Deeper than primes

[doronshadmi]

You are the only one here who cannot see just how trivial it is.

Now you are proud of your trivial reasoning

Please pay attention. It is not a bijection between the natural numbers and {},{A}. The natural numbers include more than just 1 and 2.

Pay attention. Bijection between two sets is a mapping between two sets, such that each object of one set is paired with exactly one object of the other set and there are no members of those sets that are not mapped with each other (it is called 1-to-1 and onto). Bijection (1-to-1 and onto) can be found between any two sets, even if the two sets have finite cardinality.

 

[doronshadmi]

Please show us where Cantor provided such a method. Stop showing us your bogus posts; show us Cantor's work.

EDIT:

Can you use your trivial reasoning in order to say how many D explicit objects are constructed, as shown in http://en.wikipedia.org/wiki/Cantor's_theorem ?

By this theorem Cantor explicitly provides |P(N)| D explicit objects that can be put in one-to-one correspondence with N objects.

Cantor's mistake is based on the fact that he concluded general conclusions by using only a one case of his theorem.

 

[jsfisher]

Now you are proud of your trivial reasoning

You are the only one here rejoicing in the trivial. The rest of us see it as, well, trivial.

Pay attention. Bijection between two sets is a mapping between two sets, such that each object of one set is paired with exactly one object of the other set and there are no members of those sets that are not mapped with each other (it is called 1-to-1 and onto).

No one has disputed this. Why do you so prone to argue things that are not in contention?

And, as should be blatantly obvious even to you, Doron, since you just recited the basic meaning of bijection, your mapping between the natural numbers and {},{A} isn't a bijection since you provide no mapping for 3, 4, 5, etc.

Bijection (1-to-1 and onto) can be found between any two sets, even if the two sets have finite cardinality.

This part contradicts what you just posted about bijections.

 

[jsfisher]

Can you use your trivial reasoning in order to say how many D explicit objects are constructed, as shown in http://www.internationalskeptics.com/forums/showpost.php?p=6916077&postcount=14361 ?

So where, exactly, did Cantor provide this construction method for these mysterious "D explicit objects" of which you speak?

 

[jsfisher]

EDIT:

Can you use your trivial reasoning in order to say how many D explicit objects are constructed, as shown in http://en.wikipedia.org/wiki/Cantor's_theorem ?

By this theorem Cantor explicitly provides |P(N)| D explicit objects that can be put in one-to-one correspondence with N objects.

Cantor's mistake is based on the fact that he concluded general conclusions by using only a one case of his theorem.

I see you have taken the time to completely re-write a previous post. You poor reasoning skills and lack of reading comprehension ability are exceeded by your rudeness.

Be that as it may, I see you'd like to use Wikipedia for your Cantor's proof reference. You may notice that the proof, amazingly enough, appears under the heading Proof. No where in that proof is there a construction method given for any "D explicit objects".

The proof given in Wikipedia is substantially the same as the proof I gave some ways back in this thread. Neither presentation provides a construction method for any existing set.

 

[doronshadmi]

You are the only one here rejoicing in the trivial. The rest of us see it as, well, trivial.



No one has disputed this. Why do you so prone to argue things that are not in contention?

And, as should be blatantly obvious even to you, Doron, since you just recited the basic meaning of bijection, your mapping between the natural numbers and {},{A} isn't a bijection since you provide no mapping for 3, 4, 5, etc.



This part contradicts what you just posted about bijections.

No jsfisher, this part simply does not fit to your trivial reasoning about 1-to-1 and onto mapping between two sets, because your reasoning can't get the fact that also two finite sets have a bijection, by using Cantor's construction method.

 

[jsfisher]

No jsfisher, this part simply does not fit to your trivial reasoning about 1-to-1 and onto mapping between two sets, because your reasoning can't get the fact that also two finite sets have a bijection, by using Cantor's construction method.

You have yet to show a bijection between the members of {A} and P({A}), with or without the use of the non-existent construction method. Substituting a trivial mapping between two sets of two elements each isn't even close.

 

[doronshadmi]

I see you have taken the time to completely re-write a previous post. You poor reasoning skills and lack of reading comprehension ability are exceeded by your rudeness.

Be that as it may, I see you'd like to use Wikipedia for your Cantor's proof reference. You may notice that the proof, amazingly enough, appears under the heading Proof. No where in that proof is there a construction method given for any "D explicit objects".

