Deeper than primes

[epix]

epix, it is well known that there is a bijection between a set and its proper subset, if the set is infinite, exactly as shown in http://www.internationalskeptics.com/forums/showpost.php?p=6985470&postcount=14579 (which is a fact that is ignored by you).

Is there any conclusion limping behind your reply, like that if your statement is true, the power set of natural numbers must have greater cardinality than the set of the natural numbers?

Since by definition the cardinality of a proper subset must be less than the cardinality of the set, your statement is rather strange. Consider the infinite set of counting numbers and one of its proper subsets, such as A = {3, 1, 2}.

C..........A
1 <---> 3
2 <---> 1
3 <---> 2
4 <--->
5 <--->
6 <--->
.
.
.

As you see, there can be only a partial bijection due to aleph0 > |A|

 

[epix]

EDIT:

jsfisher, if B (the member of P(S)) does not exist, Cantor's theorem can't provide a P(S) member, which is not mapped with some S member, and therefore it can't conclude that |P(S)| > |[S|

B does exist, but no member of of N is onto B, given the axiomatic framework that this particular edition of the set theory runs on. Look again at the conclusion in Wiki (Cantor's Theorem) where B is defined as D.

Therefore, there is no natural number which can be paired with D, and we have contradicted our original supposition, that there is a bijection between N and P(N).

 

[doronshadmi]

No, it does not. Even if I translate your gibberish into something meaningful like "an explicit P(S) member that is not in any mapping from S to P(S) must be shown", it is still wrong.

No explicit member is shown. All that is shown, all that needs to be shown is that for any mapping from S to P(S), any mapping at all, there must exist at least one element of P(S) not mapped by an element of S.

In other words, this one element of P(S) that is not mapped by an element of S, is en explicit P(S) member, and by using Cantor's construction method independently of Cantor's theorem ( as shown in http://www.internationalskeptics.com/forums/showpost.php?p=6986611&postcount=14585 ) one enables to construct explicit P(S) members without exceptional and define mapping between them and S members, again, without exceptional.

The result is a bijection between the infinite members of P(S) and S, exactly as shown in http://www.internationalskeptics.com/forums/showpost.php?p=6986611&postcount=14585 , which you simply ignore and as a result you (the community of current formalists, Platonists and, so called, logicians of Mathematicsת, which accepts the existence of infinite sets) are closed under your own ignorance.

Shell we continue to celebrate such ignorance? I do not think so!

Real Mathematics is leaking between your ignorant fingers.

 

[doronshadmi]

Is there any conclusion limping behind your reply, like that if your statement is true, the power set of natural numbers must have greater cardinality than the set of the natural numbers?

Since by definition the cardinality of a proper subset must be less than the cardinality of the set, your statement is rather strange. Consider the infinite set of counting numbers and one of its proper subsets, such as A = {3, 1, 2}.

C..........A
1 <---> 3
2 <---> 1
3 <---> 2
4 <--->
5 <--->
6 <--->
.
.
.

As you see, there can be only a partial bijection due to aleph0 > |A|

Look at this:

N..........E
1 <---> 2
2 <---> 4
3 <---> 6
4 <---> 8
5 <---> 10
6 <---> 12
.
.
.

If both sets are infinite, then there is a bijection between a set (in this case, the set of natural numbers) and (in this particular case) the set of even numbers, which are a proper subset of the set natural numbers.

 

[epix]

Look at this:

N..........E
1 <---> 2
2 <---> 4
3 <---> 6
4 <---> 8
5 <---> 10
6 <---> 12
.
.
.

If both sets are infinite, then there is a bijection between a set (in this case, the set of natural numbers) and (in this particular case) the set of even numbers, which are a proper subset of the set natural numbers.

I know what you meant by

epix, it is well known that there is a bijection between a set and its proper subset, if the set is infinite, exactly as shown in http://www.internationalskeptics.com/forums/showpost.php...ostcount=14579 (which is a fact that is ignored by you).

but your assertion didn't distinguish between the finite and infinite case of a proper subset. In this type of contention, you need to be really specific otherwise goulash.

