Confusion with finite sets vs an infinite set of natural numbers

[Boltzmannbrain]

18 hours ago, Genady said:

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

I give up.  I can not see any relevant reason why your proof works for the first list but not the second list.  Why do the extra elements change anything about your proof?

[Genady]

3 minutes ago, Boltzmannbrain said:

I give up.  I can not see any relevant reason why your proof works for the first list but not the second list.  Why do the extra elements change anything about your proof?

Here it is with a technical modification to fit your new list.

We want to show that the set

S = {x∈N | x≥1}

is not in your list.

Let's assume that the set S is in your list. Then there is a row, r, in the list with the set S.

The row r has the set

R = {x∈N | x=r}

Evidently, 

R ≠ S

which contradicts the assumption.

Thus, S in not in the list.

[Boltzmannbrain]

18 hours ago, studiot said:

 

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

 

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.

 

 

Okay, I will use those terms for now on.

5 hours ago, studiot said:

 

 

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.

 

Thanks for this +1

10 minutes ago, Genady said:

Here it is with a technical modification to fit your new list.

We want to show that the set

S = {x∈N | x≥1}

is not in your list.

Let's assume that the set S is in your list. Then there is a row, r, in the list with the set S.

The row r has the set

R = {x∈N | x=r}

Evidently, 

R ≠ S

which contradicts the assumption.

Thus, S in not in the list.

Oh, then according to your proof, the second list is not complete either.  I thought that we both thought it was?

[Genady]

7 minutes ago, Boltzmannbrain said:

the second list is not complete either.  I thought that we both thought it was?

I don't know what you mean by the list being "complete." I didn't use this word.

[Boltzmannbrain]

1 minute ago, Genady said:

I don't know what you mean by the list being "complete." I didn't use this word.

I thought that the new list had all natural numbers in it.  I thought it would be a complete list of all natural numbers where each is contained in its own set.

[Genady]

11 minutes ago, Boltzmannbrain said:

I thought that the new list had all natural numbers in it. 

Yes, this list has all natural numbers in it.

It does not have a set of natural numbers in it.

"Set of natural numbers" and "all natural numbers" are different entities.

Edited April 30, 2023 by Genady

[Boltzmannbrain]

1 hour ago, Genady said:

Yes, this list has all natural numbers in it.

It does not have a set of natural numbers in it.

"Set of natural numbers" and "all natural numbers" are different entities.

I thought your new proof for the new list was saying that every natural number (in their own individual sets) could not be listed.

[Genady]

1 hour ago, Boltzmannbrain said:

I thought your new proof for the new list was saying that every natural number (in their own individual sets) could not be listed.

Well, it is not what it was saying.

Rather, every natural number is listed, but the set of natural numbers is not.

[studiot]

4 hours ago, Boltzmannbrain said:

thought that the new list had all natural numbers in it.  I thought it would be a complete list of all natural numbers where each is contained in its own set.

 

Let's revisit your opening post from a new point of view.

 

The sets

{1}
{1, 2}
{1, 3}
etc

Are all finite sets.

This means that if I write out their contents as a series,

The partial sums are all finite.

 

Do you know what a partial sum is ?

For the three example sets they are

1

1, 3

1,3, 6

 

The last sum always arrives at a finite number.

In other words the series is always convergent or unconditionally convergent.

 

Now look at what happens with infinite series.

 

The 1, 3, 6  pattern goes on forever, getting larger and large at every partial sum.

That is the infinite series is divergent.

 

You can add or subtract or do other more complicated arithmetic with any of these finite series, by replacing the series with its final partial sum.

So {1+2}  + {1+2+3}  = 3 + 6 = 9

 

but what happens if you try to perform these tasks with an infinite series ?

 

{1+2}   + {1 +2 + 3+ 4...}  =  {1+2} + N  =  3  +  ???

This is the problem lying at the base of simple set theory

 

Note some infinite series are convergent for example the series  1/n².

 

So as soon as you try to introduce N into your list of sets, you loose all the set operations  -Union, sum, difference etc.

 

 

 

 

[Boltzmannbrain]

6 hours ago, Genady said:

Rather, every natural number is listed, but the set of natural numbers is not.

What do you mean by the set of natural numbers?  Each number in the second list is alone.  

