Deeper than primes - Continuation

Status
Not open for further replies.
Actually Doron, I think you may have better luck in redefining set theory instead of trying to bend logic.

Maybe you can have different 'memberships', that should work better in what you are trying to do.

Something like:
'normal membership' -> has an element in the set AND counts in cardinality
'special membership' -> has an element in the set BUT does not count in cardinality

Because no matter how you try, what you say, the thing you keep on showing *is* an extra member in the set.
 
Last edited:
EDIT:

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|

and

1 ↔ 1
2 →
3 ↔ 2
4 ↔ 3
5 ↔ 4
...

is an example of such function.

The left side of the example above is defined by (for each a in A there is at most one a in A) which enables also functions without any output.

The right side of the example above is defined by (for each a in A there is exactly one a in A) which enables only functions with input AND output.

(there is at least one a in A that does not have any a in A) exactly because (for each a in A there is at most one a in A) enables also functions without any output.

A is one and only one set, and in this case, it is the set of all natural numbers.
 
Last edited:
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...
 
realpaladin said:
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:


But this can be shown invalid by using a single set (Doron's set of all natural numbers, for example):
Click to expand...
You are wrong, as shown in http://www.internationalskeptics.com/forums/showpost.php?p=9763587&postcount=3202.

Also why did you cut and paste the entire http://www.internationalskeptics.com/forums/showpost.php?p=9763538&postcount=3200, this is spam.
 
Last edited:
doronshadmi said:
Also why did you cut and paste the entire http://www.internationalskeptics.com/forums/showpost.php?p=9763538&postcount=3200, this is spam.
Click to expand...

This made me laugh out loud! Are you that afraid that anyone might notice your blatant logic errors?

But let's take your next bit of nonsense:

doronshadmi said:
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|

and

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

is an example of such function.

The left side of the example above is defined by (for each a in A there is at most one a in A) which enables also functions without any output.

The right side of the example above is defined by (for each a in A there is exactly one a in A) which enables only functions with input AND output.

(there is at least one a in A that does not have any a in A) exactly because (for each a in A there is at most one a in A) enables also functions without any output.

A is one and only one set, and in this case, it is the set of all natural numbers.
Click to expand...

Excellent! Now you finally come clean about your sneakily adding stuff.

Just in case you thought I forgot about your weak language skills: also means in addition to ( http://www.merriam-webster.com/dictionary/also )

So you now have proven that you add members but don't count them!
 
realpaladin said:
This made me laugh out loud! Are you that afraid that anyone might notice your blatant logic errors?

But let's take your next bit of nonsense:



Excellent! Now you finally come clean about your sneakily adding stuff.

Just in case you thought I forgot about your weak language skills: also means in addition to ( http://www.merriam-webster.com/dictionary/also )

So you now have proven that you add members but don't count them!
Click to expand...
The function 2 → did not add any member to the set of all natural numbers.

Thank for your remark about "also", we actually do not need it as follows:

The left side of the example above is defined by (for each a in A there is at most one a in A) which enables functions without any output.
 
Last edited:
doronshadmi said:
The function 2 → did not add any member to the set of all natural numbers.

Thank for your remark about "also", we actually do not need it as follows:

The left side of the example above is defined by (for each a in A there is at most one a in A) which enables functions without any output.
Click to expand...

Obtuse as always: nobody ever, in any post, said that.

2 →

Added a member to the right side. Namely, "no output".

Hence, the set of the left side (which might or might not be the set of natural numbers, since you fail to complete anything after the ellipsis) is a different set than the set on the right side.

Again. It does not matter what you say, it matters what you show. And you keep showing that you add members to the right side but refuse to count them because you think you can babble them away.

"No output", "empty", "non existence", "whatchamacallit"... no matter what you call it, it is a member of the right side.

And therefore the right side is not (but we could not know anyhow because you fail to complete after the ellipsis) the set of natural numbers.
 
doronshadmi said:
Thank for your remark about "also", we actually do not need it as follows:

The left side of the example above is defined by (for each a in A there is at most one a in A) which enables functions without any output.
Click to expand...

The also is implicit because of the word 'enables'. I pointed out the also because you finally showed your true colours without anywhere to hide.

