Deeper than primes - Continuation

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doronshadmi said:
In case of sets with bounded amount of elements, the answer is trivial, because we can count the elements of both sets, and if they satisfy Statement #1 we know that indeed A = B.
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You might want to reconsider that answer. Statement #1 is a biconditional, an if-and-only-if construct. <left part> if and only if <right part>.

The entire statement being true doesn't make the left part true.


(By the way, it seems obvious, but for the sake of completeness, you meant |A| = |B| where you wrote A = B.)
 
jsfisher said:
You might want to reconsider that answer. Statement #1 is a biconditional, an if-and-only-if construct. <left part> if and only if <right part>.

The entire statement being true doesn't make the left part true.


(By the way, it seems obvious, but for the sake of completeness, you meant |A| = |B| where you wrote A = B.)
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Thank you for the remark about A = B.

OK, if will be used instead of if-and-only-if, as follows:

|A|^=|B| is not clear since it is equivalent to |A|>|B| OR |A|<|B|

|A|^>|B| is not clear since it is equivalent to |A|=|B| OR |A|<|B|

|A|^<|B| is not clear since it is equivalent to |A|=|B| OR |A|>|B|


So, there are only 3 clear options: |A|=|B| OR |A|>|B| OR |A|<|B|


Also there are only 3 definitions for each case above, as follows:


The definition for option |A|=|B| is:

Given two sets, A and B, the doron-cardinality of the set A = the doron-cardinality of set B 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).


The definition for option |A|>|B| is:

Given two sets, A and B, the doron-cardinality of the set A > the doron-cardinality of set B if there is doron-function from A to B such that (for each a in A there is at most 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).


The definition for option |A|<|B| is:

Given two sets, A and B, the doron-cardinality of the set A < the doron-cardinality of set B 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 at most one a in A) AND (there is at least one b in B that does not have any a in A).

I can clearly conclude about the relationship between the doron-cardinalities of sets A and B, only if they satisfy one and only one of the 3 definitions above.
 
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Another correction which uses if...then.

|A|^=|B| is not clear since it is equivalent to |A|>|B| OR |A|<|B|

|A|^>|B| is not clear since it is equivalent to |A|=|B| OR |A|<|B|

|A|^<|B| is not clear since it is equivalent to |A|=|B| OR |A|>|B|


So, there are only 3 clear options: |A|=|B| OR |A|>|B| OR |A|<|B|


Also there are only 3 definitions for each case above, as follows:


The definition for option |A|=|B| is:

Given two sets, A and B, 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) then the doron-cardinality of the set A = the doron-cardinality of set B.


The definition for option |A|>|B| is:

Given two sets, A and B, if there is doron-function from A to B such that (for each a in A there is at most 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) then the doron-cardinality of the set A > the doron-cardinality of set B.


The definition for option |A|<|B| is:

Given two sets, A and B, 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 at most one a in A) AND (there is at least one b in B that does not have any a in A) then the doron-cardinality of the set A < the doron-cardinality of set B.

I can clearly conclude about the relationship between the doron-cardinalities of sets A and B, only if they satisfy one and only one of the 3 definitions above.
 
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doronshadmi said:
OK, if will be used instead of if-and-only-if
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So, you want to change everything you've provided up until now? Does it not give at least a little pause that all that you believed in up until an hour ago now gets a full make-over? And it is not like this happened only once and only a little tweak was required. Your revisions often strike at the fundamentals.

Nevertheless....

doronshadmi said:
Another correction which uses if...then.
Click to expand...

Of course there's another correction...and another...and another, because you just make it up as you go along, trying to protect your serious misunderstanding of all things Mathematics.

|A|^=|B| is not clear since it is equivalent to |A|>|B| OR |A|<|B|
Click to expand...

What does this mean?

...
So, there are only 3 clear options: |A|=|B| OR |A|>|B| OR |A|<|B|
Click to expand...

More, actually, but let's leave that until later.

Also there are only 3 definitions for each case above, as follows
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No, it is one definition with three conditions.

...
The definition for option |A|>|B| is:

Given two sets, A and B, if there is doron-function from A to B such that (for each a in A there is at most 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) then the doron-cardinality of the set A > the doron-cardinality of set B.
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Are you ever going to fix this? There is no doron-function possible for any A and B that satisfies those conditions. None. It was wrong before you inserted a third AND'ed condition; it remains equally wrong after the insertion.