So there are two subsets of N: one is made of even numbers and the other of odd numbers. In your example above, you started to map the subset made of even numbers. How long would it take before the mapping involves the subset made of odd numbers. Any idea?

 

[doronshadmi]

but your assertion didn't distinguish between the finite and infinite case of a proper subset. In this type of contention, you need to be really specific otherwise goulash.

1) It was demonstrated between finite P(S) and its finite proper subset S, by using finite amount of members taken from infinite S ( as shown in http://www.internationalskeptics.com/forums/showpost.php?p=6963549&postcount=14542 ).

2) But now we are talking about the case where both P(S) and S are infinite sets.

Any way, in both cases the existence of infinite S is involved, where in (1) case only a finite amount, taken from infinite S, is used.

So there are two subsets of N: one is made of even numbers and the other of odd numbers. In your example above, you started to map the subset made of even numbers. How long would it take before the mapping involves the subset made of odd numbers. Any idea?

At no time.

It simply doesn't happen, because we are talking about a bijection between two infinite sets of different elements, and in this case it does not matter if one of the infinite sets of different elements is a proper subset of the other infinite set of different elements.

 

[doronshadmi]

Not all circles are concentric Doron, you have simply limited the circles you are considering to the set of all concentric circles ...

I am talking exactly about the set of all circles with different curvatures.

In this case, each circle's curvature exists exactly once as property of a member of this set.

By using the members of this set along an infinitely long straight line, their intersections with this line do not completely cover that line, no matter how the circles are organized along this line.

The reason for that remains the same (at any organized way) which is:

The set of all circles with different curvatures does not have the smallest circle or the biggest circle, as shown in:

and any attempt to be (totally curved (being a point)) AND (totally straight (being a line)) is possible only by discontinuity, and resulted by an uncovered line.

Furthermore, let the center point be any arbitrary real number along the infinitely long straight line, it does not change the facts described above.

 

[jsfisher]

In other words, this one element of P(S) that is not mapped by an element of S, is en explicit P(S) member

Add "explicit" to the list of terms you don't understand.

You also continue to render you thoughts as gibberish. Elements of S to not do any mapping. It is nonsense to say "mapped by an element of S", yet you persist.

The mapping would be done by some mapping function. It is this function that would map elements of S to elements of P(S). The proof for Cantor's Theorem shows that there exists no bijection (a particular type of mapping function) between elements of S and P(S). No such bijection exists.

You cannot then talk about explicit elements the mapping function does or does not map because the mapping function does not exist.

Similarly, you cannot construct any examples of the set B used in the proof, since to do that, you'd need a bijection and that simply does not existent.

 

[epix]

Similarly, you cannot construct any examples of the set B used in the proof, since to do that, you'd need a bijection and that simply does not existent.

I'm not 100% sure how that set B was defined by either you or Doron, but I guess that it is identical to set B in Wiki and set D used in the attempt to translate the formal logic of the proof. Set B must exist in the hypothetical form, otherwise the contradiction would not materialize. The construction of set B defines the set.

For all x the sets B and f(x) cannot be the same because B was constructed from elements of A whose images (under f) did not include themselves.

The construction is similar to the construction of infinite sequences where function f=n² constructs the infinite set of square numbers, so the hypothesis that the set includes a prime number is false due to the function. If Doron asks for an example of B, he gets its definition and not a particular or "explicit" sample of the sequence, unless a particular example is shown, like in Wiki. In that case B = {1, 3, 4, . . .}, which is pretty much a meaningless exhibit, unless its construction/definition is known.

It's a devilish stuff, coz the sight of the cases of bijection between two infinite sources of elements that obviously head for infinity is pretty convincing to junk Cantor's proof. LOL.

 

[jsfisher]

I'm not 100% sure how that set B was defined by either you or Doron, but I guess that it is identical to set B in Wiki and set D used in the attempt to translate the formal logic of the proof. Set B must exist in the hypothetical form, otherwise the contradiction would not materialize. The construction of set B defines the set.

