Cont: Deeper than primes - Continuation 2

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doronshadmi said:
You are not using also your visual_spatial brain skills (in addition to your verbal_symbolic skills) in order to define |x1-x2| as a totally smooth endlessly increasing value.
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|x1-x2| [sic] doesn't need to be defined by you. From the page you referred to (http://math.tutorvista.com/geometry/distance-formula.html), |x1-x2| [sic] is the absolute value of a difference of two points distance from the x - axis (or the difference of the abcissae).
 
Little 10 Toes said:
|x1-x2| [sic] doesn't need to be defined by you. From the page you referred to (http://math.tutorvista.com/geometry/distance-formula.html), |x1-x2| [sic] is the absolute value of a difference of two points distance from the x - axis (or the difference of the abcissae).
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This definition is about fixed-only lengths of the sides of a given right triangle.

My definition is also about non-fixed lengths of the sides of a given right triangle, and also in this case c2 = a2 + b2, such that (in this particular case) the non-fixed length c2 > a2 exactly by the fixed length b2 .

There are more than just fixed points in my mathematical framework, which can't be defined by any Cantorean set theory, since such theory is a fixed-only mathematical framework, because a given mathematician or a community of mathematicians using their verbal_symbolic-only brain skills in order to do math.

But as clearly seen by my fixed non-fixed argument of c2 = a2 + b2, it is not limited to verbal_symbolic-only brain skills mathematical frameworks.

Conclusion: Cantorean set theories (which are the result of using verbal_symbolic-only brain skills for deduction) can't be considered as the one and only one foundation of every possible interesting mathematical framework (as the majority of mathematicians currently claim, exactly because they are a community of people that have been trained to use only their verbal_symbolic-only brain skills in order to do math).
 
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In mathematics, a multiset (aka bag or mset) is a generalization of the concept of a set that, unlike a set, allows multiple instances of the multiset's elements. For example, {a, a, b} and {a, b} are different multisets although they are the same set. However, order does not matter, so {a, a, b} and {a, b, a} are the same multiset.
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https://en.wikipedia.org/wiki/Multiset

Let {1,2,3,4} be a set of four natural numbers.

Let multiset A = {{1,2,3,4}, {3,1,2,4}}

In terms of multiset A {1,2,3,4} and {3,1,4,2} are not the same member even if they have the same members (where order does not matter), which means that they are independent of each other under A.

So under A we can make a shift of the mapping between {1,2,3,4} and {3,1,4,2} (for example) as follows:

Code: 
1,2,3,4
    ↓ ↓    
    3,1,4,2

But it does not matter in case of finite number of members, since the number of the shifted members is equal in both sides, so the mapping above is actually the following mapping:

Code: 
3,4,2,1
↓ ↓ ↓ ↓ 
3,1,4,2

Is this fact holds also among infinite sets, which are independent members of a given multiset?

Let {1,2,3,...} be the set of all natural numbers (order does not matter).

Let multiset S = {{1,2,3,4,...}, {8,2,3,5,...}} (S has two independent copies of the set of all natural numbers, where order does not matter).

So under S we can make a shift of the mapping between {1,2,3,4,...} and {8,2,3,5,...} (for example) as follows:

Code: 
1,2,3,4,5,6,...
    ↓ ↓ ↓ ↓
    8,2,3,5,...

As we have seen above, this kind of shift does not matter in case of finite number of members, but in this case the shift is done among members of two independent infinite sets under multiset S, and one discovers that there can be also no bijection between two infinite sets under a given multiset even if they have the same members.

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

In this example |{8,2,3,5,...}| < |{1,2,3,4,...}| by 2, which means that the notion that finite cardinality has no impact on infinite cardinality, does not necessarily hold in case of multisets, or more generally, the notion that the set of all natural numbers must have a fixed cardinality (notated as ℵ0) does not necessarily hold in this context.
 
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doronshadmi said:
https://en.wikipedia.org/wiki/Multiset

So under A we can make a shift of the mapping between {1,2,3,4} and {3,1,4,2} (for example) as follows:

Code: 
1,2,3,4
    ↓ ↓    
    3,1,4,2

But it does not matter in case of finite number of members, since the number of the shifted members is equal in both sides, so the mapping above is actually the following mapping:

Code: 
3,4,2,1
↓ ↓ ↓ ↓ 
3,1,4,2
Click to expand...
Please explain this. Where does 3,4,2,1 come from? It isn't a permutation of 1,2,3,4. And why do you think it's the same mapping as the one above? :confused:
 
Hevneren said:
Please explain this. Where does 3,4,2,1 come from? It isn't a permutation of 1,2,3,4. And why do you think it's the same mapping as the one above? :confused:
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You have probably noticed that Doronshadmi is given to a lot of extraneous complexity in his presentations. Simple and straightforward are not his forte, nor is well-defined. Putting aside the whole multiset excursion and the ordered sets (after telling us repeatedly that order does not matter), Doronshadmi's point could more easily be made like this:

- Let N be the set of whole numbers.
- Let f be a mapping from N -> N such that f(a) = a+2.

