Argument from Motion

182 posts / 0 new

Fri, 02/03/2017 - 10:33

Dave Matson

If one infinity were truly bigger, there would be no way to pair the elements one-to-one with nothing left over. That you can do other pairings that leave something left over doesn't change that fact. (Read up on Cantor for some background here.) Call it the subtle side of infinity! There are actually different sizes of infinities, and they can't be paired up as above. The lowest infinity would consist of all the integers and all the fractions. Next up is the real number line and solid geometrical space. (The last two can be put into a one-to-one correspondence!!) I think the next higher up is all the possible configurations a straight line can be twisted into.

Thus, all the integers is an infinity of the same size as all the fractions! However, if you treat them as sets, then you can obviously point to infinitely many fractions that aren't in the integer set. But that doesn't make the fractions any bigger an infinity since they can be put into a one-to-one correspondence! So, there are two concepts here, the cardinality of the infinity and the idea of a set. They are not contradictory, being different things.

Mon, 01/30/2017 - 09:14

(Reply to #92) #93

Nyarlathotep

skd6348 - The girl removes a

skd6348 - The girl removes a red, green, blue and yellow bar from A. She also removes one of each of these from B as well. If there are still bars left in B, but no bars in A, B has more bars than A.

Likewise, if:

(N-2) - (N-6) != {},

I thought about what you said here over breakfast and I think now understand a little more about what you are saying, and what the problem is with it. I'll try to rephrase your argument then address it (and if it isn't your argument, well then I'm sorry):

N2 = { 2,4,6,8,10,12,...}
N6 = { 6,12,18,... }
You have noticed that every entry in N6 exists in N2 so you are suggesting a 1:1 mapping from N6 to N2 by just pairing each value with its matching value. In other words you match the 6 (in the first position) of N6 with the 6 (in the 3rd position) of N2. Then you do this with the 2nd value in N6 which is 12. Then you conclude that since there are leftovers in N2 it must have more entries. That is the mistake. This is only half of the bijection. What we can conclude from your mapping is that |N2| >= |N6|; where the || just means cardinality. This is because there is more than one way to do the mapping, and some methods leave extra elements in N2 and some do not. So we have to use "greater than or equal" instead of just "equal".

Now we have to do the mapping the other way from N2 to N6. We can do this with f(n) = 3n, at which point we can conclude that |N6| >= |N2|. Look what that leaves us with:

|N2| >= |N6|
|N6| >= |N2|

. There is only one way both of those statement can be true: |N2| = |N6|. They have the same cardinality.I know it seems kind of like a gotcha, and for that I'm sorry but I did mention it, even though it probably got buried quickly in the forest of all that has been said:

Nyarlathotep - There exists a 1:1 mapping from every element in N2 to a unique element in N6. There also exists a mapping from every element in N6 to a unique element in N2. This demands that they have the same exact number of elements.

Nyarlathotep - If every element in A can be paired with a unique element in B (and vice-versa) they MUST have the same number of elements and therefore the same cardinality.

Tue, 01/31/2017 - 00:20

(Reply to #93) #94

Salman Durrani

You said:

You said:
" N2 = { 2,4,6,8,10,12,...}
N6 = { 6,12,18,... }
You have noticed that every entry in N6 exists in N2 so you are suggesting a 1:1 mapping from N6 to N2 by just pairing each value with its matching value. In other words you match the 6 (in the first position) of N6 with the 6 (in the 3rd position) of N2. Then you do this with the 2nd value in N6 which is 12. Then you conclude that since there are leftovers in N2 it must have more entries. That is the mistake. This is only half of the bijection. "

My method isn't a bijection.
(N-2) - ( N-6) does not refer to a mapping, but that we eliminate the SAME members of the entire series ( N-6) from (N-2).

Tue, 01/31/2017 - 01:28

(Reply to #94) #95

Salman Durrani

"What we can conclude from

YOU SAID:
///
"What we can conclude from your mapping is that |N2| >= |N6|; where the || just means cardinality. This is because there is more than one way to do the mapping, and some methods leave extra elements in N2 and some do not. So we have to use "greater than or equal" instead of just "equal".

