Deeper than primes - Continuation

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doronshadmi said:
EDIT:

Consider A={1,2} and B={X,Y,Z}

f1(1) = X
f1(2) = Y

that is also can be written as
Code: 
{1,2}
 ↓ ↓
{X,Y,Z}

So, there is one case of no function from all the elements of A to some element of B (element Z, in this case), and we can conclude that |A|<|B|.
Click to expand...

No. That does not follow the very definition you provided. This is a case where we have a doron-function which maps every element of A to some element of B. There was a doron-mapping for 1 and a doron-mapping for 2; all the elements of A have a doron-mapping.

Therefore, |A| = |B|. That's according to your doron-cardinality definition.


By the way, "there is one case of no function" is gibberish. You cannot have "one case" of "no function". Cases show examples of things, not the lack of any possible example. Saying "no function" would mean there is no possible example that meets whatever criterion is under consideration.
 
Some correction of the previous reply:

B.T.W, if "being beyond the range" of some collection (whether it has bounded or unbounded amount of elements) is shown by "no function at all" or by "function with input that does not have any output", then the traditional and the non-traditional definitions are equivalent.

Both cases enable to conclude that |A| > or < |B|, and in this case we actually use only the traditional definition of function on Hilbert's Hotel by using the term "no function at all" in order to show that |N| > |N| or |N| < |N|.
 
jsfisher said:
No. That does not follow the very definition you provided. This is a case where we have a doron-function which maps every element of A to some element of B. There was a doron-mapping for 1 and a doron-mapping for 2; all the elements of A have a doron-mapping.

Therefore, |A| = |B|. That's according to your doron-cardinality definition.


By the way, "there is one case of no function" is gibberish. You cannot have "one case" of "no function". Cases show examples of things, not the lack of any possible example. Saying "no function" would mean there is no possible example that meets whatever criterion is under consideration.
Click to expand...

jsfisher, |A| = |B| is not the considered case in your example, where A = {1,2} and B = {X,Y,Z}.

In other words, this part of the definition
Give two sets, A and B, the doron-cardinality of the set A = the doron-cardinality of set B if and only if there is doron-function from A to B such that for all a in A there is exactly one b in B.
Click to expand...
has nothing to do with your given example.
 
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doronshadmi said:
jsfisher, |A| = |B| is not the considered case in your example, where A = {1,2} and B = {X,Y,Z}.

In other words, this part of the definition
Given two sets, A and B, the doron-cardinality of the set A = the doron-cardinality of set B if and only if there is doron-function from A to B such that for all a in A there is exactly one b in B.
Click to expand...
has nothing to do with your given example.
Click to expand...

Let's take this a step at a time.

Given two sets, A and B
Got that: A = {1, 2} and B = {X, Y, Z}

if and only if there is a doron-function from A to B
Got that: f1(1) = X and f1(2) = Z

such that for all a in A there is exactly one b in B
Got that: f1 is defined for all members of A and in each case, the a in A is mapped to exactly one b in B. (Well, we must assume your really meant "for each" and not "for all", otherwise this would be even more bizarre.)

So, the entire right side of this if and only if proposition is true: There is doron-function from A to B such that for all each a in A there is exactly one b in B.

Therefore, the left side must be true as well: The doron-cardinality of the set A = the doron-cardinality of set B.

Ergo, |A| = |B|.
 
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doronshadmi said:
Wrong, |N| < |R| exactly because "there is one case of no function" beween all N members and at least one R member.
Click to expand...

The gibberish was with respect to your use of English, not the mathematical conclusion.

And, by the way, there are many, many functions (even doron-functions) between all members of N and at least one R member. G(n) = 3.14159, for example, where the domain of G is N.
 
doronshadmi said:
----------------------------

B.T.W, if "being beyond the range" of some collection (whether it has bounded or unbounded amount of elements) is shown by "no function at all" or by "function with input that does not have any output", then the traditional and the non-traditional definitions

Both cases enable to conclude that |A| > or < |B|, and in this case we actually use only the traditional definition of function on Hilbert's Hotel by using the term "no function at all" in order to show that |N| > |N| or |N| < |N|.
Click to expand...

Actually it only shows that your definitions are no better than traditional definitions.

Also, it does not automatically follow from the above that |N|>|N| exists.