The proof given in Wikipedia is substantially the same as the proof I gave some ways back in this thread. Neither presentation provides a construction method for any existing set.

Call Cantor's theorem whatever you like.

It does not change the fact that by using this theorem, one enables to explicitly define |P(N)| Ds AND a bijection between N members and these Ds.

Once again, Cantor missed his own construction method, because, like you, he wrongly was focused on some single and particular case of that construction.

 

[jsfisher]

Cantor missed his own construction method, because, like you, he wrongly was focused on some single and particular case of that construction.

Please, point out for all of us this construction method as it appears in Cantor's proof. Surely, it must appear right there under that Proof heading in Wikipedia. We anxiously await your revelation of that that isn't there.

 

[doronshadmi]

You have yet to show a bijection between the members of {A} and P({A}), with or without the use of the non-existent construction method. Substituting a trivial mapping between two sets of two elements each isn't even close.

Are you kidding?

Do you really unable to understand that the same construction method also provides |P(P(N))|, |P(P(P(N)))|, ... ,|...P(P(P(N)))...| explicit D's that are explicitly paired (1-to-1 and onto) with N members, such that N ↔ P(N) ↔ P(P(N)) ↔ P(P(P(N))) ↔ ... ↔ ...P(P(P(N)))... ?

 

[jsfisher]

Are you kidding?

Nope, not kidding. You have been completely unable to show a bijection between the members of {A} and {{},{A}}.

Use any construction method you like, real or imaginary, then just show us the final the result: a bijection between the members of {A} and {{},{A}}.

Leave out the natural numbers (or any part of them); leave out all your fabulous multi-round step-by-step processing. All we need to see is the conclusion: a bijection between the members of {A} and {{},{A}}.

Just so you know, we are all predicting you will fail again.

 

[doronshadmi]

Please, point out for all of us this construction method as it appears in Cantor's proof. Surely, it must appear right there under that Proof heading in Wikipedia. We anxiously await your revelation of that that isn't there.

EDIT:

Cantor's construction method constructs explicit |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members as follows:

1) Some defined explicit {{},...,{1,2,3,...}} member is the result of a 1-to-1 correspondence between {1,2,3,...} members and the same amount of members taken from {{},...,{1,2,3,...}}.

2) This explicit constructed {{},...,{1,2,3,...}} member (notated as D) includes a {1,2,3,...} member only if this {1,2,3,...} member does not exist as one of the members of the {{},...,{1,2,3,...}} member that is mapped with it.

By using this construction method |{{},...,{1,2,3,...}}l times, one enables to define a bijection between {1,2,3,...} members
and {{},...,{1,2,3,...}} members, where D is actually a placeholder for |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members.

 

[doronshadmi]

Nope, not kidding. You have been completely unable to show a bijection between the members of {A} and {{},{A}}.

Use any construction method you like, real or imaginary, then just show us the final the result: a bijection between the members of {A} and {{},{A}}.

Leave out the natural numbers (or any part of them); leave out all your fabulous multi-round step-by-step processing. All we need to see is the conclusion: a bijection between the members of {A} and {{},{A}}.

Just so you know, we are all predicting you will fail again.

I am not a participator of your "running in circles on a closed box" game.

Because of your game you are unable to get, for example, the following:

http://www.internationalskeptics.com/forums/showpost.php?p=6955135&postcount=14519

http://www.internationalskeptics.com/forums/showpost.php?p=6955416&postcount=14521

http://www.internationalskeptics.com/forums/showpost.php?p=6955576&postcount=14530

http://www.internationalskeptics.com/forums/showpost.php?p=6955625&postcount=14532

 

[jsfisher]

EDIT:

Cantor's construction method constructs explicit |{{},...,{1,2,3,...}}l {{},...,{1,2,3,...}} members as follows:

Where is that in the Proof?

 

[jsfisher]

I am not a participator of your "running in circles on a closed box" game.

So, you have given up trying to show a bijection between the members of {A} and P({A}), then?

Just as well, since there isn't one.

 

[epix]

Again you demonstrate your trivial reasoning by get this bijection as trivial.