Yes, given a bijection, f(), from A to P(S), B is the set, { x ∈ A : x ∉ f(x) }.

And yes, under the hypothetical that f() exists, the set B also exists. However, since the hypothetical condition leads to a contradiction, f() does not in fact exist, and by extension, neither does B.

The problem that comes up is Doron's insistence B not only exists in fact, but exists in a multitude of forms that can be used to build up a bijection already proven to not exist.

Doron is oblivious the fact what he builds is (1) not a bijection and (2) not according to any construction method attributable to Cantor. While both (1) and (2) are trivial and obvious facts, Doron remains deliberately, willfully ignorant.

 

[doronshadmi]

Add "explicit" to the list of terms you don't understand.

You also continue to render you thoughts as gibberish. Elements of S to not do any mapping. It is nonsense to say "mapped by an element of S", yet you persist.

Thank you for the correction (which actually is nothing but a wrong use in English).

This is a minor English problem, instead of "by" use "with", that's all.

Again you are focused on trivial and minor mistakes that have no impact on the considered subject in order to avoid the real challenge for you as written in http://www.internationalskeptics.com/forums/showpost.php?p=6986611&postcount=14585 .

In other words, you have no case and therefore no one has to take your reasoning seriously.

 

[doronshadmi]

Yes, given a bijection, f(), from A to P(S), B is the set, { x ∈ A : x ∉ f(x) }.

And yes, under the hypothetical that f() exists, the set B also exists. However, since the hypothetical condition leads to a contradiction, f() does not in fact exist, and by extension, neither does B.

Another nonsense made by jsfisher.

B is a member of P(S) that is not mapped with any S member, by Cantor's theorem.

If B does not exist, as jsfisher claims, then there is no way to conclude that |P(S)| > |S| by Cantor's theorem.

Look how jsfisher contradicts himself about B:

However, since the hypothetical condition leads to a contradiction, f() does not in fact exist, and by extension, neither does B

All that is shown, all that needs to be shown is that for any mapping from S to P(S), any mapping at all, there must exist at least one element of P(S) not mapped by an element of S.

This "at least one element of P(S) not mapped by an element of S" is exactly B or D as written in http://en.wikipedia.org/wiki/Cantor's_theorem

Here is the relevant part taken from http://en.wikipedia.org/wiki/Cantor's_theorem, which explicitly says that B is a member of P(S) (and it is exactly a member of P(S) that is not mapped with any S member, by Cantor's theorem):

Thus there is no x such that f(x) = B; in other words, B is not in the image of f. Because B is in the power set of A, the power set of A has a greater cardinality than A itself.

 

[doronshadmi]

Add "explicit" to the list of terms you don't understand.

Add "reasoning" to the list of terms that you don't understand.

 

[doronshadmi]

I'm not 100% sure how that set B was defined

Maybe because you ignored http://www.internationalskeptics.com/forums/showpost.php?p=6986611&postcount=14585 (please see the link at the top of it (in this link B is called F)).

 

[jsfisher]

Another nonsense made by jsfisher.

B is a member of P(S) that is not mapped with any S member, by Cantor's theorem.

If B does not exist, as jsfisher claims, then there is no way to conclude that |P(S)| > |S| by Cantor's theorem.

Proof by contradiction is another concept that escapes you, Doron.

Nonetheless, since you are so convinced this set, B, must exist in all cases, construct one for us using S = {A} in the example. All you need do, Doron, is provide a bijection, f(), from S to P(S). From that, the set B will follow.

Proceed, please.

 

[jsfisher]

B is a member of P(S) that is not mapped with any S member, by Cantor's theorem.

By the way, Doron, this is another demonstration of your muddled thought processes. You've been very reliable about confusing theorems, proofs, and examples, and conflating them at inappropriate opportunities.

Cantor's Theorem does nothing of the kind. It does not map elements of P(S), nor does it construct sets named B. Cantor's Theorem simply states that |S| < |P(S)|.