Now we have to do the mapping the other way from N2 to N6. We can do this with f(n) = 3n, at which point we can conclude that |N6| >= |N2|. Look what that leaves us with:

|N2| >= |N6|
|N6| >= |N2|
. There is only one way both of those statement can be true: |N2| = |N6|. They have the same cardinality."
///

This becomes more interesting when we create a 1:2 function from N-2 to N-6...

f(n)= 6n-6, 6n

for n=2, this gives f(n) = 6,12
for n=4, this gives f(n) = 18,24
for n=6, this gives f(n) = 30,36
and so on...

Every member in N-2 gives 2 members in N-6. Furthermore, there are no overlaps in the mapping, and the mapping is complete.

Here is code for this in python programming language:

def N2to6(n):
B=[]
for x in range(2,2*n+1,2):
B.append(6*x-6)
B.append(6*x)
print(B)

n is the input. What it does is, it takes the first n members from N-2, and then lists 2 corresponding for each of them from N-6.

When I run it:

>>> N2to6(3)
[6, 12, 18, 24, 30, 36]
>>> N2to6(10)
[6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120]
>>> N2to6(15)
[6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180]
>>>

It successfully generates twice the members, without any overlaps or gaps. This looks like N-6 has double members, which does not make sense. [ It should be the other way around... N-2 has thrice the members ]

This is why I don't find bijection to be a valid justification for cardinality of infinite sets.

A word about statements like |A|>=|B|...
When we say A is greater than OR equal to B, we either mean that:
1) At least one of them are variables, and depending on when we measure A and B, we might get a different comparative result.
2) It's exactly one of the two ( equal or greater) , but we don't know.

The problem is, N-2 and N-6 are constants. Their members never change, with time or otherwise. So exactly one is true IF cardinality applies to infinite set comprehensions:
1) |N-2| = |N-6|

2) |N-2| > |N-6|

3) |N-2| < |N-6|

When you concluded |N-2| = |N-6| from:

|N2| >= |N6|
|N6| >= |N2|

The method seems fully valid, AS LONG AS it is provided that exactly one of them are true.

[ (A or B)^(A or C ) ] => A is a valid theorem, IF not more than of A,B and C are true.

The problem seems to be with the way you established:
|N2| >= |N6|
|N6| >= |N2|
In the first place.

If we map, 1:1, "all" members of N-6 to "some" members of N-2, then if it justifies anything, it justifies:

|N6|<|N2|

BUT NOT:

|N6|<=|N2|

If we repeat some mappings we get atomic propositions, rather than disjunctive propositions:

|N6|=|N2|

|N6|<|N2|

|N6|>|N2|

These are CONTRARY propositions. This, along with the idea that:
|(N-2)-(N-6)| != 0

Does not satisfy me as using bijection to determine relative cardinality.

Tue, 01/31/2017 - 01:48

(Reply to #95) #96

Salman Durrani

There seems to be some

There seems to be some problem... Some of my expressions are not showing up fully in the above reply...

Tue, 01/31/2017 - 02:22

(Reply to #96) #97

Nyarlathotep

skd6348 - Some of my

skd6348 - Some of my expressions are not showing up fully in the above reply...

Yeah, I see that some stuff is missing but don't worry about it since what you posted is madess anyway:

skd6348 - we create a 1:2 function from N-2 to N-6

This is the madness, no function can map one element in A to two elements in B. It's like talking about a square circle.

Tue, 01/31/2017 - 04:10

(Reply to #97) #98

Salman Durrani

You said:

You said:
" This is the madness, no function can map one element in A to two elements in B. It's like talking about a square circle. "
--------------------

I just did. You run the code: ( In the Python interpreter )

def N2to6(n):
B=[]
for x in range(2,2*n+1,2):
B.append(6*x-6)
B.append(6*x)
print(B)

And it literally generates two members of N-6 for every member of N-2. It uses two equations to do that. There are no overlaps, and no gaps ( i.e., the mapping is complete ).

