite, count of orders.

This situation lead, in turn, to Godel's famous theorems about the subject.

 

The best plain explanation of all this I have come across is put forwards in Hofstadter's award winning book

Godel Escher Back

Oh interesting!  I will have to read up on this. +1

21 hours ago, pzkpfw said:

Then you simply need to make your mind up on whether n is infinite or not.

I had always intended it to be infinite.  I just did not use the proper notation.

Quote

 

If it's not, then your list is finite and has an end:

8888 { 1, 2, 3, ..., 8887, 8888 }

If it is, then your list is also infinite and has no end:

... { 1, 2, 3, ... }

The only "issue" comes in limiting n to some finite number, then wondering why the list doesn't contain all of N. Plainly, if you limit n to 1,000, then 1,001 won't be in your list.

 

I don't think you are seeing my issue for infinite n. 

If you look at the list I made in the OP, you will see that everything is very "equivalent".  Number of objects in each set = greatest element in each set = nth row (nth set).  Looks good and everything is fine, until we use every n (where every n in N is mapped to a row (as Genady pointed out are proper terms that I want to convey)).  Now the equalities create a problem.  They seem to cause an n to exist that does not end. 

Infinite sets do not have a greatest element, but it also seems logical that an n exists with no end somewhere in a set in the list.  And this is a problem of course because the naturals have to be finite.

Edited April 29, 2023 by Boltzmannbrain

[Genady]

32 minutes ago, Boltzmannbrain said:

Does every natural number exist in the list of each set? 

I'm afraid I don't understand what you mean by "list of each set". What I see in your last presentation is a list of rows.

Each row contains a number and a set. Each set contains one number.

For example, the third row contains the number 3 and the set {3}. The set {3} contains one number, namely, 3. There is nothing else in this set.

An arbitrary n-th row contains the number n and the set {n}. The set {n} contains one and only one number, the number n, and nothing else.

[Boltzmannbrain]

24 minutes ago, Genady said:

I'm afraid I don't understand what you mean by "list of each set". What I see in your last presentation is a list of rows.

Each row contains a number and a set. Each set contains one number.

For example, the third row contains the number 3 and the set {3}. The set {3} contains one number, namely, 3. There is nothing else in this set.

An arbitrary n-th row contains the number n and the set {n}. The set {n} contains one and only one number, the number n, and nothing else.

I meant to ask if every natural number exists in the list.  This is a similar situation as in my OP.    

[pzkpfw]

1 hour ago, Boltzmannbrain said:

...

I don't think you are seeing my issue for infinite n. 

If you look at the list I made in the OP, you will see that everything is very "equivalent".  Number of objects in each set = greatest element in each set = nth row (nth set).  Looks good and everything is fine, until we use every n (where every n in N is mapped to a row (as Genady pointed out are proper terms that I want to convey)).  Now the equalities create a problem.  They seem to cause an n to exist that does not end. 

...

If n is infinite, then it doesn't end. That's not a "problem". Why would it be?

nth row (where n is infinite) -> {1, 2, 3, ... } -> N

  

1 hour ago, Boltzmannbrain said:

...

Infinite sets do not have a greatest element, 

...

Correct. (Edit: for your list of distinct Naturals)

  

1 hour ago, Boltzmannbrain said:

...

but it also seems logical that an n exists with no end somewhere in a set in the list. 

...

Not quite sure what you mean here.

  

1 hour ago, Boltzmannbrain said:

...

And this is a problem of course because the naturals have to be finite.

What?? Why do the naturals have to be finite? If that's not a typo, it may be the cause of your confusion.

 

 

Edited April 29, 2023 by pzkpfw

[studiot]

1 hour ago, Boltzmannbrain said:

Infinite sets do not have a greatest element, but it also seems logical that an n exists with no end somewhere in a set in the list.  And this is a problem of course because the naturals have to be finite.

Some infinite sets do, some do not.

For instance the set of all values of the function f(x) = exp(-x)  for all x >= 0 have a greatest element exactly = to 1

This particular infinite set has no least element.

A different function, eg the sine function has both a greatest and a least element.

4 minutes ago, pzkpfw said:

Correct.

Sorry, no.

[Genady]

33 minutes ago, Boltzmannbrain said:

I meant to ask if every natural number exists in the list.  This is a similar situation as in my OP.    

The answer is in this statement:

1 hour ago, Boltzmannbrain said:

Every n in N is mapped to a row

This statement means that for every natural number there is a row in the list. That row, by construction, contains that number. Consequently, every natural number is present in some row in the list.

[studiot]

15 minutes ago, pzkpfw said:

What?? Why do the naturals have to be finite? If that's not a typo, it may be the cause of your confusion.

Every natural number is finite.

It is just that the count of them does not terminate, since there is no last natural number.

The set, N of all natural numbers is infinite because it has no limite to the count of its elements.