The definition for option |A|<|B| is...
Click to expand...

And this remains completely unnecessary. If |A| < |B|, then |B| > |A|, and that can be established using the previous condition.

...
I can clearly conclude about the relationship between the doron-cardinalities of sets A and B, only if they satisfy one and only one of the 3 definitions above.
Click to expand...

No, all three must be satisfied. You really don't understand conditionals at all, do you?
 
jsfisher said:
Are you ever going to fix this? There is no doron-function possible for any A and B that satisfies those conditions. None. It was wrong before you inserted a third AND'ed condition; it remains equally wrong after the insertion.
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There is such a function, as follows:

Given two sets, A and B, if there is doron-function from A to B such that (for each a in A there is at most 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) then the doron-cardinality of the set A > the doron-cardinality of set B.

Here is such function from A to B among N members:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…

If you have something to say about it, then please do it in details according what is written in this definition.
 
jsfisher said:
So, you want to change everything you've provided up until now? Does it not give at least a little pause that all that you believed in up until an hour ago now gets a full make-over? And it is not like this happened only once and only a little tweak was required. Your revisions often strike at the fundamentals.

Nevertheless....



Of course there's another correction...and another...and another, because you just make it up as you go along, trying to protect your serious misunderstanding of all things Mathematics.
Click to expand...
jsfisher, non-traditional work does not have enough background to be based on, so expressions are redefined more than once with a lot of errors during the work.

I redefine things again and again in order to formally address, for example, the following case:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…

and, at least for me, it is not an easy task, but the criticism of you and BenjaminTR helps me a lot, so thank you for your criticism.

jsfisher said:
What does this mean?
Click to expand...

I mean that first one has to choose one of two options (|A|>|B| OR |A|<|B|) in case of |A|^=|B|, in order to deal with a given relation among |A| and |B|.

jsfisher said:
No, it is one definition with three conditions.
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I agree, but these three conditions are used for 3 different cases which are |A|=|B| OR |A|>|B| OR |A|<|B| (where |A|>|B| is not the same as |A|<|B|, unless only equivalency or non-equivalency is considered).

jsfisher said:
No, all three must be satisfied.
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I agree, but it is done for each given case, exactly as the expression |A| ≥ |B| is used in terms of |A|=|B| OR |A|>|B|.
 
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...delete for now...

Missed a classic Doron-edit.
 
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doronshadmi said:
jsfisher, non-traditional work does not have enough background to be based on, so expressions are redefined more than once with a lot of errors during the work.
Click to expand...

Your non-traditional work has no background, just wishful thinking. You make it up as you go along, and only when finally faced with the crash into a brick wall do you redefine things, yet again.

You start with your conclusion than work to establish it. That's why your stuff continues to be broken and requires constant repair.

...
I mean that first one has to choose one of two options (|A|>|B| OR |A|<|B|) in case of |A|^=|B|, in order to deal with a given relation among |A| and |B|.
Click to expand...

So you need to know the answer before you can determine the answer?

I agree, but these three conditions are used for 3 different cases which are |A|=|B| OR |A|>|B| OR |A|<|B| (where |A|>|B| is not the same as |A|<|B|, unless only equivalency or non-equivalency is considered).
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You are assuming without any foundation that the Law of Trichotomy applies to doron-cardinality, aren't you?

I agree, but it is done for each given case, exactly as the expression |A| ≥ |B| is used in terms of |A|=|B| OR |A|>|B|.
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Nope. When |A|>|B| can be established by applying your second statement (i.e. If <condition> then |A| > |B|), your first statement will still be true.

Conditionals, simple stuff. The conditional can be true even when the consequent is false. Please stop making the same mistake again and again.
 
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jsfisher said:
Your non-traditional work has no background, just wishful thinking. You make it up as you go along, and only when finally faced with the crash into a brick wall do you redefine things, yet again.

<SNIP>

Conditionals, simple stuff. The conditional can be true even when the consequent is false. Please stop making the same mistake again and again.
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^^^ This, Doron, this! ^^^
 
doronshadmi said:
<snip>

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…

<snip>
Click to expand...

Now, let's look at the ellipsis here (the ...).

In traditional mathematics, we would write:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…
n → n-1

Which enables us to see, even if n = ∞, that the lefthand side is never equal to the righthand side.