Let me run it again for a larger value:

>>> N2to6(100)
[6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66, 72, 78, 84, 90, 96, 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198, 204, 210, 216, 222, 228, 234, 240, 246, 252, 258, 264, 270, 276, 282, 288, 294, 300, 306, 312, 318, 324, 330, 336, 342, 348, 354, 360, 366, 372, 378, 384, 390, 396, 402, 408, 414, 420, 426, 432, 438, 444, 450, 456, 462, 468, 474, 480, 486, 492, 498, 504, 510, 516, 522, 528, 534, 540, 546, 552, 558, 564, 570, 576, 582, 588, 594, 600, 606, 612, 618, 624, 630, 636, 642, 648, 654, 660, 666, 672, 678, 684, 690, 696, 702, 708, 714, 720, 726, 732, 738, 744, 750, 756, 762, 768, 774, 780, 786, 792, 798, 804, 810, 816, 822, 828, 834, 840, 846, 852, 858, 864, 870, 876, 882, 888, 894, 900, 906, 912, 918, 924, 930, 936, 942, 948, 954, 960, 966, 972, 978, 984, 990, 996, 1002, 1008, 1014, 1020, 1026, 1032, 1038, 1044, 1050, 1056, 1062, 1068, 1074, 1080, 1086, 1092, 1098, 1104, 1110, 1116, 1122, 1128, 1134, 1140, 1146, 1152, 1158, 1164, 1170, 1176, 1182, 1188, 1194, 1200]
>>>

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

Also, this 1:2 map is only part of my post. If you'd like, we can simply create a " 2:1 function from N-6 to N-2 ", and we will still get the same problem.

There is also my note on your disjunctions, and that I've pointed out that bijection wasn't my method, but rather it was subtraction.

Tue, 01/31/2017 - 04:39

(Reply to #98) #99

Nyarlathotep

I just did. You run the code:

I just did. You run the code: ( In the Python interpreter )

What you have written is a function in python, not a mathematical function. It's madness.

Tue, 01/31/2017 - 05:05

(Reply to #99) #100

Salman Durrani

A "mathematical" function?

A "mathematical" function? Are you referring that the output, by definition, has to be one for each input, in order for your bijection method to apply?

Tue, 01/31/2017 - 05:45

(Reply to #100) #101

Nyarlathotep

skd6348 - A "mathematical"

skd6348 - A "mathematical" function?

Of course!

skd6348 - in order for your bijection method to apply

This isn't my method. This is the method used by mathematicians, math departments, and textbooks the world over. And a bijection is a function, so your question is rather odd.

Tue, 01/31/2017 - 06:28

(Reply to #101) #102

Salman Durrani

You said:

You said:
"This isn't my method. This is the method used by mathematicians, math departments, and textbooks the world over. And a bijection is a function, so your question doesn't make sense. "

I'm not referring to simply taking a bijection, or the definition of a mathematical function, but that:

" If we have a 1:1 function from a set A to B, and B to A as well, then A and B have the same cardinality, where A and B are INFINITE sets. "

Also, I didn't say that bijection isn't a function...

But I think I need to clarify a few points:

1) First, conventions don't mean proof to me. If you use axioms with which I disagree, and I provide reason for disagreeing, please answer my criticism ( by yourself or a link, but the link must really justify ) and provide some justification for them as well...
The same applies to me as well. This is a discussion, not a dictation.

2) I didn't take the word " function" in your words to be strictly " mathematical ". I'd mistaken it to mean something else, and I apologize for that.

3) You seem to answer only parts of many of my posts, which isn't helping...
I often have to repeat.
I'll repeat again in my next post.

4) I do not strictly hold everything in the modern ZFC or equivalent set theory to be true... Likewise for other literature.

5) Maybe if I'd state the relevant axioms which I hold, and ask for yours, we could use the common ground to discuss the matter. Although this could be cumbersome...

Tue, 01/31/2017 - 06:33

(Reply to #102) #103

Salman Durrani

1) So, my first criticism

1) So, my first criticism arises as:

(N-2)-(N-6) != {}, AND
(x is in N-6) => ( x is in N-2)

N-6 satisfies axioms to be a proper subset of N-2. It should have a lower cardinality than N-2...
You haven't answered this. AND this is not a mapping. A-B doesn't refer to a mapping.

Now if you'd ignore this conflict, by calling it " fascinating " and dismissing it, that is not a justification...

I could go the other way around say that,....."due to the above, a bijection is invalid to determine cardinality for infinite sets", and dismiss it...