 

There is a difference between the allowable size of a set and the allowable size of any element of that set.

 

8 minutes ago, Genady said:

This statement means that for every natural number there is a row in the list. That row, by construction, contains that number. Consequently, every natural number is present in some row in the list.

But this also means that the list does not terminate.

[Genady]

2 minutes ago, studiot said:

this also means that the list does not terminate.

Absolutely. I think that has been established some time ago, hasn't it?

3 minutes ago, studiot said:

But

Why but? It doesn't contradict anything, does it?

[studiot]

2 hours ago, Boltzmannbrain said:

I don't think you are seeing my issue for infinite n. 

If you look at the list I made in the OP, you will see that everything is very "equivalent".  Number of objects in each set = greatest element in each set = nth row (nth set).  Looks good and everything is fine, until we use every n (where every n in N is mapped to a row (as Genady pointed out are proper terms that I want to convey)).  Now the equalities create a problem.  They seem to cause an n to exist that does not end.

 

It is an axiom that if n exists then (n+1) exists and so on.

 

so our sequence should be written

 

1, 2, 3, ...(n-2), (n-1), n, (n+1), (n+2),...

 

This is consistent with n being finite, but never the largest element and yet the sequence does not terminate.

 

[pzkpfw]

52 minutes ago, studiot said:

Every natural number is finite.

It is just that the count of them does not terminate, since there is no last natural number.

The set, N of all natural numbers is infinite because it has no limite to the count of its elements.

 

There is a difference between the allowable size of a set and the allowable size of any element of that set.

...

By "the naturals" I took him to mean "the set of natural numbers", not "any given number in that set".

 

(Edit: I've seen people in forums claim that "natural" numbers must have some limit, as they over-think what "natural" means and make claims like there can't be higher number than the count of atoms in the Universe (seriously, have seen stuff like that). So I did want to double-check that the OP did not have some kind of "N is finite" thinking.)

Edited April 29, 2023 by pzkpfw

[Boltzmannbrain]

1 hour ago, pzkpfw said:

If n is infinite, then it doesn't end. That's not a "problem". Why would it be?

nth row (where n is infinite) -> {1, 2, 3, ... } -> N

  

Correct.

  

Not quite sure what you mean here.

  

What?? Why do the naturals have to be finite? If that's not a typo, it may be the cause of your confusion.

 

 

Each n has to finite, not infinite.  That's what I mean.

1 hour ago, studiot said:

Some infinite sets do, some do not.

For instance the set of all values of the function f(x) = exp(-x)  for all x >= 0 have a greatest element exactly = to 1

This particular infinite set has no least element.

A different function, eg the sine function has both a greatest and a least element.

Ah, of course! +1

[pzkpfw]

6 minutes ago, Boltzmannbrain said:

Each n has to finite, not infinite.  That's what I mean.

Cool. So maybe it's still all just a notation problem. Personally I think it's awkward to be using "n" when thinking of "infinite". Are you happy with the following? Does this have a "problem"?

1 { 1 }

2 { 1, 2 }

n { 1, 2, ..., n }

... { 1, 2, ... }

(I don't claim this is formal notation.)

Edited April 29, 2023 by pzkpfw

[Boltzmannbrain]

1 hour ago, Genady said:

The answer is in this statement:

Then when I try to apply your proof by contradiction argument to the new list, it doesn't seem to work anymore.  And the relevant properties of this simple list are quite parallel to the properties in the list in my OP.  This seems to be a problem.   

49 minutes ago, studiot said:

 

It is an axiom that if n exists then (n+1) exists and so on.

 

so our sequence should be written

 

1, 2, 3, ...(n-2), (n-1), n, (n+1), (n+2),...

 

This is consistent with n being finite, but never the largest element and yet the sequence does not terminate.

 

Okay, but a set with all natural numbers, namely the set N, there is no longer that equivalence from the number of rows to an n existing in the set that equals the number of rows (that we called greatest element in the finite sets).  In other words, each n is finite and cannot match the infinite number of rows, so the equivalence breaks.  In addition to my OP, that related issue is also what is driving me crazy.

12 minutes ago, pzkpfw said:

Cool. So maybe it's still all just a notation problem. Personally I think it's awkward to be using "n" when thinking of "infinite". Are you happy with the following? Does this have a "problem"?

1 { 1 }

2 { 1, 2 }

...

n { 1, 2, ..., n }

... { 1, 2, ... }

(I don't claim this is formal notation.)

Yeah, I understand that.  But I am not sure why you are posting this.

Edited April 29, 2023 by Boltzmannbrain

[Genady]

44 minutes ago, Boltzmannbrain said:

Then when I try to apply your proof by contradiction argument to the new list, it doesn't seem to work anymore.  And the relevant properties of this simple list are quite parallel to the properties in the list in my OP.  This seems to be a problem.   