[Genady]

5 minutes ago, Boltzmannbrain said:

What do you mean by the set of natural numbers?

I mean the same that it means in math.

 

5 minutes ago, Boltzmannbrain said:

Each number in the second list is alone.

I see that.

[Boltzmannbrain]

5 hours ago, studiot said:

 

Let's revisit your opening post from a new point of view.

 

The sets

{1}
{1, 2}
{1, 3}
etc

Are all finite sets.

This means that if I write out their contents as a series,

The partial sums are all finite.

 

Do you know what a partial sum is ?

For the three example sets they are

1

1, 3

1,3, 6

 

The last sum always arrives at a finite number.

In other words the series is always convergent or unconditionally convergent.

 

Now look at what happens with infinite series.

 

The 1, 3, 6  pattern goes on forever, getting larger and large at every partial sum.

That is the infinite series is divergent.

 

You can add or subtract or do other more complicated arithmetic with any of these finite series, by replacing the series with its final partial sum.

So {1+2}  + {1+2+3}  = 3 + 6 = 9

 

but what happens if you try to perform these tasks with an infinite series ?

 

{1+2}   + {1 +2 + 3+ 4...}  =  {1+2} + N  =  3  +  ???

This is the problem lying at the base of simple set theory

 

Note some infinite series are convergent for example the series  1/n².

 

So as soon as you try to introduce N into your list of sets, you loose all the set operations  -Union, sum, difference etc.

I understand partial sums, and I think I understand the rest of what you are saying here.  But I do not understand how this addresses the post you quoted.  The list I made on page 2 is suppose to have every natural number as every row maps to every n in N.  At least that's what I wanted.

My main question is whether or not every natural number would be listed alone in its own set.

2 minutes ago, Genady said:

I mean the same that it means in math.

 

I see that.

Why would I think that the set of natural numbers would be there?  I don't understand what you are getting at.

[Genady]

3 minutes ago, Boltzmannbrain said:

Why would I think that the set of natural numbers would be there?  I don't understand what you are getting at.

That is what that proof was about. If it is not relevant anymore, fine.

The other question,

4 minutes ago, Boltzmannbrain said:

whether or not every natural number would be listed alone in its own set

has been answered, too. The answer is, yes.

[Boltzmannbrain]

1 hour ago, Genady said:

That is what that proof was about. If it is not relevant anymore, fine.

 

I thought the proof (second proof) proved that every natural number cannot be listed alone in its own set.

 

[Genady]

3 hours ago, Boltzmannbrain said:

 

I thought the proof (second proof) proved that every natural number cannot be listed alone in its own set.

 

It certainly did not. 

All is clear now, I hope.

[joigus]

At the root of it all, I think, is @Boltzmannbrain's remarkable inability --or stubbornness to not recognise-- the limit operation, which in common language is captured by the words "and so on."

That is,

1

1, 2

1, 2, 3

and so on.

Don't look now BB, but these are the words you're having a problem with.

Embrace infinity.

[Boltzmannbrain]

7 hours ago, Genady said:

It certainly did not. 

All is clear now, I hope.

My mistake, I did not look at your second proof close enough.  My mind was going in a completely different direction than yours, apparently.  So, I only assumed a different type of analogous proof.  I hope I come across a lot more direct this time.  Let's rewind a little. 

Your first proof shows that S = {x| x∈N & x≥1} cannot be listed since each row can be listed as  L = {x| x∈N & x≥1 & x≤l} and L ≠ S.  Or in other words, for any row in my original list, there is always at least one more row.  Moreover, your proof also shows that the greatest element in each of the sets in my original list is its limiting factor that disallows S, from your proof, to exist in my list.  These seem to be the principles of your proof that are relevant to explaining how my list does not have every set, and thus resolving my issue in the OP.

The original list was:

1 {1}

2 {1, 2}

3 {1, 2, 3}

4 {1, 2, 3, 4}  

.

.

.

 

Your proof made perfect sense to me, and I was happy ... but then I thought of an analogous situation.

Since the relevant part of your proof really only concerns the nth set (lth to be exact) and the greatest element in each set in my list, I thought of a list that only shows the greatest element in each set. 

1 {1}

2 {2}

3 {3}

4 {4}

.

.

.

 

So I hope you can see where I am going with this.  If we use this same relevant principles from your original proof, even though it still seems totally sound, you say that every natural number is in the second list.