2) I have said:

"///
The problem seems to be with the way you established:
|N2| >= |N6|
|N6| >= |N2|
In the first place.

If we map, 1:1, "all" members of N-6 to "some" members of N-2, then if it justifies anything, it justifies:

|N6|<|N2|

BUT NOT:

|N6|<=|N2|

If we repeat some mappings we get atomic propositions, rather than disjunctive propositions:

|N6|=|N2|

|N6|<|N2|

|N6|>|N2|

These are CONTRARY propositions. This, along with the idea that:
|(N-2)-(N-6)| != 0

Does not satisfy me as using bijection to determine relative cardinality.
///"

You haven't answered this. How did you conclude those disjunctions from partial or incomplete 1:1 mappings?

Tue, 01/31/2017 - 06:58

(Reply to #103) #104

Nyarlathotep

skd6348 - N-6 satisfies

skd6348 - N-6 satisfies axioms to be a proper subset of N-2. It should have a lower cardinality than N-2

That statement is clearly false. Subsets are not required to have a smaller cardinality than the sets they are built from.
-------------------------

If we map, 1:1, "all" members of N-6 to "some" members of N-2, then if it justifies anything, it justifies
|N6|<|N2|

BUT NOT:

|N6|<=|N2|

This statement is also false. Clearly if I have 5 candy bars and I preform a 1:1 map of my 5 to 5 of yours, we can only be sure that you have 5 or more. Now if we do a bijection from mine to yours, we can be sure you have exactly 5. This is why the bijection is important.
-------------------------

skd6348 - You seem to answer only parts of many of my posts

It is pointless to continue further into one of your arguments when the first part is false. Garbage in, garbage out. For example why should I pay any attention to your "function" that isn't a function?
-------------------------

skd6348 - |(N-2)-(N-6)| != 0

Yeah, newflash: apples don't equal oranges. Not sure what the point of that is.

Tue, 01/31/2017 - 07:37

(Reply to #104) #105

Salman Durrani

Ummm.... OK.

I said:

"If we map, 1:1, "all" members of N-6 to "some" members of N-2, then if it justifies anything, it justifies... "

Didn't you notice the words "all" and "some"? This isn't just that you mapped 5 to 5...
But that if, AFTER you mapped all 5 of yours with 5 of mine, all of yours are mapped, but I have still some unmapped, then you would have less bars.

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

You said:

" That statement is clearly false. Subsets are not required to have a smaller cardinality than the sets they are built from. "

Describe one finite set, that satisfies both axioms in relation to another finite set, but does not have a lower cardinality.

My axiom is:

[ [ (A-B) != 0 ] ^ [ (x is in B) => (x is in A) ] ] => |B|<|A|

This is true for, at least, ALL finite sets.

Yours is:

[There exists 1:1 from A to B] ^ [There exists 1:1 from B to A] => |A| = |B|

This is ALSO true for all finite sets.

But in infinite sets, holding both often leads to contrary propositions.
On what grounds does your axiom apply to infinite sets, but mine doesn't?

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

You said:

" Yeah, newflash: apples don't equal oranges. Not sure what the point of that is. "

That is just to emphasize my subset problem.

You said:
"It is pointless to continue further into one of your arguments when the first part is false. Garbage in, garbage out. For example why should I pay any attention to your "function" that isn't a function?"

My "function" which isn't a " mathematical function".
No need to pay attention to it anymore, but you seem to pay it a lot of attention ( and not the rest ).
Of course, I'm not obliged to know what term definitions you have in mind, without you defining them in the first place. But I still apologized for that...

Tue, 01/31/2017 - 09:48

(Reply to #105) #106

Nyarlathotep

Nyarlathotep - Subsets are

Nyarlathotep - Subsets are not required to have a smaller cardinality than the sets they are built from.

The simplest example I can think of:
A = {1,2,3,4}
A ⊆ A
|A| = |A|
---------------------------------

skd6348 - But that if, AFTER you mapped all 5 of yours with 5 of mine, all of yours are mapped, but I have still some unmapped, then you would have less bars.

I've already explained why this is doomed, and gave an example. Because there can be more than one way to map from A to B, and some of the ways leave elements in B unmapped while others do not. This is why with only a one way mapping you can only ensure a non strict inequality. But if you map both ways you get two non strict inequalities which you can build a strct inequality with (or an equality).