This follows from the fact that there is no case (at least not defined) where:
→ m such that at some point n → n

This in turn enables us to conclude that whatever doron-cardinality means, it can never hold equality.
 
jsfisher said:
...delete for now...

Missed a classic Doron-edit.
Click to expand...

The silent alteration was to change "exactly one" to "at most one" in a rather strategic place.

Here's where we now stand:

Doron-cardinality, symbolized as |A| for the doron-cardinality of set A, is a relative measure of set "size" between sets defined by the following three conditionals for any two sets, A and B:
  1. 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 then the |A| = |B|.
  2. If there is doron-function from A to B such that for each a in A there is at most 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 then |A| > |B|.
  3. 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 at most one a in A AND there is at least one b in B that does not have any a in A then |A| < |B|.
Before progressing, let's at least compare clarity of what doronetics has to offer versus what is found in accepted Mathematics:
Cardinality, symbolized as |A| for the cardinality of set A, is a relative measure of set "size" between sets defined as follows: Given any two sets, A and B, |A| ≤ |B| if and only if there exists a one-to-one mapping from A to B.​
Additionally, all the necessary proofs exist to establish the remaining comparative relations from the lone less-than-or-equal relation.
 
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jsfisher said:
Doron-cardinality, symbolized as |A| for the doron-cardinality of set A, is a relative measure of set "size" between sets defined by the following three conditionals for any two sets, A and B:
  1. 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 then the |A| = |B|.
  2. If there is doron-function from A to B such that for each a in A there is at most 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 then |A| > |B|.
  3. 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 at most one a in A AND there is at least one b in B that does not have any a in A then |A| < |B|.
Click to expand...

I still fail to see why Doron could not have modelled this in AGDA (heck, Prolog would suffice. Should be only a few lines of code). It looks like TurboPascal to me....
 
realpaladin said:
I still fail to see why Doron could not have modelled this in AGDA (heck, Prolog would suffice. Should be only a few lines of code). It looks like TurboPascal to me....
Click to expand...

Well, for one, Doron wasn't able to express himself in English (and, I will assume, Hebrew as well). It took a lot of prodding and shoving just to get what he now has, and it is still a convoluted mess.
 
jsfisher said:
Well, for one, Doron wasn't able to express himself in English (and, I will assume, Hebrew as well). It took a lot of prodding and shoving just to get what he now has, and it is still a convoluted mess.
Click to expand...
TBH I think you are doing an excellent job in getting this far with regards to clarity of what he thinks he is proposing.
 
jsfisher said:
The silent alteration was to change "exactly one" to "at most one" in a rather strategic place.

Here's where we now stand:

Doron-cardinality, symbolized as |A| for the doron-cardinality of set A, is a relative measure of set "size" between sets defined by the following three conditionals for any two sets, A and B:
  1. 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 then the |A| = |B|.
  2. If there is doron-function from A to B such that for each a in A there is at most 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 then |A| > |B|.
  3. 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 at most one a in A AND there is at least one b in B that does not have any a in A then |A| < |B|.
Before progressing, let's at least compare clarity of what doronetics has to offer versus what is found in accepted Mathematics:
Cardinality, symbolized as |A| for the cardinality of set A, is a relative measure of set "size" between sets defined as follows: Given any two sets, A and B, |A| ≤ |B| if and only if there exists a one-to-one mapping from A to B.​
Additionally, all the necessary proofs exist to establish the remaining comparative relations from the lone less-than-or-equal relation.
Click to expand...
By using cardinality as found in accepted Mathematics, please define the following case:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…
 
jsfisher said:
So you need to know the answer before you can determine the answer?
Click to expand...
No, the answer is X OR Y, and not both of them.


jsfisher said:
You are assuming without any foundation that the Law of Trichotomy applies to doron-cardinality, aren't you?
Click to expand...

Again if only equivalency or non-equivalency is considered, then X=Y OR X>Y is enough.


jsfisher said:
Nope. When |A|>|B| can be established by applying your second statement (i.e. If <condition> then |A| > |B|), your first statement will still be true.
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In case that the <condition> is discontented (there is no Cause and Effect) from the <result>.

In informal language if ... then is understood in terms of Cause and Effect.

In formal language if ... then a given logical result depends on the used truth values.
 
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doronshadmi said:
By using cardinality as found in accepted Mathematics, please define the following case:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…
Click to expand...

Since that is not an injection from one set (presumably N) to another set (also N), the case is not in any way directly applicable in determining a comparative relationship with respect to cardinality.

