Deeper than primes - Continuation

[dafydd]

Why does it surprise you so that a set of 16 elements will have some elements that do not appear in a set of only 4?

Because he's Doron. Anything to do with real maths is a surprise to him.

 

[jsfisher]

So what? I think that Cantor mentioned that.

You realize you've just cued Doron to rattle on about how Cantor was wrong, don't you?

 

[dafydd]

You realize you've just cued Doron to rattle on about how Cantor was wrong, don't you?

Humour an old man and let me have my fun!

 

[laca]

No need for all that binary nonsense nor pretending things to be power sets when they really aren't. It is just a very simple idea you seem hell-bent on presenting in the most convoluted manner possible. The question remains, though: Why does it surprise you so that a set of 16 elements will have some elements that do not appear in a set of only 4?

I assume that's a rhetorical question or you're looking for a more complicated answer than because he's batcrap crazy. 'Cause that's obvious by now.

 

[zooterkin]

Ok, I see that you still do not get the generality of 0;1 form.

Please be aware that in this particular case we are using here distinct members under powersets with cardinality ...2^(2^(2^(2^1)))... and the number of 0;1 symbols must grow in order to save their distinction w.r.t each other.

By using it, the powerset of S with cardinality 4 is:

0 0 1 1
0,1,0,1
-------
P(S)=
{
0 0 0 0,
1 0 0 0,
0 1 0 0,
1 1 0 0,
0 0 1 0,
1 0 1 0,
0 1 1 0,
1 1 1 0,
0 0 0 1,
1 0 0 1,
0 1 0 1,
1 1 0 1,
0 0 1 1,
1 0 1 1,
0 1 1 1,
1 1 1 1,
}

such that each member is made of 4 0;1 symbols and the cardinality of S is 4, where the cardinality of P(S) is 2^4=16

Now please use Diagonalization among the 4 forms that are based on 0;1 symbols as follows:

S=
{
1001,
0111,
0111,
0001,
}

such that 0000 is an explicit member of P(S) that is not a member of S, etc., as shown in http://www.internationalskeptics.com/forums/showpost.php?p=8144371&postcount=942.

In other words, even if the cardinality of P(S) 2^∞, there is S with ∞ members and both S AND P(S) have members with 0;1 form, where each member has ∞ symbols.

Since (P(S) and S have the same construction of forms, but there are members of this construction in P(S), but not in S) AND (the largest P(S) does not exist) we can conclude that no collection with distinct members is complete (whether the cardinality > 0 is finite or not).

No, Doron, that's not the set I asked about. I asked about

S=
{
00,
01,
10,
11
}

As jfisher points out, you restrict the number of elements allowed in a set so you can then apply your favourite diagonal technique. By restricting the number of values you can have, then of course you'll never be able to have them all.

The set of possible two digit binary values is {00, 01, 10, 11}. Please explain how this is not complete. Which value is missing?

 

[doronshadmi]

No, Doron, that's not the set I asked about. I asked about

S=
{
00,
01,
10,
11
}

As jfisher points out, you restrict the number of elements allowed in a set so you can then apply your favourite diagonal technique. By restricting the number of values you can have, then of course you'll never be able to have them all.

The set of possible two digit binary values is {00, 01, 10, 11}. Please explain how this is not complete. Which value is missing?

In the case of the cardinality 2^|S| (which is the cardinality of P(S)) and cardinality |S| (which is the cardinality of S) it is always shown that S and P(S) can share the same type of forms that are based of 0;1 symbols, such that there is always an explicit element of P(S), which is not an element of S, and by following this fact (and the fact that the largest P(S) does not exist), we can conclude that no non-empty set, that its elements are based on 0;1 symbols, is complete, whether its cardinality is finite or not.

Your restriction of the elements of S to particular cardinality of 0;1 symbols each, is irrelevant, because according to my argument the number of 0;1 symbols of each element of a given S is at least the same as the number of such elements as members of that S, and under this condition Diagonalization and its results are always shown.

But as I show in http://www.internationalskeptics.com/forums/showpost.php?p=8148489&postcount=958 , we do not need the comparison between S and P(S), in order to know that no collection that its elements are based on 0;1 symbols (such that the amount of symbols of each element and the amount of the elements of that collection is the same, whether this amount is finite or infinite) is complete, because we are always can show an explicit element, which is not one of the elements of the given collection.

 

[laca]

But as I show in http://www.internationalskeptics.com/forums/showpost.php?p=8148489&postcount=958 , we do not need the comparison between S and P(S), in order to know that no collection that its elements are based on 0;1 symbols (such that the amount of symbols of each element and the amount of the elements of that collection is the same, whether this amount is finite or infinite) is complete, because we are always can show an explicit element, which is not one of the elements of the given collection.

Excellent, then it's not problem for you to show explicitly which two digit binary value is not in the set {00, 01, 10, 11}. Unless you don't know what you're talking about.

 

[zooterkin]

Let's simplify the argument.

Given a collection of distinct elements that are constructed by 0;1 symbols, such that the size of each element is the same as the size of the collection, there is always an explicit element of 0;1 symbols that is constructed by using Diagonalization, which is not an element of the given collection, no matter if the considered size is finite or infinite.

Artificial restriction highlighted.

By following this fact, we can conclude that no such collection of distinct elements is complete.

(Weasel words highlighted.)
Well, duh! If you limit the number of values to less than the full possible range, then you won't be able to use the full possible range.

What on earth do you think this proves? What can you do with it?
What does this have to do with anything general?

 

[doronshadmi]

So what? I think that Cantor mentioned that.