The only thing the quoted text shows, as I have demonstrated earlier is that you implicitly redefine the > operator so that it allows for a new domain which is separated from the domain before the operator.
 
realpaladin said:
Actually it only shows that your definitions are no better than traditional definitions.
Click to expand...
Only if "no function at all" is also considered, but according to jsfisher (which is a professional traditional mathematician) "no function at all" is gibberish.

realpaladin said:
Also, it does not automatically follow from the above that |N|>|N| exists.
Click to expand...
It follows if the considered case is, for example:

((1),1)
((),2) (this is the case of "no function at all", which is also considered)
((2),3)
((3),4)
((4),5)
...


realpaladin said:
The only thing the quoted text shows, as I have demonstrated earlier is that you implicitly redefine the > operator so that it allows for a new domain which is separated from the domain before the operator.
Click to expand...
No, you did not demonstrate it since your argument is about:

(1) --> 1
() --> 2
(2) --> 3
(3) --> 4
(4) --> 5

or

1 --> (1)
2 --> ()
3 --> 2
4 --> 3
5 --> 4
...

and not about

1 --> 1
--> 2 (this is the case of "no function at all", which is also considered)
2 --> 3
3 --> 4
4 --> 5

or

1 --> 1
2 -->
3 --> 2
4 --> 3
5 --> 4
...

Again, you wrongly use the function from the names of the rooms to the rooms, or from the rooms to the names of the rooms,
instead of using the function from the names of the name of the visitors, or from the names of the visitors to the names of the rooms (as I do).
 
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jsfisher said:
Let's take this a step at a time.

Given two sets, A and B
Got that: A = {1, 2} and B = {X, Y, Z}

if and only if there is a doron-function from A to B
Got that: f1(1) = X and f1(2) = Z

such that for all a in A there is exactly one b in B
Got that: f1 is defined for all members of A and in each case, the a in A is mapped to exactly one b in B. (Well, we must assume your really meant "for each" and not "for all", otherwise this would be even more bizarre.)

So, the entire right side of this if and only if proposition is true: There is doron-function from A to B such that for all each a in A there is exactly one b in B.

Therefore, the left side must be true as well: The doron-cardinality of the set A = the doron-cardinality of set B.

Ergo, |A| = |B|.
Click to expand...

jsfisher, this definition is not satisfied exactly because there is at least one of B that is not mapped to any a of A, or in other words |A| = |B| does not hold.
 
doronshadmi said:
Only if "no function at all" is also considered, but according to jsfisher (which is a professional traditional mathematician) "no function at all" is gibberish.

It follows if the considered case is, for example:

((1),1)
((),2) (this is the case of "no function at all", which is also considered)
((2),3)
((3),4)
((4),5)
...

No, you did not demonstrate it since your argument is about:

(1) --> 1
() --> 2
(2) --> 3
(3) --> 4
(4) --> 5

or

1 --> (1)
2 --> ()
3 --> 2
4 --> 3
5 --> 4
...

and not about

1 --> 1
--> 2 (this is the case of "no function at all", which is also considered)
2 --> 3
3 --> 4
4 --> 5

or

1 --> 1
2 -->
3 --> 2
4 --> 3
5 --> 4
...
Click to expand...

Blatant nonsense Doron, and you know it.

What I am saying is that according to *your * definitions, as stated in this thread, your logic shows that it is a redefinition operator.

I would be able to concur with you if you would define the operator differently.

At the moment it says something like 'there is a set A for which <some conditions > in a set B.'

This means, and all of this is only from your definitions, that the set before the operator does not need to be the same as after.
Therefore, if we name the sets the same and use the inequality operator, we *must* have different sets before and after.

Use a different symbol and define it properly, then, and only then, you might have a case.
 
doronshadmi said:
jsfisher, this definition is not satisfied exactly because there is at least one of B that is not mapped to any a of A, or in other words |A| = |B| does not hold.
Click to expand...

Nowhere in your definition is there such a requirement. Either your definition is incomplete and still needs more work to cover all the necessary criteria, or |A| = |B| in doronetics.
 
jsfisher said:
The gibberish was with respect to your use of English, not the mathematical conclusion.

And, by the way, there are many, many functions (even doron-functions) between all members of N and at least one R member. G(n) = 3.14159, for example, where the domain of G is N.
Click to expand...
jsfisher, by Cantor's proof by contradiction, there is an element of P(X), such that the attempt to define function from some element of X to this element of P(X), leads to logical contradiction.