However, if I interpret your "2 →" gibberish to mean 2 is not in the domain of the partial mapping from N to N and then consider the inverse mapping, the inverse mapping is an injection and therefore I can conclude from this case alone that |N| ≤ |N|.

Moreover, relying on the Cantor–Bernstein–Schroeder theorem, I can then conclude |N| = |N|.

Therefore, |N| = |N|.

Any questions?
 
doronshadmi said:
No, the answer is X OR Y, and not both of them.
Click to expand...

I'll save this for another post.

Again if only equivalency or non-equivalency is considered, then X=Y OR X>Y is enough.
Click to expand...

What has this to do with assuming the Law of Trichotomy applies? Proof is needed, not bare assertion.

In case that the <condition> is discontented (there is no Cause and Effect) from the <result>.
Click to expand...

Seriously? There's nothing worse than a disgruntled conditional (except maybe one that is gruntled).

In informal language if ... then is understood in terms of Cause and Effect.
Click to expand...

"IF you go outside THEN you will catch a cold." -- Can you not still catch a cold even though you stayed inside?

In formal language if ... then a given logical result depends on the used truth values.
Click to expand...

Informal language may be looser with some of its terminology, but the basic rules of reasoning are not abandoned, as the cold-catching example shows.
 
jsfisher said:
Since that is not an injection from one set (presumably N) to another set (also N), the case is not in any way directly applicable in determining a comparative relationship with respect to cardinality.

However, if I interpret your "2 →" gibberish to mean 2 is not in the domain of the partial mapping from N to N and then consider the inverse mapping, the inverse mapping is an injection and therefore I can conclude from this case alone that |N| ≤ |N|.

Moreover, relying on the Cantor–Bernstein–Schroeder theorem, I can then conclude |N| = |N|.

Therefore, |N| = |N|.

Any questions?
Click to expand...

"2 →" is gibberish by accepted Mathematics exactly because
Cardinality, symbolized as |A| for the cardinality of set A, is a relative measure of set "size" between sets defined as follows: Given any two sets, A and B, |A| ≤ |B| if and only if there exists a one-to-one mapping from A to B.
Click to expand...

can't address

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…

On the contrary, the non-traditional definition
Given two sets, A and B, if there is function from A to B such that (for each a in A there is at most 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) then |A| > |B|
Click to expand...

where in the case of Hilbert's Hotel

|N| > |N|

exactly because the names of the rooms and the names of the visitors are actually one and only one thing, which is the set of all natural numbers (notated as N).

Any questions?
 
EDIT:

jsfisher said:
What has this to do with assuming the Law of Trichotomy applies?
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Since you claim that only |A|=|B| or |A|>|B| is enough, then please show how only '>' is used in the case of A={1,3}, B={7,6,9} (where |A|<|B|), without replacing between |A| and |B| in the expression |A|<|B|.



jsfisher said:
"IF you go outside THEN you will catch a cold." -- Can you not still catch a cold even though you stayed inside?
Click to expand...
"IF you stay inside THEN you will catch a cold." (also cause and effect)

"IF you stay inside or go outside THEN you will catch a cold." (also cause and effect)
 
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doronshadmi said:
"2 →" is gibberish by accepted Mathematics exactly because
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...because you made it up to cover up a lapse in your understanding of Mathematics.

...
[The definition of cardinality] can't address

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…
Click to expand...

It addresses it just fine. It is a partial function N -> N in which 2 is not in the domain of the function. That really is easily grasped. As such, it doesn't provide any direct insight into a cardinality relationship, and the definition I proved states that.

Bananas are yellow. How does your definition for doron-cardinality address that case?

On the contrary, the non-traditional definition

where in the case of Hilbert's Hotel

|N| > |N|
Click to expand...

No where in the Hilbert Hotel presentation is it ever demonstrated, alleged, or even hinted that |N| > |N|. That is a figment of doronetics, not established Mathematics. Mathematics concludes |N| = |N|. Doronetics concludes contradictory relationships.

You really should stop trying to discredit Mathematics by pointing out how contradictory doronetics is.
 
doronshadmi said:
Since you claim that only |A|=|B| or |A|>|B| is enough, then please show how only '>' is used in the case of A={1,3}, B={7,6,9} (where |A|<|B|), without replacing between |A| and |B| in the expression |A|<|B|.
Click to expand...

