Boltzmannbrain

9 minutes ago, Genady said:

Yes, it is. But there is no such set in your construction. IOW, there is no set in your construction "that starts at 1, increases by 1 and has infinite elements."

I think you are saying that we can't glean anything from the finite sets of the forementioned type (start at 1 and increase by 1), let's call type T, to the infinite sets of type T.  Is that accurate?  

8 minutes ago, Genady said:

I understand. There is no set equal to N in your construction.

 

There is nothing to compare to because there is no infinite set in your construction that starts at 1 and increases by 1.

A set that starts at 1, increases by 1 and has infinite elements has the properties of N.  Isn't that sufficient to construct N?

35 minutes ago, Genady said:

There are never infinite elements and so no such n is needed.

I was referring to a set equal to N.

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There is no such jump.

I meant a jump in comparing the finite vs infinite sets that start at 1 and increase by 1. 

13 hours ago, Genady said:

Because it never goes from R to N.

This may be the best way that I can state my confusion.

What I am trying to show is that the finite sets of natural numbers, that start at 1 and increase by 1, seem quite logical when comparing both sides (the number of elements versus the input n). 

For example 5 elements in the set implies an input of n = 5.  Infinite elements implies an input of n = ???

How do we make this asymmetrical jump to an infinite number of elements with finite inputs??? 

11 hours ago, pzkpfw said:

That starts with "n∈N". So n is a member of the set of Natural numbers. That set is infinite. Any given selected value for n will be finite, and can be plugged into the "1≤x≤n" to make a finite list of 1 to n; but that n can be selected from any of the Naturals ... and there are infinite of them.

I am not sure what you are saying that I don't agree with.

My issue is really quite simple.  

Take a set of natural numbers that start at 1 and increase by 1.  When this kind of set is finite, we know there must be a finite n that also equals the number of elements that it has.  When this kind of set has an infinite number of elements, we know there must be an infinite n that also equals the number of elements that it has.

That is where my logic is leading me (for better or worse).  It leads me to believe that there must be an infinitely large n in the set of all natural numbers. 

It is perfectly symmetrical and perfectly proportional.  Why should it even be wrong?

14 minutes ago, Genady said:

This does not exist. It never happens. There is no such symmetry breaking.

There are only finite numbers of elements in R on the left side. As has been said above, R(n) is always finite.

There are only R's on the left, never N.

I just wanted to show how the symmetry breaks when going from R (finite) to N (infinite)

Why can't I do this?

27 minutes ago, wtf said:

To the best of my understanding of this thread, the heart of your issue seems to be that you don't quite get the following idea:

Each of the individual natural numbers 0, 1, 2, 3, 4, 5, ... is itself a finite quantity.

And there are infinitely many of them. 

In other words there are infinitely many finite things. And for some reason you have trouble going back and forth between those two levels. The finitude of each of the natural numbers, and the endlessness, or infinitude, of the procession of all of them via the process of endlessly adding 1.

There are infinitely many natural numbers, and each of them are finite.

Likewise each R(n) = {1, 2, 3, ..., n} is a finite set; and there are infinitely many of the finite sets R(n), namely R(1), R(2), R(3), etc.

Hope this is helpful.

@Genady just went over this with me.  What keeps me confused is how the symmetry below gets broken.  Just going from some set R to the set of natural numbers N does not make sense to me.  Let's continue to use each n∈N is mapped to set R(n)= {x∈N | 1≤x≤n} 

1 element in R --> n = 1

2 elements in R --> n = 2

3 elements in R --> n =3

4 elements in R --> n =4

.

.

.

infinite elements in N --> n = a finite number

Or put more generally,

finite --> finite

finite --> finite

finite --> finite

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.

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infinite --> finite

How this symmetry between the left side and the right side gets broken is my main issue.

59 minutes ago, Genady said:

R(n) does not change to infinite. 

R(n) and {R(n)} are different things.

The former contains numbers in the range [1, n]. The latter contains sets R(n) for all n's. 

The former is finite, the latter is not.

 

Okay, thanks for your patience.  I forgot to take into account the set symbols.  +1

My same issue still lingers if you care to continue.  

You say, "R(n) is finite. {R(n) | n∈N} is not".  Is every R(n) finite in LIST?

21 minutes ago, Genady said:

Correct. R(n) is finite. {R(n) | n∈N} is not.

Why does R(n) change to infinite when n is an element of N instead of just an n?

This is not a facetious question.  I think we have come to the absolute heart of my issue.  

49 minutes ago, pzkpfw said:

 

Just a few posts ago you wrote (my bold) "I forgot to put more ellipsis under the n for indefinite rows in the OP."

This is what I tried to show with:

1 { 1 }

2 { 1, 2 }

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

There is no single finite n in N that gives a set with no end. The list ( { 1, 2, ... } ) is infinite, and the row number is also infinite.

Try thinking of the list ( { 1, 2, ... } ) as X on a graph and the list of lists as the Y. It's unbounded on both axis.