This would give us 0 -> 2, 1 -> 3, 2 -> 4, and so on.

- Nothing maps to 0 or 1, so therefore the cardinality of N isn't fixed.

Or so we are supposed to believe. If Doronshadmi actually understood how cardinality is defined, he might reach a different conclusion.
 
Hevneren said:
Please explain this. Where does 3,4,2,1 come from? It isn't a permutation of 1,2,3,4.
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Let multiset A = {{1,2,3,4}, {3,1,2,4}}

The mapping is between the members of sets {1,2,3,4} and {3,1,2,4}, which are the two independent members of multiset A.

Hevneren said:
And why do you think it's the same mapping as the one above? :confused:
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1) Since the two sets ({1,2,3,4} and {3,1,2,4}) are independent of each other under multiset A (even if they have the same members, where order does not matter) their mapping can be shifted with respect to each other, as seen in the first example.

2) But since they are finite sets, their shifted sides are actually mapped with each other, as seen in the second diagram.

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

This is not necessarily the case if the two independent members of a given multiset (called S in this example) have infinitely many members.

In other words

Code: 
1,2,3,4,...
↓ ↓ ↓ ↓ 
8,2,3,5,...

is not the one and only one possible mapping between the two independent Ns under multiset S.
 
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doronshadmi said:
Code: 
1,2,3,4,5,6,...
    ↓ ↓ ↓ ↓
    8,2,3,5,...

As we have seen above, this kind of shift does not matter in case of finite number of members, but in this case the shift is done among members of two independent infinite sets under multiset S, and one discovers that there can be also no bijection between two infinite sets under a given multiset even if they have the same members.
Click to expand...
The mapping you show here isn't a bijection, that's correct. But one could easily come up with one, for example this:
Code: 
1,2,3,4,...
↓ ↓ ↓ ↓
8,2,3,5,...
In other words, there exists a bijection between the two sets, so they have the same cardinality.
 
doronshadmi said:
Wrong, there are two independent Ns under multiset S, so the rest of your reply does not hold.
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Your introduction of multisets is red herring. Also, there is only one N; it just happens to be a member of your unnecessary multiset twice.
 
jsfisher said:
Your introduction of multisets is red herring. Also, there is only one N; it just happens to be a member of your unnecessary multiset twice.
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Multiset is a generalization of the concept of set with redundancy degree = or > 1.

The, so called, set is simply a multiset with redundancy degree = 1, and in this particular case N --> N can't be but a bijective function.

If redundancy degree > 1, then there are at least two independent Ns under some given mutiset, such that bijective function between the independent Ns is not the one and only one option, as very simply observed in http://www.internationalskeptics.com/forums/showpost.php?p=12158706&postcount=2864.

Since all your mathematical notions on the considered subject are closed under a multiset with redundancy degree = 1, anything that is not a multiset with redundancy degree = 1, is necessarily observed by you as red herring.
 
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doronshadmi said:
The, so called, set is simply a multiset with redundancy degree = 1, and in this particular case N --> N can't be but a bijective function.
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The example mapping I cited, f: N --> N where f(a) = a+2, is not a bijection.
 
Little 10 Toes said:
Unsubstantiated assertion and made up definitions.

The rest of your post is based off that unsubstantiated assertion.
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The concept of a multiset is not made up. It's a standard mathematical concept.

It is used here in a nonsensical argument.
 
jsfisher said:
The example mapping I cited, f: N --> N where f(a) = a+2, is not a bijection.
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What do you know, you have just solved CH;)

More seriously, since (as you say) "there is only one N" f(a) = a+2 is no more than gibberish.

Since in your mathematical framework {{1,2,3,...}, {1,2,3,...}} = {{1,2,3,...}} such that |{{1,2,3,...}, {1,2,3,...}}|=1, you have no argument.
 
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Little 10 Toes said:
Sorry. I meant the terms that Doronshadmi was using. Specifically, "redundancy degree".
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For example, {a, a, b} and {a, b} are different multisets although they are the same set
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https://en.wikipedia.org/wiki/Multiset

Since multiset is a generalization of the concept of set
In mathematics, a multiset (aka bag or mset) is a generalization of the concept of a set
Click to expand...
https://en.wikipedia.org/wiki/Multiset

then {a, a, b} is an example of a multiset with redundancy degree > 1, where {a, b} is a multiset with redundancy degree = 1, known by the name "set".

So, you have no argument.
 
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doronshadmi said:
What do you know, you have just solved CH;)

More seriously, since (as you say) "there is only one N" f(a) = a+2 is no more than gibberish.

Since in your mathematical framework {{1,2,3,...}, {1,2,3,...}} = {{1,2,3,...}} such that |{{1,2,3,...}, {1,2,3,...}}|=1, you have no argument.
Click to expand...

The set of all natural numbers is one set. You can, however, include it multiple times in a multiset, like you did here.
 
Hevneren said:
The set of all natural numbers is one set.
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Not under a multiset with redundancy degree > 1 (where multiset is a generalization of the concept of set, such that a set is actually a multiset with redundancy degree = 1, known by the name "set").
 
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