The part I struck out is an artificial constraint. A and B are free variables in the definition. Of course the bindings can be swapped around.

"IF you stay inside THEN you will catch a cold." (also cause and effect)

"IF you stay inside or go outside THEN you will catch a cold." (also cause and effect)
Click to expand...

Both of which missed my point. Whether the if-then construct is used to express a causal relationship or a logical proposition, if the antecedent is not satisfied (the cause is absent or the truth value is false, as you prefer), then there is no constraint on the consequent.
 
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Well, Doron already jumped to the end.

He started out this little arc trying to point out inconsistencies in cardinality for infinite sets. In his own unique non-communicative way, he alleged that |N| was both equal to and greater than itself, and he continually muttered "Hilbert's Hotel" as if that was necessary and sufficient to make his point.

Upon pressing him as to the meaning of his gibberish, it became clear Doron needed to alter the meaning of 'function', 'domain', 'nothing', and 'cardinality' to cover his misunderstanding.

Now that he is done, we are left with the normal meaning for cardinality, function, domain, and nothing, still producing consistent, familiar results, while doronetics with all the arbitrarily shifted meanings does not fare so well.

By the way, extra points to all who noticed that doron-cardinality cannot establish when the doron-cardinality of two sets are not equal. (It was a self-inflicted wound, too. Damn those biconditionals.) Extra-extra points to anyone who can explain why if the Law of Trichotomy does hold, only Doron's second conditional and no other is needed.
 
I may as well give Doron one more thing to wake up to.

Consider just this one conditional:
If there is a doron-function from A to B such that for each a in A there is at most 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 then |A| > |B|.​
The antecedent has three requirements for a doron-function from A to B:
  1. for each a in A there is at most one b in B
  2. for each b in B there is exactly one a in A
  3. there is at least one a in A that does not have any b in B
The first of the three is redundant with the basic definition of a doron-function. For any doron-function from X to Y, every element of X corresponds to either zero or one element of Y. So, we can discard the explicit statement of this requirement.

The second requirement states that every element in the range (aka codomain) is the image of exactly one element in the doron-function's domain. That's a slightly stronger condition than "onto"; more importantly, it means the inverse of the doron-function is an injection, a one-to-one function (and a real function, too, not a doron-function).

We can restate the conditional, then, thusly:
If there is a function from B to A such that the function is an injection and there is at least one a in A that is not the image of any element of B then |A| > |B|.​
Or, more simply:
If there is an injection from B to A that is not surjective then |A| > |B|.​

A couple things to note: First, a redefinition of what was meant by function was totally unnecessary. Doron tends to work through things backwards, and the redefinition was a outgrowth of that, but it was unnecessary (as was mentioned a few times).

Second, this version of the conditional is very similar to the real definition. It is a straight conditional instead of a biconditional, and it prohibits surjection, but the former came about because of Doron's misconception of biconditionals (I'd steered him towards framing his definitions in that form, but he later rebelled), and the latter is a more a matter of Doron trying to force <, =, and > as explicit results rather than produce a consistent result. With surjection permitted, the result becomes |A| >= |B|.

It is the surjection prohibition that makes doron-cardinality misbehave for non-finite sets.
 
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jsfisher said:
The part I struck out is an artificial constraint. A and B are free variables in the definition.
Click to expand...
Or in other words, only equality or non-equality are considered in this case.


jsfisher said:
Both of which missed my point. Whether the if-then construct is used to express a causal relationship or a logical proposition, if the antecedent is not satisfied (the cause is absent or the truth value is false, as you prefer), then there is no constraint on the consequent.
Click to expand...

"If water is liquid then your name is jsfisher".

This kind of statement is not commonly used by informal language.
 
doronshadmi said:
Or in other words, only equality or non-equality are considered in this case.

"If water is liquid then your name is jsfisher".

This kind of statement is not commonly used by informal language.
Click to expand...

Informal language consists of grunts and growls.

We grown-ups use normal language.

'Formal language' is another doronetic attempt to segue the discussion away from the obvious crushing defeat.
 
jsfisher said:
With surjection permitted, the result becomes |A| >= |B|.

It is the surjection prohibition that makes doron-cardinality misbehave for non-finite sets.
Click to expand...
The surjection prohibition is exactly a given function with an input that does not have any output (2 → in the example below) exactly as seen in

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…

A function without any output is not defined by accepted mathematics.

As a result |N| > |N| can't be deduced by accepted mathematics.
 