Or to phrase it more simply, if you only map it one way you can get contradictions (I gave an example of this already).
---------------------------------

skd6348 - [ [ (A-B) != 0 ] ^ [ (x is in B) => (x is in A) ] ] => |B|<|A|

Sidenote: what you have written here is probably not what you really want. I will give you the benfit of the doubt and address what I think you want:

A = {1,2,3,...}
Now, lets create a new set B, each value of B is just the corresponding value of A but times 2:
B = {2,4,6,...}
Clearly |A| = |B| because of the way we constructed B (it is just A with each element being doubled in size). Now lets try your rule and see what it says:

(A-B) != 0; well that is clearly True.
Is every element of B in A? True
Therefore |B|<|A| and |A| = |B|; a contradiction.

Notice if you had used "less than or equal" we wouldn't have a contradiction.

Which answers your question:

skd6348 - On what grounds does your axiom apply to infinite sets, but mine doesn't?

Tue, 01/31/2017 - 10:46

(Reply to #106) #107

Nyarlathotep

Nyarlathotep - A = {1,2,3,...

Nyarlathotep - A = {1,2,3,...}
Now, lets create a new set B, each value of B is just the corresponding value of A but times 2:
B = {2,4,6,...}
Clearly |A| = |B| because of the way we constructed B (it is just A with each element being doubled in size).

If you think about that for a while you might start to suspect the following (present to try to give some insight):
[handwaving]
This very special way we used to create B (to ensure |A| = |B|), smells suspiciously like a bijection already. Which might lead you to notice the following:

If there is a bijection from A to B, then you could have created B in this special way
If B can be created in this special way, we know that |A| = |B|.

[/handwaving]

Tue, 01/31/2017 - 11:40

(Reply to #107) #108

Salman Durrani

You said:

You said:
///
"The simplest example I can think of:
A = {1,2,3,4}
A ⊆ A
|A| = |A|"
///

This is not a "PROPER" subset.
A-A = {}
It doesn't satisfy the first condition.

I had said:
" N-6 satisfies axioms to be a proper subset of N-2. It should have a lower cardinality than N-2..."

And:
" Describe one finite set, that satisfies BOTH axioms in relation to another finite set, but does not have a lower cardinality. "

Read my previous posts. I am only talking about proper subsets...

-----------------------------------------------------------
You said:
" I've already explained why this is doomed, and gave an example. Because there can be more than one way to map from A to B, and some of the ways leave elements in B unmapped while others do not. This is why with only a one way mapping you can only ensure a non strict inequality. But if you map both ways you get two non strict inequalities which you can build a strct inequality with (or an equality).

Or to phrase it more simply, if you only map it one way you can get contradictions (I gave an example of this already)."

No. For all finite sets, if you have 1:1 function from A to B, which maps "all of A" onto only "some of B", B cannot have the same cardinality.

I think you are referring to a 1:1 only. A 1:1 function A to B, ALONE would mean:
|A|<=|B| ( in all finite sets, no two members of A are mapped to the same member in B )

BUT if, in addition, we are told that this leaves some members in B without being mapped, then:
|A|<|B|

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

You Said:

Sidenote: what you have written here is probably not what you really want. I will give you the benfit of the doubt and address what I think you want:

"A = {1,2,3,...}
Now, lets create a new set B, each value of B is just the corresponding value of A but times 2:
B = {2,4,6,...}
Clearly |A| = |B| because of the way we constructed B (it is just A with each element being doubled in size). Now lets try your rule and see what it says:

(A-B) != 0; well that is clearly True.
Is every element of B in A? True
Therefore |B|<|A| and |A| = |B|; a contradiction.
Notice if you had used "less than or equal" we wouldn't have a contradiction."

Side Note: Yeah, its should be [ (A-B) != {} ]

This is not a justification, but a different way of expressing the same axiom of yours...

You simply showed if it satisfies your axiom, it contradicts mine... Which I had said also.

I already said that I understand where you're coming from with the bijection, but that it contradicts my axiom is still a problem.

If we write |B|<=|A|, it wouldn't help.