 

5 hours ago, joigus said:

At the root of it all, I think, is @Boltzmannbrain's remarkable inability --or stubbornness to not recognise-- the limit operation, which in common language is captured by the words "and so on."

That is,

1

1, 2

1, 2, 3

and so on.

Don't look now BB, but these are the words you're having a problem with.

Embrace infinity.

I don't agree.  The more I look into all of this the more strange and complicated it is.  

[Genady]

28 minutes ago, Boltzmannbrain said:

in other words, for any row in my original list, there is always at least one more row.  Moreover, your proof also shows that the greatest element in each of the sets in my original list is its limiting factor that disallows S, from your proof, to exist in my list.  These seem to be the principles of your proof that are relevant to explaining how my list does not have every set, and thus resolving my issue in the OP.

No, you read way too much in this proof. It is quite a trivial one. It just shows that whatever set a row in your list has, is not a set of natural numbers. I don't see that digging into this proof will help to clear out the confusion.

In fact, by now I don't know what the confusion is and thus don't know how to help with it. Maybe you can clearly state it from scratch.

[Boltzmannbrain]

29 minutes ago, Genady said:

In fact, by now I don't know what the confusion is and thus don't know how to help with it. Maybe you can clearly state it from scratch.

I thought I just put in a very thorough review of our conversation from scratch.  But we can move on if you want, and I will attempt to explain my confusion in a very brief and direct way.

 

The confusion stems from my OP.  Instead of asking whether or not the list has every set of natural numbers, what happens if I just define the list to have every set (starting from 1 and increasing by 1)?  Can I do this?  If I can, then what about your proof? 

 

Quote

No, you read way too much in this proof. It is quite a trivial one. It just shows that whatever set a row in your list has, is not a set of natural numbers. I don't see that digging into this proof will help to clear out the confusion.

I tried to explain how your proof does not seem to work for what I think is a directly analogous example.

Edited May 1, 2023 by Boltzmannbrain

[Genady]

2 minutes ago, Boltzmannbrain said:

what happens if I just define the list to have every set (starting from 1 and increasing by 1)?  Can I do this?

Not clear. Are these sets that you define, finite or infinite?

[Boltzmannbrain]

1 minute ago, Genady said:

Not clear. Are these sets that you define, finite or infinite?

I mean the list in my OP.  The sets are natural numbers that start at 1 and increase by 1.  So I can only suppose they would all be finite (which is the crux of the issue).

[Genady]

2 minutes ago, Boltzmannbrain said:

I mean the list in my OP.  The sets are natural numbers that start at 1 and increase by 1.  So I can only suppose they would all be finite (which is the crux of the issue).

If you define them as "starting at 1 and increasing by 1", then they are infinite.

But if you define them as "starting at 1, increasing by 1, and stopping when the row number is reached", then they are finite.

What is your definition?

[studiot]

This rather hinges on what property of infinity is being invoked.

What sort of mathematical object do you think n is ?

 

5 minutes ago, Genady said:

If you define them as "starting at 1 and increasing by 1", then they are infinite.

But if you define them as "starting at 1, increasing by 1, and stopping when the row number is reached", then they are finite.

What is your definition?

Surely you can't have two different definitions of the same thing ?

[Boltzmannbrain]

9 minutes ago, Genady said:

If you define them as "starting at 1 and increasing by 1", then they are infinite.

 

By "they, I meant each set is finite.

If what you mean by "they" is the amount of sets, then I agree, the amount of sets are infinite.  

This is the heart of my confusion.  Somehow, it seems that the amount of sets can be infinite while each particular set is not (that is if we want every element in the set to be a finite natural number).

 

[Genady]

Just now, Boltzmannbrain said:

 

By "they, I meant each set is finite.

If what you mean by "they" is the amount of sets, then I agree, the amount of sets are infinite.  

This is the heart of my confusion.  Somehow, it seems that the amount of sets can be infinite while each particular set is not (that is if we want every element in the set to be a finite natural number).

 

By "they" I mean each set. So, I need to rephrase my question:

If you define each set as "starting at 1 and increasing by 1", then it is infinite.

But if you define each set as "starting at 1, increasing by 1, and stopping when the row number is reached", then it is finite.

What is your definition?