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.  

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.

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.

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?

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?

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.

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.

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.

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

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.    

Yeah, these are fun.  

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.

48 minutes ago, Genady said:

If we will continue the discussion of my proof that the set S is not in the list, the following clarification of the list might help:

l=1 L={1}

l=2 L={1, 2}

l=3 L={1, 2, 3}

l=4 L={1, 2, 3, 4}

...

Or, equivalently:

l=1, L = {x| x∈N & x≥1 & x≤1}

l=2, L = {x| x∈N & x≥1 & x≤2}

l=3, L = {x| x∈N & x≥1 & x≤3}

l=4, L = {x| x∈N & x≥1 & x≤4}

...

Yeah, I think I understand now. 

When we put something like 1, 2, 3,  ... n ... as n goes to infinity, is that the same as saying "for all n element of N"? 

How would have I wrote it if I wanted every n in the set of N to be assigned a row, or is this not possible?   +4

1 hour ago, pzkpfw said:

Boltzmannbrain, now knowing n in your OP is not meant to itself be infinity ("Note: I forgot to put another downward ellipsis under the n in the OP."), then its line would be:

n {1, 2, 3, 4, ..., n-1, n }

n is not infinity, the set does not have infinite numbers in it. If n = 1000, then there are 1000 numbers in the set.

Happy so far?

Now add 1 to n. And again, and again. Each time you add another number to the set.

1000 { 1, 2, 3, ..., 999, 1000 }

1001 { 1, 2, 3, ..., 1000, 1001 }

1002 { 1, 2, 3, ..., 1001, 1002 }

When/why do you stop? Is there a last n?

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

8 minutes ago, Genady said:

L does not have to be in S, and it not in S. L is a set on a line in your list.

Each line in your list has a set. L is one of these sets, the one on the line number l.

Just so I am clear, I will give an example.

Let's say the line l = 5.  

This would be shown in the list as

5 {1, 2, 3, 4, 5}

Then there is a set L "on it" (I put this in quotes because I assuming what it means here).

L can be something like {2, 3, 4} (and maybe in other words L is a subset of the numbers on line 5?)

Is this the idea so far?

2 minutes ago, Genady said:

No, l is not a sequence. It is a number. It is a number of some line in your list. We don't know which line it is, thus we call it l.

Okay, I understand.  And now why does L have to be in S?  And why doesn't S = L?

3 minutes ago, Genady said:

L is not a line in S. L is a set on a line number l in your list.

If this is clear, I'll go to the second question.

Okay, so l is a sequence of digits like 1, 2, 3, ...  and L is the set of those digits?

19 minutes ago, Genady said:

It is not 1, it is l (the letter):

L = {x| x∈N & x≥1 & x≤l}

Right.  But I am still confused.  Why does this set L have to be a line in S?  And why doesn't S = L?

2 hours ago, Genady said:

I don't know what you mean in the second statement, but here is a proof that this set is not there.

Let's define the set:

S = {x| x∈N & x≥1}

Let's assume that the set S is in the list. Then there is a line, l, in the list with this set on it.

The set on line l is:

L = {x| x∈N & x≥1 & x≤l}

But 

L ≠ S

which contradicts the assumption.

Thus, S in not in the list.

I am lost.  It looks like S is the set of natural numbers, or S = N.  Then the part that confuses me is that the set L has x < or = 1 in it.  Where is that coming from?

Note: I forgot to put another downward ellipsis under the n in the OP.

5 minutes ago, Genady said:

This is wrong, i.e., not every possible set of increasing natural numbers (that increase by 1 starting from 1) is there.

The set {1, 2, 3, 4, ...} is a possible set of increasing natural numbers (that increase by 1 starting from 1), and it is not there.

Okay, I am taking this to mean that the set of all natural numbers is not there.  This is good.  But I don't understand why it isn't there.  

Maybe you say this because I forgot the extra vertical ellipsis after the n?  

40 minutes ago, studiot said:

 

I did actually answer your question since your list was an incomplete representation of the stated infinite set N.

Of course N does not and can not appear on your list since N itself is not a natural number.

To put it another way the problem is confusing a set, N which is not finite, with its elements (natural numbers)  which are all finite.