Enabling something means it is disabled before and so not available. If you enable something you add the functionality.

Man, your language skills are almost on par with your logic.
 
realpaladin said:
Obtuse as always: nobody ever, in any post, said that.

2 →

Added a member to the right side. Namely, "no output".
Click to expand...
EDIT:

Dear realpaladin,

The term at most enables function whether it has an output (that can be counted) or it does not have any output (nothing can be counted).

In both cases no member was added to the set of all natural numbers, as can be seen in the following case:

1 ↔ 1
2 →
3 ↔ 2
4 ↔ 3
5 ↔ 4
...

Sincerely yours,

doronshami

b.t.w time is not involved here, so your "before\after" argument does not hold, in this case.
 
Last edited:
doron, regardless of your failings in logic, language and all things maths, you still don't get that even your retarded definitions for cardinality include there exists (doron) function such that ... - in other words, if there is a function that satisfies the conditions, relative cardinality is decided. Period. Just cooking up an example of a function where the conditions are not satisfied does not allow you to make any conclusions whatsoever regarding the relative cardinality of two sets. Just FYI.
 
doronshadmi said:
EDIT:

Dear realpaladin,

The term at most enables function whether it has an output (that can be counted) or it does not have any output (nothing can be counted).

In both cases no member was added to the set of all natural numbers, as can be seen in the following case:

1 ↔ 1
2 →
3 ↔ 2
4 ↔ 3
5 ↔ 4
...

Sincerely yours,

doronshami

b.t.w time is not involved here, so your "before\after" argument does not hold, in this case.
Click to expand...

Just to preserve this for posterity so you don't go editing again.

First, again... YOUR LANGUAGE SKILL SUCKS!!!!

Before and after are words that designate, on a line, the relative position with regards to a point.
Where the heck do you get the idea from that time is involved? That only happens if the designated point is a point in time. Classic Doron.

Second, just by rewriting your concoction to:

1 ↔ 1
2 →
3 ↔ 2
4 ↔ 3
5 ↔ 4

you show that in fact, we have three sets!

The set on the left (which may or may not be the natural numbers or the collection of toes Doron has on his left foot, since we never know as he can not complete his notation) then we have the set of switches that enable the 'function' and then on the right we have a set which might be the collection of numbers that Doron can count to without using his fingers.

This 'magical middle' set is what makes your proof so utterly worthless; it has no relation to either the set on the left nor to the set on the right. It is an arbitrary set of boolean switches which is not defined.

Therefore, if you insist on this notation, you can not conclude anything at all from what you write. Not |A| = |B|, not |A| > |A|, not anything at all.

Mind you, the *functions* may be defined, but the set of switches is not, so therefore no conclusion at all is possible.
 
Last edited:
Current status of Doron's Misconceptions (as of post 3209):



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...



Removing all relations with the operands:

Doron now writes his concoction as:

1 ↔ 1
2 →
3 ↔ 2
4 ↔ 3
5 ↔ 4

Where the arrows are arbitrarily enabled/disabled functions.

It shows that in fact, we have three sets!

The set on the left (which may or may not be the natural numbers or the collection of toes Doron has on his left foot, since we never know as he can not complete his notation) then we have the set of switches that enable the 'function' and then on the right we have a set which might be the collection of numbers that Doron can count to without using his fingers.

The 'magical middle' set is what makes any of Doron's proof so utterly worthless; it has no relation to either the set on the left nor to the set on the right. It is an arbitrary set of boolean switches which is not defined.

Because this 'magical middle' is not defined no conclusion can be made about any of Doron's claims. The functions might be defined, but the set that toggles them is not.
 
Last edited:
realpaladin said:
Current status of Doron's Misconceptions (as of post 3209):

...<snip>...
Click to expand...
EDIT:

Simply nonsense.

1 ↔ 1
2 →
3 ↔ 2
4 ↔ 3
5 ↔ 4

are two different finite sets |{1,2,3,4,5}| > |{1,2,3,4}| such that
Code: 
{2,1,3,4,5}
 ↓ ↕ ↕ ↕ ↕ 
  {1,2,3,4}
where

1 ↔ 1
2 →
3 ↔ 2
4 ↔ 3
5 ↔ 4
...

is the same set with unbounded amount of members such that

Code: 
{2,1,3,4,5,...}
 ↓ ↕ ↕ ↕ ↕ 
  {1,2,3,4,...}
 
Last edited:
Are you absolutely, 100% sure that you want to put it like this, Doron?