In other words, we consider at least one case of "no function at all" from all the elements of X to this this element of P(X), which enables us to conclude that |X| < |P(X)|.

If you disagree with me than please formally show how you can conclude that |X| < |P(X)|, without considering at least one case of "no function at all" from all the elements of X to some element of P(X).
 
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jsfisher said:
Nowhere in your definition is there such a requirement. Either your definition is incomplete and still needs more work to cover all the necessary criteria, or |A| = |B| in doronetics.
Click to expand...

EDIT:

Ok:

Given two sets, A and B, the doron-cardinality of the set A = the doron-cardinality of set B if and only if there is doron-function from A to B such that (for each a in A there is exactly one b in B) AND (for each b in B there is exactly one a in A).
 
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doronshadmi said:
In other words, we consider at least one case of "no function at all"....
Click to expand...

No. The proof you referenced starts by assuming there is an appropriate function, that at least such one case exists. The assumption leads to a contradiction, thereby proving there is no case in which such a function exists. No case in which a function exists.

One case of no function at all means something quite different and certainly not the thought you were trying to express in the post I flagged as gibberish.
 
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This is getting ridiculous, doron. Just stop digging, you're obviously way in over your head.

ETA: I know, I know... It's been ridiculous since page one, but still. New levels of fail are being achieved with every post.
 
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doronshadmi said:
Ok:

Given two sets, A and B, the doron-cardinality of the set A = the doron-cardinality of set B if and only if there is doron-function from A to B such that the term 'for each a in A there is exactly one b in B' is satisfied.
Click to expand...

"Is satisfied"? What do you think that adds to your definition. "The term...is satisfied" can only mean the quoted clause must be true, in which case the added text is completely extraneous. Your change does not alter the meaning.
 
jsfisher said:
No. The proof you referenced starts by assuming there is an appropriate function, that at least such one case exists. The assumption leads to a contradiction, thereby proving there is no case in which such a function exists. No case in which a function exists.

One case of no function at all means something quite different and certainly not the thought you were trying to express in the post I flagged as gibberish.
Click to expand...

EDIT:

At least one case of no function at all (from X to P(X)) is the same as No case in which a function exists (from X to P(X)).

In both cases we provide some P(X) element that is not mapped with any element of X.
 
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So according to jsfisher's remarks, we currently have:

Given two sets, A and B, the doron-cardinality of the set A = the doron-cardinality of set B if and only if there is doron-function from A to B such that (for each a in A there is exactly one b in B) AND (for each b in B there is exactly one a in A).

For example:

1 ↔ 1
2 ↔ 2
3 ↔ 3
4 ↔ 4
5 ↔ 5
…



OR

Given two sets, A and B, the doron-cardinality of the set A > the doron-cardinality of set B if and only if there is doron-function from A to B such that for each a in A there is at least one a in A that does not have any b in B.

For example:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…



OR

Given two sets, A and B, the doron-cardinality of the set A < the doron-cardinality of set B if and only if there is doron-function from B to A such that for each b in B there is at least one b in B that does not have any a in A.

For example:

1 ← 1
← 2
2 ← 3
3 ← 4
4 ← 5
…


So, we do not need the "no function at all".
 
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doronshadmi said:
http://www.internationalskeptics.com/forums/showpost.php?p=9754385&postcount=3093 was edited during your reply, please look at it again.
Click to expand...

I see. It is an improvement, but your definition is getting increasingly complicated and convoluted.

The conjunction between the two statements that make up the definition is and, not OR. Please stop using the wrong word.

Now, before we tear apart the second statement of the definition, how about you go back and read my original objection to the two statements. I provided counterexamples for both statements.

You asked that I reread your most recent post on this subject. I ask that you reread my original one.
 
@jsfisher Fun! He declares that 'there exists one with <negated condition>' is the same as 'there are none with <condition>'.

@Doron, your logic is faulty my friend.
 
jsfisher said:
The conjunction between the two statements that make up the definition is and, not OR. Please stop using the wrong word.
Click to expand...

EDIT:

It is = OR > OR <.

Please look at http://www.internationalskeptics.com/forums/showpost.php?p=9754527&postcount=3099 to see how they work among sets with unbounded amount of members.