It's either |A|<|B|,
OR |A| = |B|,
BUT not both.
My axiom doesn't just remove |A|>|B|. It also removes |A|=|B|

My axiom strictly uses |A|<|B| for a reason. It is true for ALL finite "proper" subsets, and it is highly counter-intuitive to suggest that |A|=|B| even if:
1) (A-B) != {}, and all elements of B
2) All elements of B are all also elements of A.

The way you create a set B using a 1:1 function on A, and then reason that |A|=|B|, is merely "using" your axiom, not justifying it over mine for infinite sets.

Tue, 01/31/2017 - 12:17

(Reply to #108) #109

Nyarlathotep

skd6348 - You simply showed

skd6348 - You simply showed if it satisfies your axiom, it contradicts mine... Which I had said also.

No, its worse: yours lead to a contradiction in this specific example. No amount of gymnastics will change that. "Mine" (not really mine of course) does not lead to a contradiction in this specific example. Now if you could only find an example where "mine" leads to contradiction (good luck!) then we could discard both ideas; but currently yours is the one on the chopping block!
--------------------------------

skd6348 - and it is highly counter-intuitive to suggest that |A|=|B|

Well then I suppose it is a good thing mathematics doesn't depend on your intuition.
--------------------------------
The funny thing is that "my" (again, not mine) idea is very similar to your idea, if you were to apply yours once in each direction.

Tue, 01/31/2017 - 23:36

(Reply to #109) #110

Salman Durrani

The reason mine leads to a

The reason mine leads to a contradiction is because you use yours in the first place.

When you construct:
B= { 2,4,6,...}

From :
A={1,2,3,...}

And then conclude, due to the 1:1 construction, that |A|=|B|, then of course you are "using your axiom". And of course it would contradict |B| < |A|.

I could go the other way and subtract:

C={1,3,5,7,...}

From A, to get:

B= A-C = {2,4,6,8...}

And conclude |B| < |A|, ( since it resulted from subtracting a non-empty proper subset out of A )

And it would imply:
1) ~(|B|=|A|)

This contradicts yours,since:
1) [There exists 1:1 from A to B] ^ [There exists 1:1 from B to A] => |A| = |B|
2) [There exists 1:1 from A to B]
3) [There exists 1:1 from B to A]
4) Therefore, [There exists 1:1 from A to B] ^ [There exists 1:1 from B to A]
5) Therefore, |A|=|B|

But I'm merely using my axiom this way to show it contradicts yours.

As far as intuition is there, where do you think we get fundamental axioms anyways, if not intuition or inductive reasoning ( and of course, books and such) ? How did you conclude "your" axiom else wise? Maybe you derived it from other axioms, but if you did, could you give me this derivation? ( And where did you get those axioms... )

Our two axioms can lead to contradictions for infinite sets "as they are".

Wed, 02/01/2017 - 01:02

(Reply to #110) #111

Nyarlathotep

You added a dubious axiom to

You added an axiom that we already know to be false, to get a contradiction using my method: ("And conclude |B| < |A|, ( since it resulted from subtracting a non-empty proper subset out of A )"). Listen: I'm done with your mental gymnastics.

Wed, 02/01/2017 - 02:02

(Reply to #111) #112

Salman Durrani

You said:

You said:
" You added an axiom that we already know to be false "

////

You mean this one: (?)

[ [ (A-B) != {} ] ^ [ (x is in B) => (x is in A) ] ] => |B|<|A|

This axiom is true for all finite sets.

If you are talking about infinite sets, UNDER WHAT JUSTIFICATION is it false, but yours isn't?

So far, what I can see in your argument is:

1) ( My axiom ) => |A| = |B|
2) ( Your axiom) => |B|<|A|
3) Therefore, (Your axiom ) is false.

This is circular reasoning. You assume yours is true over mine, to show that yours is true over mine.

Wed, 02/01/2017 - 02:11

(Reply to #112) #113

Salman Durrani

You said:

You said:
to get a contradiction using my method: ("And conclude |B| < |A|, ( since it resulted from subtracting a non-empty proper subset out of A )")"

If you read my whole post, it says:
" But I'm merely using my axiom this way to show it contradicts yours. "

I already stated this isn't a proof for my axiom. It simply says that:

Proving that A contradicts B, is NOT a proof for A.