Trivial reasoning cannot be used to establish the cardinality of a set through bijection. Please get familiar with the meaning of various prefixes, such as "bi-" and "tri-" in this particular case:

bi = "two" => bicycle
tri = "three" => tricycle

Since 3 ≠ 2, trivial reasoning cannot contribute to the construction of bijection, as opposed to the case where erotic reasoning facilitates erection.

 

[The Man]

Again your limited reasoning airs its view.

The set of natural numbers, is some particular case of a collection of different objects.

“objects” that are all specifically natural numbers. Are you claiming that some member(s) of your “set of natural numbers” is/(are) not (a) natural number(s)? If that is not your claim then all the members of your “set of natural numbers” are natural numbers. Thus you have no actual problems with “all” in reference to the natural numbers or any set of natural numbers.

Are all of your natural numbers natural numbers, Doron, or are some of your natural numbers not natural numbers?

Perhaps you just don’t know or just don’t want to know? Whichever, it really doesn’t matter.

You can propose no argument to all natural numbers being natural numbers that is not simply self contradictory (some natural numbers aren’t natural numbers) or simply an argument from ignorance (I don’t know if all natural numbers are natural numbers). The former simply and directly refutes itself while the latter asserts a specific lack of knowledge about whether all natural numbers are natural numbers or not and thus just professes its own irrelevance.

Given any collection of different objects, succession (in terms of present continuous, where also parallel reasoning is used) is its essential property, whether it is finite (and in this case the succession is external w.r.t the collection) or infinite (and in this case the succession is internal w.r.t the collection).

“external w.r.t the collection”? So by your own assertion you’re just talking about extraneous elements that you specifically claim are not included in that collection.

“succession is internal w.r.t the collection”? So the successor of any member of the set is also a member of that set? Perhaps we could come up with an expression to assert this amazing notion? How about “the set is closed under an operation of succession”? Oh wait that is already being used but, hey, by some strange coincidence that is exactly what it means. If we didn’t know any better we might start thinking that you’re learning some math (or at least set theory).

You still do not understand the meaning of succession +1 in terms of cardinality as a present continuous, because parallel reasoning is beyond your only step-by-step serial reasoning.

You are still deliberately trying to confuse a list with a set. “succession +1”? So the successor of a successor? By your own assertion above it too can be a member of the set.





















Oh and again…

By all means please explain to us the difference between increasing and decreasing with “no past (before) and no future (after)”?

 

[epix]

|S| = |P(S)| is right, where both |S| and |P(S)| are equally ever increasing (which is present continuous).

LOL. Considering a finite case: if |S| = 3, then |P(S)| = 2³. According to you, 3 = 2³.

Considering infinite case S = N: The consequence of |N| = |P(N)| is that your "parallel reasoning" holds 2^aleph0 equivalent to aleph0. In other words, if there is a 1-on-1 correspondence between the members of N and the members of P(N), then the cardinality of the power set P(N) is strictly aleph0. But any construction that proves the previous statement true must apply to the finite case. So show us the jolly-good construction when S = {1, 2, 3}.

 

[epix]

Pay attention.

Released from its prison of absolute truth, mathematics was free to roam and to develop any kind of system it wanted -- with the one restriction that these systems must be internally consistent. Whether they are useful in describing the world does not matter, for modern mathematicians are no longer concerned with depicting what they see in the most realistic way possible. Like modern painters, they have become more interested in the techniques and methods of their art than in mere description.

 

[doronshadmi]

So, you have given up trying to show a bijection between the members of {A} and P({A}), then?

Just as well, since there isn't one.

Not at all.

If the powerset of set A is {{},A} it means that |A|=1

It means that also |{A}|=1, and in the case P({A}) = {{},{A}}.

By using Cantor's construction method we get a 1-to-1 correspondence between 1,2 and {},{A}, which is not different than the 1-to-1 correspondence between 1,2 and {},A, such that:

1 ↔ {} ↔ {}
|
2 ↔ A ↔ {A}

and we get the same 1-to-1 correspondence for both cases, which are A and P(A) or {A} and P({A}).

Here it is again, by using Cantor's construction method:

A ↔ A | Provides 1 ↔ {}, which is equivalent to {A} ↔ {A} | Provides 1 ↔ {}
A ↔ { } | Provides 2 ↔ A, which is equivalent to {A} ↔ {} | Provides 1 ↔ {A}