The reference proof for Cantor's Theorem, on the other hand, does rely on mapping functions, sets named B, and such. And an example used to explain the proof includes a set named D.

 

[epix]

Maybe because you ignored http://www.internationalskeptics.com/forums/showpost.php?p=6986611&postcount=14585 (please see the link at the top of it (in this link B is called F)).

I can see it now. The fans of Cantor's theorem usually define that particular subset with B (stands for Bijection), but that dude in your reference used F.

Speaking of ignorance . . . The symbolism

# <---> @

does not automatically refer to Bijection. That function is a composition of Injection AND Surjection. In Cantor's Theorem proof, there is no bijection due to the absence of surjection. That may confuse the proof when the insufficient symbolism # <---> @ enters the intuitive logic that makes the proof improbable.

Rewrite the Wiki informal proof in such a way that the D set, which is B in the formal definition, is made of odd or even numbers. Then hit an exhaustive reference to the mapping functions. After that, it maybe easier for you to see the absence of surjection in the proof.

You need to adjust the axiomatic framework that allows the proof to exist in order to toss the proof into the garbage. But you may fall into the garbage as well, coz setting up a consistent set of axioms that could run a complex environment such as the set theory is akin to designing a bug-free computer operating system. I don't like that Cantor's proof -- it reminds me the case when a judge let a criminal go free on technicalities involving a loophole in the penal code. But unless you change the law/axioms, you need to abide by it.

 

[doronshadmi]

Cantor's Theorem does nothing of the kind. It does not map elements of P(S), nor does it construct sets named B. Cantor's Theorem simply states that |S| < |P(S)|.

It shows (by using a contradiction, which is demonstrated by the inability of S member to be both (a member) AND (not a member) of P(S) member) that there is B, which is a member of P(S) that is not in the range of all S members.

Independently of Cantor's theorem, by using B as a placeholder of all P(S) members

Such B is {} if:

a ↔ {a}
b ↔ {b}
c ↔ {c}
d ↔ {d}
...

where a ↔ {}


Such B is {a,b,c,d,...} if:

a ↔ { }
b ↔ {a}
c ↔ {b}
d ↔ {c}
...

where b ↔ {a,b,c,d,...}


Such B is also all the subsets of S between {} and {a,b,c,d,e,...} (in addition to {} and {a,b,c,d,...}), such that

S.....B
a ↔ {}
b ↔ {a,b,c,d,...}
c ↔ some explicit P(S) member, which is different than the previous mapped P(S) members
...

etc. ... ad infinitum, or in other words, there is a bijection between P(S) and its proper subset S.


You simply can't look at http://www.internationalskeptics.com/forums/showpost.php?p=6986611&postcount=14585 , which is not Cantor's theorem, isn't jsfisher?

 

[doronshadmi]

I don't like that Cantor's proof --

If you look at http://www.internationalskeptics.com/forums/showpost.php?p=6996924&postcount=14618 (including the link that is found in it) you are also able to understand the reason of why you don't like Cantor's proof.

You need to adjust the axiomatic framework that allows the proof to exist in order to toss the proof into the garbage.

I show that the axiomatic framework that allows the proof to exist, is itself garbage.

 

[jsfisher]

It shows (by using a contradiction, which is demonstrated by the inability of S member to be both (a member) AND (not a member) of P(S) member) that there is B, which is a member of P(S) that is not in the range of all S members.

You continue with your willful gibberish. Cantor's Theorem is |S| < |P(S)|, nothing more.

Independently of Cantor's theorem, by using B as a placeholder of all P(S) members

With this gibberish are you trying to say that the set B, is defined in the reference proof, must be an element of P(S)?

Such B is {} if:

a ↔ {a}
b ↔ {b}
c ↔ {c}
d ↔ {d}
...

where a ↔ {}

...except that there are two problem with this. First, your list of mappings is not a bijection, as required in the reference proof, so it cannot be used in constructing B. Second, you begin with the claim, a ↔ {a}, but then change it was something else, a ↔ {}.

Double Doron fail.