That proof has to be modified for the new list. When modified appropriately, it works just fine. I don't see a problem.

[Boltzmannbrain]

27 minutes ago, Genady said:

That proof has to be modified for the new list. When modified appropriately, it works just fine. I don't see a problem.

I don't understand why it has to modified at all.

[Genady]

Just now, Boltzmannbrain said:

I don't understand why it has to modified at all.

Because the sets which appear on the rows in the new list are different from the old list.

[Boltzmannbrain]

4 minutes ago, Genady said:

Because the sets which appear on the rows in the new list are different from the old list.

I am thoroughly confused and tired.  I have to think about this more.

[studiot]

1 hour ago, pzkpfw said:

By "the naturals" I took him to mean "the set of natural numbers", not "any given number in that set".

 

1 hour ago, Boltzmannbrain said:

Each n has to finite, not infinite.  That's what I mean.

This is why it is so important to be precise enough in maths.

 

1 hour ago, Boltzmannbrain said:

Okay, but a set with all natural numbers, namely the set N, there is no longer that equivalence from the number of rows to an n existing in the set that equals the number of rows (that we called greatest element in the finite sets).  In other words, each n is finite and cannot match the infinite number of rows, so the equivalence breaks.  In addition to my OP, that related issue is also what is driving me crazy.

The correct expression is not 'equivalence' but ' into ' or 'injective'  or perhaps stronger 'into and onto' or 'one-to-one' or 'bijective'

We also come back to the sine function I mentioned earlier.

For 0>= x < 90  the set of all values of sin(x) has an infinite 'count' of values, all of which are finite.

But worse, the count of this set is uncountable if x is taken from the real numbers.
So your correspondence in this case is merely injective, but not bijective.

 

 

[studiot]

17 hours ago, Boltzmannbrain said:

On 4/28/2023 at 6:06 PM, studiot said:

I am sorry you felt like that since nothing could be further from the truth.

You are most definitely not an idiot since greater minds that yours or mine have been baffled by this question.

I have noticed since that I mistakenly assumed your set S to be a set of number when in fact I see now that you stated clearly a set of sets.

 

This brings us to the crux of greater minds since this is exactly the situation brought about by Russel's Paradox.

That is when you try to apply Cantor - ZFC naive set theory to infinite sets that cannot be members of themselves.

 

This is why Russell and Whitehead introduced the theory of types or classes, which is basically a reclassification of sets introducing a hierarchy of set types.

This also paved the way for 'orders of logic'  so ZFC is first order, infinite sets of sets is second order which is needed to correctly analyse  infinite first order sets of numbers.

In general you need a higher order of logic than the one you want to analyse and there are an unknown, perhaps indefinite or infinite, count of orders.

This situation lead, in turn, to Godel's famous theorems about the subject.

 

The best plain explanation of all this I have come across is put forwards in Hofstadter's award winning book

Godel Escher Back

Expand  

Oh interesting!  I will have to read up on this. +1

 

13 hours ago, Boltzmannbrain said:

am thoroughly confused and tired.  I have to think about this more.

 

Taking a tiredness break and then thinking out it a bit more is exactly the right way to go. +1

 

I have thrown away most of my old maths notes now, but came across this piece that you might find helpful about my comments on set theory, Godel, and the natural numbers.

 

[Genady]

33 minutes ago, studiot said:

One technical question here. The last paragraph refers to "RCL", but I didn't see it defined above, unlike the "RCF". What does the "RCL" stand for?

Edited April 30, 2023 by Genady

[studiot]

36 minutes ago, Genady said:

One technical question here. The last paragraph refers to "RCL", but I didn't see it defined above, unlike the "RCF". What does the "RCL" stand for?

 

The explanation it a bit tougher I'm afraid.

a0c62f4626de2e369d6d43aab0099057c283.pdf (semanticscholar.org)

 

 

[Genady]

30 minutes ago, studiot said:

 

The explanation it a bit tougher I'm afraid.

a0c62f4626de2e369d6d43aab0099057c283.pdf (semanticscholar.org)

 

 

So, the "RCL" in the text you've posted stands for the R_CL as defined in this article. Is it correct?

[studiot]

1 hour ago, Genady said:

So, the "RCL" in the text you've posted stands for the R_CL as defined in this article. Is it correct?

Some sets are open, some are closed 

But some are neither open nor closed

and yet others are both open and closed.

That much has been known for a long time.

 

cl stands for clopen , and is the subject of study in current maths.
As such terminology varies a good deaql.

R stands for real

You will also references to lambda or theta and supercontinuous functions.

 

So the article I linked to has some of these, rather formally,

There is plenty more especially if you look for supercontinuity.

[Genady]

1 hour ago, studiot said:

cl stands for clopen , and is the subject of study in current maths ...

Thank you. I didn't know this.