The generalization of the three cases in http://www.internationalskeptics.com/forums/showpost.php?p=9754527&postcount=3099 is:

X AND Y are placeholders for domain OR codomain , such that (if X=domain then Y=codomain) OR (if X=codomain then Y=domain).

x in X has at most one y in Y.
 
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jsfisher said:
I see. It is an improvement, but your definition is getting increasingly complicated and convoluted.

The conjunction between the two statements that make up the definition is and, not OR. Please stop using the wrong word.

Now, before we tear apart the second statement of the definition, how about you go back and read my original objection to the two statements. I provided counterexamples for both statements.

You asked that I reread your most recent post on this subject. I ask that you reread my original one.
Click to expand...
Please provide the link to your original one.
 
realpaladin said:
This means, and all of this is only from your definitions, that the set before the operator does not need to be the same as after.
Click to expand...
I agree with you, the definitions in http://www.internationalskeptics.com/forums/showpost.php?p=9754527&postcount=3099 hold also between different sets.

realpaladin said:
Therefore, if we name the sets the same and use the inequality operator, we *must* have different sets before and after.
Click to expand...
I disagree with you.


For example, this case:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…

is

|N| > |N|


and this case:

1 ← 1
← 2
2 ← 3
3 ← 4
4 ← 5
…

is

|N| < |N|

exactly because both names (of rooms and visitors in Hilbert's Hotel http://www.internationalskeptics.com/forums/showpost.php?p=9742045&postcount=3022) are actually one and only one set, which is the set of all natural number (which is notated as N).
 
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jsfisher said:
I see. It is an improvement, but your definition is getting increasingly complicated and convoluted.
Click to expand...

EDIT:

I disagree with you, because
Given two sets, A and B, the doron-cardinality of the set A = the doron-cardinality of set B if and only if there is doron-function from A to B such that (for each a in A there is exactly one b in B) AND (for each b in B there is exactly one a in A).
Click to expand...
is actually a bijection, as follows:

Every element of one set is paired with exactly one element of the other set, and every element of the other set is paired with exactly one element of the first set.
Click to expand...
and also
A bijection from the set X to the set Y has an inverse function from Y to X
Click to expand...

(http://en.wikipedia.org/wiki/Bijection)
 
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jsfisher said:
No. The definition of doron-cardinality is statement #1 AND statement #2.
Click to expand...
Wrong. The definition of doron-cardinality is statement #1 (A=B) OR statement #2 (A>B) OR statement #3 (A<B) as follows:

Given two sets, A and B, the doron-cardinality of the set A = the doron-cardinality of set B if and only if there is doron-function from A to B such that (for each a in A there is exactly one b in B) AND (for each b in B there is exactly one a in A).

For example:

1 ↔ 1
2 ↔ 2
3 ↔ 3
4 ↔ 4
5 ↔ 5
…



OR

Given two sets, A and B, the doron-cardinality of the set A > the doron-cardinality of set B if and only if there is doron-function from A to B such that (for each a in A there is exactly one b in B) AND (also there is at least one a in A that does not have any b in B).

For example:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…



OR

Given two sets, A and B, the doron-cardinality of the set A < the doron-cardinality of set B if and only if there is doron-function from B to A such that (for each b in B there is exactly one a in A) AND (also there is at least one b in B that does not have any a in A).

For example:

1 ← 1
← 2
2 ← 3
3 ← 4
4 ← 5
…
 
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doronshadmi said:
Wrong. The definition of doron-cardinality is statement #1 (A=B) OR statement #2 (A>B) OR statement #3 (A<B) as follows: [...]
Click to expand...
Suppose there are three balls--call them A, B and C--and I need to know the color of each ball. Now suppose you tell me, "Ball A is red OR ball B is green OR ball C is black." Does this give me the information I need?



No. It does not. I do not know the color of even one ball. It could be that A is black, B is black and C is black. It could be that A is red, B is orange and C is purple. However, if you tell me, "A is red, AND B is green, AND C is black," then I do have all the information I need.



Going back to math, we want to know the definitions of each of these: |A| = |B|, |A| < |B|, |A| > |B|. If your three definitions are connected by 'or', that is consistent with any two of your given definitions being false. If you want to assert that all three given definitions are correct, then 'and' is the word you are looking for.
 