I am off to bed and tomorrow I have some chores until the afternoon... you have until then to make sure you really, really want to write it down like this.

Hint: unbounded.
 
realpaladin said:
Are you absolutely, 100% sure that you want to put it like this, Doron?

I am off to bed and tomorrow I have some chores until the afternoon... you have until then to make sure you really, really want to write it down like this.

Hint: unbounded.
Click to expand...
Happy dreams realpaladin.

Do you have a problem to understand that the set of all natural numbers has unbounded amount of members?

Hint: the used term is unbounded amount, not just unbounded (as you wrongly express it).
 
Last edited:
jsfisher said:
You said one relation was a "close relative" of something, but were never clear what it actually was all by itself.
Click to expand...

Here it is.

According to Dedekind infinite, there is bijection, for example, between the set of all natural numbers and its proper sub-set of all even numbers, as follows:
Code: 
{1,2,3,4,...}
 ↕ ↕ ↕ ↕   
{2,4,6,8,...}

By using 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 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...

one of the options is
Code: 
            {1,2,3,4,...}
     ↑ ↑ ↑ ↑ ↕ ↕ ↕ ↕ 
{...,7,5,3,1,2,4,6,8,...}

that is equivalent to

← 1
1 ↔ 2
← 3
2 ↔ 4
← 5
3 ↔ 6
← 7
4 ↔ 8
…

or in other words, |N| < |N|.
 
Last edited:
jsfisher said:
If there is an injection from B to A that is not surjective then |A| > |B|.​
Click to expand...

Yet we also need

If there is an injection from A to B that is not surjective then |A| < |B|​

unless only non-equality is considered (and in that case A and B are free variables).
 
Last edited:
doronshadmi said:
Happy dreams realpaladin.

Do you have a problem to understand that the set of all natural numbers has unbounded amount of members?

Hint: the used term is unbounded amount, not just unbounded (as you wrongly express it).
Click to expand...

Good morning! You are really, really sure then? :)

Good! You have about 3 to 4 hours now. But since you are really really sure and are so happy with how you phrase things, no worries eh?
 
realpaladin said:
Good morning! You are really, really sure then? :)

Good! You have about 3 to 4 hours now. But since you are really really sure and are so happy with how you phrase things, no worries eh?
Click to expand...
If you want to shoot, shoot! don't talk! (I am not impressed of your ridiculous "3 to 4 hours" ultimatum).
 
Last edited:
Look, doron, using your "logic" and "definitions" we can "prove" that |N| = |N|, |N| < |N| and (since you seem to be insisting that it's also needed), |N| > |N|.

I'd love to see where you go from there.
 
laca said:
Look, doron, using your "logic" and "definitions" we can "prove" that |N| = |N|, |N| < |N| and (since you seem to be insisting that it's also needed), |N| > |N|.

I'd love to see where you go from there.
Click to expand...

Actually I like Iaca's idea much better.

Go ahead Doron, lead the way! Let us boldly go where no sane mind has gone before!
 
doronshadmi said:
Yet we also need

If there is an injection from A to B that is not surjective then |A| < |B|​

unless only non-equality is considered (and in that case A and B are free variables).
Click to expand...


(1) No, I don't. I can rely on the correspondence between the less-than and the greater-than relationships to make your addition superfluous.

(2) Your doron-cardinality definition is incapable of establishing inequality.
 
doronshadmi said:
Here it is.
...
By using the following definition
...
or in other words, |N| < |N|.
Click to expand...

That would be for doron-cardinality, which exists only in doronetics. One can also conclude in doronetics that |N| = |N| and |N| > |N|.

All in all, rather useless.

In Mathematics, cardinality is much more sensible.
 
Some clarifications:

-------------------
A AND B are free variables, and in this case A<B is equivalent to B>A.