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doronshadmi said:
The surjection prohibition is exactly a given function with an input that does not have any output (2 -> in the example below) exactly as seen in

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

A function without any output is not defined by accepted mathematics.

As a result |N| > |N| can't be deduced by accepted mathematics.
Click to expand...

I demonstrated how, if you would use the "notation" cooked up by Doron, accepted mathematics would say that you have redefined the > operator since you can not hold that the left hand side is equal to the right hand side and therefore both sets are different sets.

I will gladly copy paste a shortened version with a link to where I first explained it to you under each and every Doron post that uses that notation until you address this.
 
realpaladin said:
I demonstrated how, if you would use the "notation" cooked up by Doron, accepted mathematics would say that you have redefined the > operator since you can not hold that the left hand side is equal to the right hand side and therefore both sets are different sets.
Click to expand...

1 → 1
2 →
3 → 2
4 → 3
5 → 4
…

is one and only one set known as the set of all natural numbers.

The void (the "right hand" of 2 →) is the absence of member in any given set.

This is the reason of why function 2 → does not return any output.

By deduce such function by the following definition
Given two sets, A and B, if there is function from A to B such that (for each a in A there is at most 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) then |A| > |B|
Click to expand...

|N| > |N| is one of the possible results.
 
doronshadmi said:
1 → 1
2 →
3 → 2
4 → 3
5 → 4
…
is one and only one set known as the set of all natural numbers.
Click to expand...

I finally figured out why *you* think what you say is logically coherent. It is a beginners mistake.

Let's start with the handwaving after the ellipsis. Since you never to bother to complete it and write it down correctly, I will do this for you:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...
n → n-1

If you do not write it out correctly one might surmise all manner of unicorn and pixie magic to happen after the ellipsis.

This should give you a clue already; the right side has one 'extra' possibility which displaces the rest of the natural numbers by one.

The right side has the 'no result' as a member of the set and is therefore *not* the set of natural numbers.
It is in fact the set of natural numbers + 'no result'.

Therefore, as is shown in simple baby steps of basic logic, the left and right are different sets.

So either you need to face the fact that only the left side is |N| and the right side is "|N| + 'no result'" or you must agree that the > operator in your |N|>|N| statement actually means |N|=(|N|-'no result') (or something horrible like that).

You simply claim that they are one and the same set, but you have never shown this. Just saying this is so, does not make it so.
 
doronshadmi said:
Given two sets, A and B, if there is function from A to B such that (for each a in A there is at most 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) then |A| > |B|
Click to expand...

This might look great to you, but let's rewrite it as you use it:

Given a set, A, if there is function from A to A such that (for each a in A there is at most one a in A) AND (for each a in A there is exactly one a in A) AND (there is at least one a in A that does not have any a in A) then |A| > |A|

This can only happen if the > operator adds or subtracts elements during the comparison (without getting the logic circuit meltdown that is before the 'then')

@JSFisher: I think this is what you meant, btw. since the last two terms are mutually exclusive.
 
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realpaladin said:
This might look great to you, but let's rewrite it as you use it:

Given a set, A, if there is function from A to A such that (for each a in A there is at most one a in A) AND (for each a in A there is exactly one a in A) AND (there is at least one a in A that does not have any a in A) then |A| > |A|

This can only happen if the > operator adds or subtracts elements during the comparison (without getting the logic circuit meltdown that is before the 'then')

@JSFisher: I think this is what you meant, btw. since the last two terms are mutually exclusive.
Click to expand...
realpaladin, this is the beautiful thing about Hilbert's Hotel.

No visitor is added or subtracted form that hotel, and yet there are more rooms than visitors only by moving all the visitors one room forward, for example:

1 →
2 → 1
3 → 2
4 → 3
5 → 4
…

Again the names of the elements of this hotel are members of one and only one set, which is the set of all natural numbers.

EDIT:

Accepted mathematics avoids the result above by using injection and surjection as follows:
Code: 
1 -.
    \   
2 -- 1

3 -- 2

4 -- 3

5 -- 4

…
 
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doronshadmi said:
realpaladin, this is the beautiful thing about Hilbert's Hotel.

No visitor is added or subtracted form that hotel, and yet there are more rooms than visitors only by moving all the visitors one room forward, for example:

1 →
2 → 1
3 → 2
4 → 3
5 → 4
…

Again the names of the elements of this hotel are members of one and only one set, which is the set of all natural numbers.
Click to expand...