So, apparently you want this:

doronshadmi said:
Statement #1: Given two sets, A and B, the doron-cardinality of the set A = the doron-cardinality of set B if and only if there is doron-function from A to B such that (for each a in A there is exactly one b in B) AND (for each b in B there is exactly one a in A).

OR AND

Statement #2: Given two sets, A and B, the doron-cardinality of the set A > the doron-cardinality of set B if and only if there is doron-function from A to B such that (for each a in A there is exactly one b in B) AND (also there is at least one a in A that does not have any b in B).

OR AND

Statement #3: Given two sets, A and B, the doron-cardinality of the set A < the doron-cardinality of set B if and only if there is doron-function from B to A such that (for each b in B there is exactly one a in A) AND (also there is at least one b in B that does not have any a in A).
Click to expand...

Well, first off, Statement #3 is extraneous. Since A < B is equivalent to B > A, Statements #2 and #3 have no difference in meaning.

As for Statement #2, the criteria, "for each a in A there is exactly one b in B" and "there is a least one a in A that does not have any b in B", are conflicting requirements. (They also lack appropriate mention they are with respect to a doron-function mapping, but that is less important than them being contradictory.)

Since the right-hand side of the if and only if expression in Statement #2 is universally false, the left-hand side must be as well, so for no sets A and B is the doron-cardinality such that |A| < |B| (nor |A| > |B|).
 
Another correction of (A>B) OR (A<B) cases (A=B is not changed):

Given two sets, A and B, the doron-cardinality of the set A = the doron-cardinality of set B if and only if there is doron-function from A to B such that (for each a in A there is exactly one b in B) AND (for each b in B there is exactly one a in A).

For example:

1 ↔ 1
2 ↔ 2
3 ↔ 3
4 ↔ 4
5 ↔ 5
…



OR

Given two sets, A and B, the doron-cardinality of the set A > the doron-cardinality of set B if and only if there is doron-function from A to B such that (for each a in A there is exactly one b in B) AND (for each b in B there is exactly one a in A) AND (there is at least one a in A that does not have any b in B).

For example:

1 ↔ 1
2 →
3 ↔ 2
4 ↔ 3
5 ↔ 4
…



OR

Given two sets, A and B, the doron-cardinality of the set A < the doron-cardinality of set B if and only if there is doron-function from A to B such that (for each a in A there is exactly one b in B) AND (for each b in B there is exactly one a in A) AND (there is at least one b in B that does not have any a in A).

For example:

1 ↔ 1
← 2
2 ↔ 3
3 ↔ 4
4 ↔ 5
…
 
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doronshadmi said:
...OR...
Click to expand...

Stop right there. It has been explained to you by more than one person why the conjunction is AND.

In the complete definition of your doron-cardinality, statement #1 establishes whether the cardinalities of two sets are equal or not equal. Statement #2 establishes whether one is greater than or not greater than the other. Both are necessary.

This is not like a word you look up in a dictionary that can have several meanings (in which case OR is the proper conjunction). For doron-cardinality, it is just one meaning, and it is established by two statements, Statement #1 AND Statement #2.

Please fix that, and we can continue.
 
BenjaminTR said:
Suppose there are three balls--call them A, B and C--and I need to know the color of each ball. Now suppose you tell me, "Ball A is red OR ball B is green OR ball C is black." Does this give me the information I need?
Click to expand...
EDIT:

Suppose there are three options of the expressions =,>,< between A and B.

Now suppose you tell me, "|A|=|B| AND |A|>|B| AND |A|<|B|". Does this give me the information I need?

(Please use the corrected version http://www.internationalskeptics.com/forums/showpost.php?p=9756363&postcount=3116).
 
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doronshadmi said:
Suppose there are three options of comparisons of the forms =,>,< between A and B.

Now suppose you tell me, "|A|=|B| AND |A|>|B| AND |A|<|B|". Does this give me the information I need?
Click to expand...
No, it does not. However, in the posts in question you are not disjoining three atomic statements such as '|A|=|B|', you are disjoining three biconditionals. This is not what you want to do. You are basically saying, "Either [definition 1] is the definition of equality, or [definition 2] is the definition of greater than, or [definition 3] is the definition of less than." If we take your use of 'or' seriously, this does not establish the definition of any of the three.
 
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