OR

A AND B are NOT free variables, and in this case A<B is NOT equivalent to A>B.
-------------------

-------------------
Some correction about Hilbert's Hotel:

I wrongly wrote that there are more rooms than visitors.

The accurate way to expression it is to state that there are more names that are related to room, than names that are related to visitors, for example:

Code: 
            {1,2,3,4,...} (names of visitors)
     ↑ ↑ ↑ ↑ ↕ ↕ ↕ ↕ 
{...,7,5,3,1,2,4,6,8,...} (names of rooms)

that is equivalent to

← 1
1 ↔ 2
← 3
2 ↔ 4
← 5
3 ↔ 6
← 7
4 ↔ 8
…

where the right side is the names of the rooms and the left side is the names of the visitors.
-------------------

-------------------
jsfisher said:
One can also conclude in doronetics that |N| = |N| and |N| > |N|.
Click to expand...

Please explicitly support your claim (you know, in details).
 
Last edited:
doronshadmi said:
Some clarifications:

-------------------
A AND B are free variables, and in this case A<B is equivalent to B>A.

OR

A AND B are NOT free variables, and in this case A<B is NOT equivalent to A>B.
-------------------

-------------------
Some correction about Hilbert's Hotel:

I wrongly wrote that there are more rooms than visitors.

The accurate way to expression it is to state that there are more names that are related to room, than names that are related to visitors, for example:

Code: 
            {1,2,3,4,...} (names of visitors)
     ↑ ↑ ↑ ↑ ↕ ↕ ↕ ↕ 
{...,7,5,3,1,2,4,6,8,...} (names of rooms)

that is equivalent to

← 1
1 ↔ 2
← 3
2 ↔ 4
← 5
3 ↔ 6
← 7
4 ↔ 8
…

where the right side is the names of the rooms and the left side is the names of the visitors.
-------------------

-------------------


Please explicitly support your claim (you know, in details).
Click to expand...

Good, we all agree that you showed that this is correct in Doronetics.

And where do we go from here? Or is it a road to actually nowhere?
 
I don't think Doron will take it any further than this.

He just loves to kibitz about what others 'dont't get'; it gives him a feeling of accomplishment and he can feel 'learned & wise'.

Do not expect any revelations or insights coming from him, even if we all would go along with his claims.

The main thing for Doron is the discussion, not the conclusion.
 
Before I go further I ask jsfisher or any other poster, to support jsfisher's following claim:
jsfisher said:
One can also conclude in doronetics that |N| = |N| and |N| > |N|.
Click to expand...

EDIT:

And please do it in details.

Thanks.
 
Last edited:
doronshadmi said:
jsfisher said:
One can also conclude in doronetics that |N| = |N| and |N| > |N|.
Click to expand...

Please explicitly support your claim (you know, in details).
Click to expand...

Seriously? You want me to explicitly show you what you have already claimed yourself? *Heavy sigh*

I'll use this statement of doron-cardinality as reference:
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 |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|.
Also, to avoid any confusion, I will take the set, N, to mean the natural numbers, including zero.

The doron-function, f : N -> N, f(i) = i, satisfies the requirements of the antecedent in conditional #1. Therefore, the consequent is true; i.e. |N| = |N|.

The doron-function, f : N -> N, f(i) = i-1 for i>0 and undefined otherwise, satisfies the requirements of the antecedent in conditional #2. Therefore, the consequent is true; i.e. |N| > |N|.

The doron-function, f : N -> N, f(i) = i+1, satisfies the requirements of the antecedent in conditional #3. Therefore, the consequent is true; i.e. |N| < |N|.

So, by applying meaning Doron has assigned to doron-cardinality, it has been shown that |N|<|N| and |N|=|N| and |N|>|N|.
 
jsfisher said:
Seriously? You want me to explicitly show you what you have already claimed yourself? *Heavy sigh*

I'll use this statement of doron-cardinality as reference:
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 |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|.
Also, to avoid any confusion, I will take the set, N, to mean the natural numbers, including zero.

The doron-function, f : N -> N, f(i) = i, satisfies the requirements of the antecedent in conditional #1. Therefore, the consequent is true; i.e. |N| = |N|.