By using Diagonalization among collection of distinct elements, which are constructed in the most general sense (by using 0;1 as two distinct symbols, as demonstrated in http://www.internationalskeptics.com/forums/showpost.php?p=8148489&postcount=958 ), we realize that no collection (whther the amount of its elements if finite or infinite) is complete.

Please be aware of the fact that no numbers of any kind and no mapping between two collections is used here, so my Diagonalization is more fundamental than Cantor's Diagonal argument.

Form this more fundamental view, one actually understands that no collection of distinct elements is complete, where according Cantor's less fundamental view this understanding is not known (instead one, by mistake, thinks that there are two complete collections with infinite elements, where one complete collection has more elements than the other complete collection).

 

[zooterkin]

By using Diagonalization among collection of distinct elements, which are constructed in the most general sense (by using 0;1 as two distinct symbols, as demonstrated in http://www.internationalskeptics.com/forums/showpost.php?p=8148489&postcount=958 ), we realize that no collection (whther the amount of its elements if finite or infinite) is complete.

No, it is not the "most general sense", you are imposing very specific limitations, which ensure that the number of elements in the set is less than the maximum number of values in the set. Your conclusion only applies to sets chosen according to your rules; you cannot state, without qualification, that "no collection is complete".

 

[jsfisher]

Please be aware of the fact that no numbers of any kind and no mapping between two collections is used here, so my use of Diagonalization is more fundamental than that Cantor's Diagonalization argument.

Well, if we ignore the fact your whole presentation assumes mere hand-waving is sufficient to get from finite to non-finite sets, and if we ignore that you are calling things power sets when they aren't, and if we ignore you never actually proved the set S (constructed via ill-defined methods and gibberish--we'll ignore that, too) was missing an element it shouldn't be missing, yeah, if we ignore all that then you may be right.

On second thought, no, even then you aren't.

 

[doronshadmi]

No, it is not the "most general sense", you are imposing very specific limitations, which ensure that the number of elements in the set is less than the maximum number of values in the set. Your conclusion only applies to sets chosen according to your rules; you cannot state, without qualification, that "no collection is complete".

What you call maximum number of values are actually P(S) elements, which are not S elements, and as a result S is incomplete.

So S is incomplete and so is P(S) since the largest P(S) does not exist, no matter if the considered amount of values is finite or infinite.

If you limit the number of values to less than the full possible range, then you won't be able to use the full possible range.

There is no such a thing like "the full possible range" if we deal with collections of distinct elements, and this is exactly my argument.

What on earth do you think this proves? What can you do with it?
What does this have to do with anything general?

One enables to actually understand the term "the power of the continuum" which is larger than the cardinality of any given collection of distinct elements.

 

[zooterkin]

What you call maximum number of values are actually P(S) elements, which are not S elements.

So S is incomplete and so is P(S) since the largest P(S) does not exist, no matter if the considered amount of values is finite or infinite.

The set of two digit binary numbers is {00, 01, 10, 11}. In what way is that set incomplete?

 

[doronshadmi]

The set of two digit binary numbers is {00, 01, 10, 11}. In what way is that set incomplete?

Why do you restrict these elements into less symbols each, than what is found as the elements of that collection?

Do you understand that by this restriction no Diagonalization can be used (including Cantor's) ?

 

[zooterkin]

Why do you restrict these elements into less symbols each, than what is found as the elements of that collection?

I didn't. It's the set of binary numbers that can be represented with two digits.

Do you understand that by this restriction no Diagonalization can be used (including Cantor's) ?

Yes. And?

Please back up your claim that the 'collection' is incomplete. Which item is missing from the set:

{00, 01, 10, 11}

 

[zooterkin]

I thought we'd been over this before; will you stop editing posts after someone has replied to them?

Make a new post if you want me to reply to the parts you've added since I replied to the post above, where I quoted your full post as it stood at the time.

What you call maximum number of values are actually P(S) elements, which are not S elements, and as a result S is incomplete.

So S is incomplete and so is P(S) since the largest P(S) does not exist, no matter if the considered amount of values is finite or infinite.


There is no such a thing like "the full possible range" if we deal with collections of distinct elements, and this is exactly my argument.


One enables to actually understand the term "the power of the continuum" which is larger than the cardinality of any given collection of distinct elements.

 

[doronshadmi]

I didn't. It's the set of binary numbers that can be represented with two digits.

{00, 01, 10, 11} of 4 distinct elements is restricted by you to the collection of two 0;1 symbols each.

Under this restriction you can't explicitly construct an element of two 0;1 symbols that is not an element of that collection.

In order to be able to construct an explicit element, which is not an element of a collection of 4 distinct elements of 0;1 symbols, the amount of symbols of each element of the given collection, must be at least the same as the amount of the elements of the given collection (only in this case Diagonalization can be used, and we can construct an explicit element, which is not an element of that collection).

 

[doronshadmi]

Zooterkin, in order to follow my reasoning, please construct the power set of {00,01,10,11} by using only elements with two 0;1 symbols each.

 

[zooterkin]

{00, 01, 10, 11} of 4 distinct elements is restricted by you to the collection of two 0;1 symbols each.

Defined.

Under this restriction you can't explicitly construct an element of two 0;1 symbols that is not an element of that collection.

So, the collection is complete, then?

 

[zooterkin]

Zooterkin, in order to follow my reasoning, please construct the power set of {00,01,10,11} by using only elements with two 0;1 symbols each.

What do you mean by 'power set'? A real power set, or a Doron 'power set'? If the latter, please supply clear instructions on what such a thing is, and how to create it.

If the former, then it should be trivial even for you to do (hint: look back at jsfisher's posts).