Yes Doron, I think I understand Hilbert's Hotel a lot better than you do.

But by adding the possibility of an empty room, you modify the set so it is no longer 'just' the set of natural numbers. You have added the 'no output'/'no result'/'under construction'/'cleaning lady' to it.

That is what makes your logic wrong.

EDIT: and from now on I will be copy/pasting the above (including your logic errors in your discourse with JSFisher) until you see the light.
 
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realpaladin said:
Yes Doron, I think I understand Hilbert's Hotel a lot better than you do.

But by adding the possibility of an empty room, you modify the set so it is no longer 'just' the set of natural numbers. You have added the 'no output'/'no result'/'under construction'/'cleaning lady' to it.

That is what makes your logic wrong.

EDIT: and from now on I will be copy/pasting the above (including your logic errors in your discourse with JSFisher) until you see the light.
Click to expand...
Once again you repeat on the mistake of using a function between the names of the rooms and the rooms , as follows:

1 → ()
2 → (1)
3 → (2)
4 → (3)
5 → (4)
…

instead of using a function between the names of the rooms and the names of the visitors, as follows:

1 →
2 → 1
3 → 2
4 → 3
5 → 4
…
 
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Now see what you made me do?

I have a .txt file on my desktop to copy/paste the following every time you keep on babbling nonsense like "using a function between..."

Doron's Misconceptions:
Handwaving consequences of his notation
Let's start with the handwaving after the ellipsis. Since Doron never bothered to do this, I did it for him:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...
n → n-1

If you do not write it out correctly one might surmise all manner of unicorn and pixie magic to happen after the ellipsis.

The right side has the 'no result' as a member of the set and is therefore *not* the set of natural numbers.
It is in fact the set of natural numbers + 'no result'.

Doron simply adds a member to the set, but does not count it. I would not want to do business with him...

Not seeing logic errors in his constructed constraints:
After a long and arduous back and forth with JSFisher Doron comes up with the following to show that you can have equality and non-equality at the same time:
Given two sets, A and B, if there is function from A to B such that (for each a in A there is at most 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) then |A| > |B|
Click to expand...

But this can be shown invalid by using a single set (Doron's set of all natural numbers, for example):

Given a set, A, if there is function from A to A such that (for each a in A there is at most one a in A) AND (for each a in A there is exactly one a in A) AND (there is at least one a in A that does not have any a in A) then |A| > |A|
Click to expand...
 
doronshadmi said:
Once again you repeat on the mistake of using a function between the names of the rooms and the rooms , as follows:

1 → ()
2 → (1)
3 → (2)
4 → (3)
5 → (4)
…

instead of using a function between the names of the rooms and the names of the visitors, as follows:

1 →
2 → 1
3 → 2
4 → 3
5 → 4
…
Click to expand...

And we need to preserve this little gem as well. Since he fails to see that what I keep telling him is applicable to both cases.
 
realpaladin said:
And we need to preserve this little gem as well. Since he fails to see that what I keep telling him is applicable to both cases.
Click to expand...
EDIT:


No, realpaladin fails to distinguish between 1 → () and 1 →

Let us use function visitor() (which is equivalent to →) :


realpaladin's case:

visitor(room 1) = ()


doron's case:

visitor(room 1) =
 
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Doron's Misconceptions:
Handwaving consequences of his notation
Let's start with the handwaving after the ellipsis. Since Doron never bothered to do this, I did it for him:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...
n → n-1

If you do not write it out correctly one might surmise all manner of unicorn and pixie magic to happen after the ellipsis.

The right side has the 'no result' as a member of the set and is therefore *not* the set of natural numbers.
It is in fact the set of natural numbers + 'no result'.

Doron simply adds a member to the set, but does not count it. I would not want to do business with him...

Not seeing logic errors in his constructed constraints:
After a long and arduous back and forth with JSFisher Doron comes up with the following to show that you can have equality and non-equality at the same time:
Given two sets, A and B, if there is function from A to B such that (for each a in A there is at most 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) then |A| > |B|
Click to expand...

But this can be shown invalid by using a single set (Doron's set of all natural numbers, for example):

Given a set, A, if there is function from A to A such that (for each a in A there is at most one a in A) AND (for each a in A there is exactly one a in A) AND (there is at least one a in A that does not have any a in A) then |A| > |A|
Click to expand...
 
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