I see what you are saying.  There are certainly many reasons to explain away my issue.  But there are still reasons that maintain my issue from being resolved.

26 minutes ago, Genady said:

This last example not only is not permitted, but it does not have any meaning in the set of natural numbers. Infinity is not an element of this set. This example does not make sense.

I know.  I was trying to make a point.

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Yes, each R(n) has n elements.

I agree.

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None. Each R(n) is finite.

But you wrote, "the set {R(n) | n∈N} is not finite" a few posts ago.

1 hour ago, Genady said:

However, if n=5 there are no 5 rows/sets, but one.

If you want to discuss a different mapping, then define it first.

Oh oops, sorry.  But this does not change anything about the point I am trying to make.  Let's still use your definition, each n∈N is mapped to set R(n)= {x∈N | 1≤x≤n}

n = 1 ---> 1 element

n = 2 ---> 2 elements

n = 3 ---> 3 elements

n = 4 ---> 4 elements

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.

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n is always finite ---> finite elements

n = infinity ---> infinite elements (of course this last example is not permitted, but it is the only thing that makes sense to me at the moment.)

Shouldn't the n value stay proportional to the number of elements? 

What n in N gives a set with no end?

42 minutes ago, Genady said:

No, for each n there is one and only one row, the row number n. This row has nothing to do with other rows.

For each n there is one set, R(n). The list is a set of these sets. Let's call it LIST.

This set, LIST is defined so that for each n∈N the set R(n)∈LIST, and for each element Q∈LIST there exists n∈N such that Q=R(n).

There is no "implies" anywhere in the definitions.

I used "implies" in the way that we would say, If n =5, then there are 5 rows/sets, or equivalently, n = 5 implies 5 rows/sets.  I don't think I am doing anything wrong here.

Furthermore, n is the input, right?  That means that the number of sets is the output.  There is no infinite number that n can be that allows an output of infinite sets. 

Doesn't the proportionality of

n = 1 = 1 set

n = 2 = 2 sets

n = 3 = 3 sets

n = 4 = 4 sets

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.

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n = infinity = infinite sets

Why not this?  But I suppose this has been thought of already.

But looking at this list, doesn't it seem wrong that the left side is always finite but the right side is not always finite?

I see your argument too.  We want infinite sets such that every n in N is mapped to a set.  But then my problem gets switched to there being an n that is infinite, which is not allowed either.

56 minutes ago, Genady said:

Yes, each set R(n) is finite.

You said something else in the previous post:

This is incorrect. The list of sets is not finite.

In other words, for each n,

the set R(n) is finite.

But

the set {R(n) | n∈N} is not finite.

Given your definition, each n∈N is mapped to set R(n)= {x∈N | 1≤x≤n}  do you agree that for each n∈N there is only a finite number of rows?

If not, here is what is in my head,

n = 1 implies 1 row

n = 2 implies 2 rows

n = 3 implies 3 rows

n = 4 implies 4 rows

n = 5 implies 5 rows

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any finite n implies a finite number of rows

Every n is finite implies there can only ever be a finite number of rows.

This is my issue.

29 minutes ago, Genady said:

No, it does not. What makes you think it does?

It says, "each n∈N", doesn't it?

Here is the definition again:

"each n∈N is mapped to set R(n)= {x∈N | 1≤x≤n}"

There is no limit on n.

Each n in N is finite.  Doesn't this imply that each set must be finite too?

Your even made a proof showing that each n is not sufficient to list every set.  But I don't agree with your proof anyways since it didn't work for the analogous list I made way back at the top of page 2 of this thread.

5 hours ago, studiot said:

As I see your argument in plain English it is this:-

 

Taking you list of sets from post#1

Set the indexing line counter aside for the moment as it is not really needed.

Consider the set which contains every set on your list.

If such  a set exists, call it W .

The listing of W then appears as in your list of sets without the indexing.

I understand up until here.

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IF you go on long enough why do you not arrive at the set {1,2,3,4...., (n-1), n (n++1)...}, why of course is N ?

I do not understand what you are saying here. 

Quote

 

Of course N is also the indexing set we have ignored up to now.

Note also that all the sets up to N are finite, but N itself is transfinite.

 

Yes, I believe that this is a part of the heart of the problem.  

Quote

 

I mentioned Russel's Paradox which queries the existence of W.

This was one of the earliest expositions of many paradoxes that appeared around the 1890s to do with the size of sets.

 

I understand Russel's paradox, but I not see how my issue is related.

Quote

 

Hints of these difficulties go right back to the Ancient Greeks and Zeno in particular,
although they did not have more modern set theory to place the questions in.

A proper course of study into the whys and wherefores of these matters takes more than a year so most folks don't attempt it but look for a quick fix explanation.

My offering to you is to consider the Greek approach, where they realised that there is more than one infinity.
They distinguished two types of infinity viz potential and actual infinity.

They believed that there are no instances of actual infinity, which we observe as for instance, the count of numbers between 1 and 2.