The doron-function, f : N -> N, f(i) = i-1 for i>0 and undefined otherwise, satisfies the requirements of the antecedent in conditional #2. Therefore, the consequent is true; i.e. |N| > |N|.

The doron-function, f : N -> N, f(i) = i+1, satisfies the requirements of the antecedent in conditional #3. Therefore, the consequent is true; i.e. |N| < |N|.

So, by applying meaning Doron has assigned to doron-cardinality, it has been shown that |N|<|N| and |N|=|N| and |N|>|N|.
Click to expand...
jsfisher, it is #1 OR #2 OR #3, where #3 is redundant if A and B are considered as free variables.

EDIT:

In other words, you did not show (yet) that there is necessarily AND condition between the definitions, simply because each definition explicitly defines the relation between A and B, independently of the other definitions.

------------------

EDIT:

To avoid any confusion, I will take the set, N, to mean the set of all natural numbers, without 0.

Code: 
{2,1,3,4,5,...}
 ↕ ↕ ↕ ↕ ↕ 
{1,2,3,4,5...}
provides |N|=|N| (definition #1 is satisfied).

OR

Code: 
{2,1,3,4,5,...}
 ↓ ↕ ↕ ↕ ↕ 
  {1,2,3,4,...}
provides |N|>|N| (definition #2 is satisfied).

In both cases no member is omitted from OR added to N.
 
Last edited:
doron, for crying out loud. Do you realize that using OR makes it even more idiotic and prone to riducule by proving insane things using them?
 
doronshadmi said:
jsfisher, it is #1 OR #2 OR #3
Click to expand...


laca's assessment is true. Your insistence on disjunction makes your position even more idiotic, but that aside, using your meaning for doron-cardinality, I provided an example that showed |N|<|N|

AND

I provided an example that showed |N|=|N|

AND

I provided an example that showed |N|>|N|

I did not show one or the other or the other other. I showed all three. I showed |N|<|N| and |N|=|N| and |N|>|N|.

...where #3 is redundant if A and B are considered as free variables.
Click to expand...

You really have no idea what that means, do you?

To avoid any confusion, I will take the set, N, to mean the set of all natural numbers, without 0.
Click to expand...

More juvenile rebellion? If you wish, we can use this version of N. A trivial adjustment would be required for my second mapping. I'll leave that as an exercise.
 
I predicted this: Doron finds a pretext to continue discussion so he can avoid running into his own errors.

We will never ever see anything worthwhile by Doron Shadmi.
 
jsfisher said:
I showed |N|<|N| and |N|=|N| and |N|>|N|.
Click to expand...

jsfisher two cases are enough to support your AND argument.

All you did is to show that your, so called, formal manipulations have nothing to do with the following straightforward case:

Here it is again.

To avoid any confusion, I will take the set, N, to mean the set of all natural numbers, without 0.

Code: 
{2,1,3,4,5,...}
 ↕ ↕ ↕ ↕ ↕ 
{1,2,3,4,5...}
provides |N|=|N| (definition #1 is satisfied).

OR

Code: 
{2,1,3,4,5,...}
 ↓ ↕ ↕ ↕ ↕ 
  {1,2,3,4,...}
provides |N|>|N| (definition #2 is satisfied).

In both cases no member is omitted from OR added to N.
 
doronshadmi said:
EDIT:

I indeed see that jsfisher is stuck with his wrong AND argument, in spite of the straightforward case (which is without loss of generality) in http://www.internationalskeptics.com/forums/showpost.php?p=9768949&postcount=3234.
Click to expand...

Who gives a hootenanny but you?

We all are waiting for you to go forward, but I make the following statement here :

Doron Shadmi is unable to actually do anything worthwhile with his so called discovery, except bicker and quarrel about non-results.

Prove me wrong!
 
Given two sets, A and B, there is doron-function from A to B such that

(

(If there is 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 |A| = |B|)

OR

(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|)

OR

(if there is 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|)

)
 
Last edited:
Doron is like an anglerfish ; dangling the bait until someone calls him wrong and then he keeps on discussing until everyone has forgotten that he has not made any progress in 7 years.

If that routine starts again I will alert a moderator to his trolling.
 
Status
Not open for further replies.

Back
Top Bottom