Then what did I say that was wrong (except for forgetting the ellipsis)? 

I made 2 premises and a conclusion (the conclusion is in the form of a question).  Please tell me which of the 3, or if all, are incorrect.  

If there is a contradiction here, I don't think it would be so much that the set of natural numbers has to finite (because it obviously can't be by its very own nature of never ending), but rather every n is not finite.  

1 hour ago, studiot said:

 

And a nice pleasant discussion to you to.

 

I get defensive when posters try to make me look like an idiot.  What you said (unless it's true of course) is not a good way to start a thread if you want things to be pleasant.  Anyway let's move on.

Quote

 

Since the list has no end,  the convention if you are going to use 'n' , is to place it between two ellipsis thus

 

1

2

3

...

n

...

 

 

The elli[psis is used mathematically as a symbol to mean 'indefinite continuation' .

 

Normally the list would be displayed horizontally, but vertically is OK so I have followed your lead in presenting it this way.

 

Ok, you're right.  I should have put another ellipses.  +1 

4 hours ago, studiot said:

Didn't you have a very long thread about this once before ?

I thought we had cleared up your misunderstanding but you are making the same mistake again.

There is no 'n' at the end of the list

 

An infinite list does not terminate,, by definition.

 

But trying to write an n at the end implies that there is a definite last number, n.

Why would putting an n there imply a last number???  That's just common notation.  If you want to help me, just read what I put, and tell me exactly what I said that is false.

And trying to say this thread is the same as the other is just garbage.  They both are clearly different threads.

For all n rows, this would be a list of all sets of natural numbers that increase by 1 starting from 1.

1 {1}

2 {1, 2}

3 {1, 2, 3}

4 {1, 2, 3, 4}

.

.

.

n

Every set listed here would have to be finite since every natural number is finite.

But if every possible set of increasing natural numbers (that increase by 1 starting from 1) is here, then how can the set of all natural numbers N be infinite?

On 3/15/2023 at 2:43 PM, wtf said:

A point in the Cartesian plane is characterized by an ordered pair of real numbers.

But just as with the line, this is misleading. In math there is no geometrical plane. Rather, there are the real numbers, and then ordered pairs of real numbers. And we call the set of ordered pairs of real numbers the "plane" as a geometrical visualization. But there aren't two things, the plane and the set of ordered pairs. There's only one thing, the set of ordered pairs, which we commonly refer to as "the plane," or, when equipped with the Euclidean metric, the Euclidean plane. But it's still just the set of ordered pairs of real numbers. The geometry is just a convenient visualization and way of speaking. 

To emphasize this point, what else can we do with ordered pairs of real numbers? We can define a "funny multiplication" on them, and then they turn into, voilà, the complex numbers C . 

So would we say that the Euclidean plane and the complex numbers are "the same?" Or would we say, rather, that the Euclidean plane and the complex numbers are two different interpretations of the same underlying set of ordered pairs of reals?

I'm mentioning all this because you are insistent in wanting to know how the pairs of real numbers "describe" or "map to" the plane. But in fact, there is no plane. There's only the set of ordered pairs of real numbers, which, as you note, is called R2 .

This is all a bit philosophical or semantic and I don't want to make too much of it. But you may be asking the wrong question. The set of ordered pairs IS the plane, it doesn't map to relate to the plane. If that helps. 

 

That makes sense.  It is what I was thinking.

Quote

 

Yes you get 0 as the limit.

To be clear, in the example of 1/2, 3/4, 7/8, ... we get 1 as the limit. I assume that's what you meant. 

So how does this work? Let's take the example of the sequence 1/2, 1/3, 1/4, 1/5, 1/6, ...

The sequence has 0 as its limit. That means that the terms of the sequence get, and stay, arbitrarily close to 0. That's formalized as the "epsilon" business that people see briefly, in passing, in a beginning calculus course, before getting on to the calculational aspects.

Note that every element of that sequence is NOT zero. This is crucial to the limit concept. The elements of the sequence never "become" zero. Zero is not an element of the sequence. Rather, the sequence has the limit of 0, where we take some pains to formally define what we mean by a limit.

 

Okay, that makes sense.