But their potential infinity does not exist' either for a different reason.

It does not exist because no finite process can ever get there.

In other words the process does not terminate or goes on forever.

Which is what I am suggesting is the reason why your list will never arrive at N.

 

Okay, I think I understand what you are saying, but can't N exhaust N?  If so, then my list of N rows, exhausts/creates a set with N elements, or vice versa.  The point being, that the 2 N's exhaust each other.  Or am I going off on a tangent?   

Quote

 

Cantor's approach considering magnitudes has run into difficulties why has yet to be fully resolved.
There are at least three different mathematical/logical schemes to try to achieve this. 
After your year and more of study you would find that none are totally satisfactory as they all wrestle with the idea that some sets are just too big to be contained in other sets.

 

Are you saying that my OP illustrates a problem in mainstream math that has yet to be resolved, or is my particular problem resolvable with mainstream math?

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Hopefully you can now sleep happy at night.

That would be nice.

3 hours ago, joigus said:

That should be your first clue. Usually, the more I look into anything, the less strange and complicated it seems.

And that's how it should be. Don't you think?

Clue to what?

36 minutes ago, Genady said:

This is needed because mapping connects elements of two sets. In your case, each element of the domain set ("from") is a natural number, and each element of the codomain set ("to") is a set of natural numbers.

Oh thanks I did not know that.  +1

Quote

 

Very well. So, following the definition,

"each n∈N is mapped to set R(n)= {x∈N | 1≤x≤n}",

please define the issue that bothers you.

 

Actually, I do see a problem with your definition.  n limits the list of sets to always be finite.  We wanted an infinite list.  We wanted every n in N.

12 hours ago, Genady said:

Let me describe how I understand your construction.

 

There is a list of numbered rows, one row for each natural number.

Each row has a set of natural numbers, which contains natural numbers between 1 and the row number (including).

 

Is this description correct?

Yes

Quote

 

Putting it more formally:

 

You construct a one-to-one (injective) map from set of natural numbers N to set of sets of natural numbers such that each n∈N is mapped to set R(n)= {x∈N | 1≤x≤n}.

 

Correct?

 

Yes I believe that's correct, except you added one small difference from your other description.  You want the sets to go inside of a set.  I guess that's fine because I can't see it changing my original description. 

7 hours ago, Genady said:

The sets shown in your OP don't fit the definition "start at 1 and increase by 1."

They fit the definition "start at 1, increase by 1, and stop at the row number."

I wanted the rows to continue indefinitely.  I forgot to put more ellipsis under the n for indefinite rows in the OP.

With that said, I still do not see how my issue of the OP is resolved.

19 minutes ago, studiot said:

Not quite.

n must be the value of a function that takes on natural numbers ( or the positive integers if you prefer) as its values.

That is the only way n can be a different natural numbers in different lines in your OP list.th

 

You are correct about (x+1) also called successor property of the natural numbers.
It is this property that invokes infinity and leads to the countable infinity property of the natural numbers.

Infinity has many properties, but the one we want in this case is that it cannot be reached.

The ancient Greeks called this the potential infinity because it is never actually realised.

Okay, I was thinking about this potential infinity vs infinity.  This might be where I am misleading myself.  Can we not talk about infinity in its totality?  For example, is it self-contradictory to say something like "all elements of N" or "every real number from 1 to 2"? 

I know I have read those kinds of terms get used, but maybe they cannot be used formally, or can they?  

Quote

 

This is imprecise and inaccurate.

Please try to be correct.

 

The list does not have a numerical value or increase by one.

Nor does a set have a numerical value.

 

I know what you mean (and your meaning is OK, but it isn't what your words are saying) Until  you  can state it correctly yourself I doubt if anyone can help you understand.

 

I don't know why I keep resorting to my own terms instead of using the proper terms.  I will try harder to use the proper terms in the future.  And if I don't know the correct term, I will say so.

2 hours ago, Genady said:

The sets shown in your OP don't fit the definition "start at 1 and increase by 1."

They fit the definition "start at 1, increase by 1, and stop at the row number."

Sorry, like I told @studiot I will try harder to use the correct terms in the future.

13 minutes ago, Genady said:

No. It refers to each individual set.

 

No. A set defined as "starting at 1 and increasing by 1" is infinite.

I wanted to talk about the list of sets that start at 1 and increase by 1.   The same list that is in my OP.

9 minutes ago, studiot said:

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

What sort of mathematical object do you think n is ?

An n is a natural number.  According to the paper you posted, we can define a natural number when it belongs to a hereditary set that is defined when x+1 is an element of F and when x is an element of F, which also has 1.  At least that is how I am understanding it. 

7 minutes ago, Genady said:

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?

The first one.  But I need to clear something up.  Your pronoun "it" must refer to the amount of sets in the list, right?  I say this because clearly each set is finite as we look down the list of sets.  Right? 

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).

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).

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.

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}  

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.

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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}

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.

.

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.  

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.

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.