Wed, 02/01/2017 - 02:54

(Reply to #113) #114

Nyarlathotep

skd6348 - You mean this one:

skd6348 - This is circular reasoning. You assume yours is true over mine, to show that yours is true over mine.

False. I didn't assume mine was true to reach a contradiction with yours. Stop making shit up. You need to resolve this contradiction before I will take anything else you have to say on the matter seriously.

skd6348 - You mean this one:
[ [ (A-B) != {} ] ^ [ (x is in B) => (x is in A) ] ] => |B|<|A|(?)

I went out of my way to list it, so you wouldn't be confused, but you managed to do it anyway. I have to suspect at this point (and I should have long ago) that you are being dumb on purpose. I can't deal with this flailing any more. Deal with the fact your idea leads to contraction if you want a response.

Wed, 02/01/2017 - 03:54

(Reply to #114) #115

Salman Durrani

You said:

You said:
"False. I didn't assume mine was true to reach a contradiction with yours. Stop making shit up. You need to resolve this contradiction before I will take anything else you have to say on the matter seriously."
///

I already explained this:

"When you construct:
B= { 2,4,6,...}

From :
A={1,2,3,...}

And then conclude, due to the 1:1 construction, that |A|=|B|, then of course you are "using your axiom". And of course it would contradict |B| < |A|."

I do NOT hold that a 1:1 "construction" for " infinite " sets means they have the same cardinality. That's the same as saying:

[There exists 1:1 from A to B] ^ [There exists 1:1 from B to A] => |A| = |B|

But from a different perspective. Bijections are used to "construct" sets. If there is a 1:1 function, that function itself represents a "construction" means.

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

You said:
"Deal with the fact your idea leads to contraction if you want a response. "

Contradiction with the idea that if B results from a 1:1 construction from A, it has the same cardinality as A?

Wed, 02/01/2017 - 04:20

(Reply to #115) #116

Nyarlathotep

If you don't think that

If you don't think that
A = {1,2,3,4,...}
has the same number of elements as
B = {1*2, 2*2, 3*2, 4*2, ...}
You are a lunatic, which would explain quite a bit. Do not expect another reply.

Wed, 02/01/2017 - 05:03

(Reply to #116) #117

Salman Durrani

Sorry, no.

A = {1, 1*2, 3, 2*2, 5, 3*2, 7, 4*2, ...}

B = {1*2, 2*2, 3*2, 4*2, ...}

Doesn't explicitly seem that they do.

Wed, 02/01/2017 - 07:15

(Reply to #117) #118

Salman Durrani

A = {1, 1*2, 3, 2*2, 5, 3*2,

A = {1, 1*2, 3, 2*2, 5, 3*2, 7, 4*2, ...}

= {1*2, 2*2, 3*2, 4*2, ...} + { 1, 3, 5, 7, ... }

= B + { 1, 3, 5, 7, ... }

Tue, 01/31/2017 - 07:23

(Reply to #118) #119

Nyarlathotep

Also, I didn't say that

skd6348 - Also, I didn't say that bijection isn't a function...

My bad; however, in this light it makes your question seem to make even less sense:

skd6348 - Are you referring that the output, by definition, has to be one for each input, in order for your bijection method to apply?

The problem is the first part of your question is contained within what a bijection is. So the only way I can make sense of this question is if you don't know that a bijection is. If you knew what a bijection is, the question is madness. So I was giving you the benefit of the doubt...

Tue, 01/31/2017 - 08:04

(Reply to #119) #120

Salman Durrani

I meant by "your bijection method" was related to what I said earlier:

" If we have a 1:1 function from a set A to B, and B to A as well, then A and B have the same cardinality, where A and B are INFINITE sets. "

I didn't mean that necessarily you'd have to take a bijection, but that since bijection is 1:1, you conclude a cardinality ratio of 1:1.

So I meant by " your bijection method " that: a ratio of m:n in a function ( m:1 for a mathematical function ) constitutes a corresponding ratio in cardinality. I was referring to the concept behind it...

Umm... let me call it " m:n " method or something.

Fri, 02/03/2017 - 10:38

(Reply to #120) #121

Dave Matson

Nyarlathotep:

Nice observation!

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