Quote

 

I did not read all the posts in this thread before arriving a few pages in. If you formerly held different beliefs, that might be generating some confusion. But even since you and I have been talking, you seem to believe that there is a largest element of [0,1), and that amounts to the same thing. 

So I cannot square this latest statement with what I've been talking to you about. 

 

I only meant that my intuition tells me recently that there is a smallest real.

Quote

 

The halving process refers (in our ongoing example of [0, 1)) to the sequence 1/2, 3/4, 7/8, 15/16, ... We can definitely do this infinitely many times, since that is an infinite sequence of rational numbers. There's an element (2^n - 1)/2^n for every natural number n.

 

I think this all starting to sink in.  

Quote

 

Yes you get 0 as the limit.

To be clear, in the example of 1/2, 3/4, 7/8, ... we get 1 as the limit. I assume that's what you meant. 

So how does this work? Let's take the example of the sequence 1/2, 1/3, 1/4, 1/5, 1/6, ...

The sequence has 0 as its limit. That means that the terms of the sequence get, and stay, arbitrarily close to 0. That's formalized as the "epsilon" business that people see briefly, in passing, in a beginning calculus course, before getting on to the calculational aspects.

Note that every element of that sequence is NOT zero. This is crucial to the limit concept. The elements of the sequence never "become" zero. Zero is not an element of the sequence. Rather, the sequence has the limit of 0, where we take some pains to formally define what we mean by a limit.

 

Okay, this makes sense too.

Quote

 

As a limit. I hope you see the ongoing theme of this post!

In the Riemann integral, we take finer and finer partitions. Say the partitions have successive widths 1/2, 1/3, 1/4, 1/5, ... Every partition has nonzero width. The limit of the widths is zero. And in fact we do NOT require a "zero width" when defining the Riemann integral. Rather, for each partition, we calculate the corresponding Riemann sum, the sum of the areas of the finitely many rectangles in that partition; and we define the integral as the limit of the sequence of Riemann sums.

As you can see, everything depends on a proper understanding of the limit concept. 

 

I see.  Thanks, I forgot about that.

Quote

That's a little word-salady to me. What we were discussing above is whether there's a largest number in [0,1), which it turns out there isn't. So whatever interest you about the Riemann integral may have been explained before I joined the thread. Feel free to catch me up. 

I was only discussing it with you.  I forgot/misunderstood the integral process.

Quote

 

I thought I explained it. You are correct that [0,1] and [1,2] have a point in common, and if you tried to rotate them past each other they'd indeed collide at the point 1.

You are wondering how they can collide if the point 1 has zero length. I tried to analogize this with a Newtonian point mass. I said that even though 1 has zero length, zero dimension, zero volume, it can "still pack a punch."

Did this analogy not resonate with you, even a little? Maybe it's kind of strained, I did admit that up front. 

But in your rotating visualization, the line segments will indeed collide at the point 1, simply because 1 is an element of each segment. 

 

This seems interesting to me in that adding one real number is significant geometrically, but I don't know right how relevant it is to this discussion.

Thank you very much for your help and incredible patience

On 3/16/2023 at 8:16 AM, studiot said:

 

So how am I meant to take this ?

You don't know anything about neighbourhoods but you don't want to consider them ?

Why not ?

Why is this not inconsiderate of others who are trying to help you ?

 

Sorry, I really appreciate your help, but I wasn't making the connection with what you were saying and my issue.  Also, I did not have enough time to figure it out.

I think I have finally understood this whole issue from my OP completely.   

Thank you for your help!

4 minutes ago, studiot said:

You clearly don't want to understand since I offered you an alternative treatment using neighbourhoods and you ignored it totally.

 

The beauty of neighbourhoods is that it does not depend upon whether or not there is a smallest real number or a nearest real number or what happens at infinity.

And yet neigherbourhood analysis arrives at the same results as the other approaches you have been arguing about.

 

I read what you said about neighborhoods and infinity.  Then I looked up neighborhoods because I have never heard of them before.  But I still have no idea how it helps me understand anything about my issue, or even how it relates to the discussion.

I am not sure if I am even ready to get into the topic of neighborhoods.  It seems a little or a lot more advanced than what I have learnt so far.