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I can't or I won't say


Best Answer bushindo, 17 October 2012 - 05:08 PM

My take on this

Spoiler for
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#11 mmiguel

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Posted 18 October 2012 - 05:07 AM

choice
Spoiler for

Edited by mmiguel, 18 October 2012 - 05:08 AM.

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#12 EventHorizon

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Posted 18 October 2012 - 09:57 AM

Spoiler for A few more thoughts

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#13 bonanova

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Posted 19 October 2012 - 04:28 AM

Spoiler for A few more thoughts


Spoiler for I would say

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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#14 mmiguel

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Posted 19 October 2012 - 07:53 AM

Spoiler for I would say



Spoiler for cool

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#15 EventHorizon

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Posted 20 October 2012 - 06:02 AM

Spoiler for I would say


Spoiler for Extra thoughts

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#16 bonanova

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Posted 20 October 2012 - 07:19 AM

Spoiler for Extra thoughts


Well we've finished the discussion of the puzzle so I won't spoiler this. You made several points and raised new questions. I'll try to reply to the extent that I know.

Inarguably the cube contains points that its edge does not. For finite sets that gives a Yes answer. For infinite sets, and these sets are uncountable, we express size by cardinality. The edge has the cardinality of the reals, so the question really asks the cardinality of the cube. Bushindo's bijection expresses Cantor's finding that geometric figures of all dimensions, arbitrarily large ones, have the cardinality of the reals.

So the cube has other points, but not more points.

The power set is the set of all subsets. Is there a bijection between the elements of a set and the elements of its power set? The answer is No, and so the cardinality of the power set is greater than the cardinality of the set. For finite sets, it can be seen by inspection. The set {1,2,3} has 23 = 8 subsets, including the empty set. No bijection exists. For infinite sets, the proof is almost as simple. It's a proof by contradiction. Suppose the elements of an infinite set have been paired with the elements of its power set. We then color the set's elements blue that are paired with a subset that contains the element, and red if they are not a member of their paired subset. Now consider the element of the set that is paired with the subset of all the red colored elements. Can it be red? No, because red elements are paired with subsets that do not contain them. Can it be blue? No, because blue elements are paired with subsets that do contain them. So the subset of all red elements cannot be paired, and this contradict the premise of a bijection.

So power sets have increasing cardinalities.

We know the integers are countably infinite, and the reals are not; and we have an understanding of both these sets. Do we have an understanding of a set that has cardinality Aleph2? One such set is the set of all functions of real numbers. Or of all curves in a plane. No one has suggested a set of cardinality Aleph3 other than the power set of the functions of a real number.

What is AlephnAlephn? I would say it lacks any recognizable similarity with a power set or a real number raised to the power Alphan because the infinite cardinals are not algebraic. So it's a symbol raised to its own power. What value or significance we could thus assign to it I do not know.
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The greatest challenge to any thinker is stating the problem in a way that will allow a solution.
- Bertrand Russell

#17 EventHorizon

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Posted 21 October 2012 - 12:43 AM

What things can I include in this post more than are contained in my previous posts?

Notice I used more in the sense of other, extra, additional, etc. We agree, except I'm harping on the various definitions of more.

Just wait until I start arguing over what the definition of is is... :wacko:

I've seen those examples of sets of various degrees of infinity, and have had to come up with various bijections including them. But it doesn't say whether bijections are always necessary or if math can be used instead. Perhaps in these cases the symbolic math is just shorthand for the bijections already given. But that seems to be the way with math in general... operators are defined by previously defined operators and/or algorithms including other previously defined operators.

Spoiler for Assuming we can assign significance to a symbol raised to its own power... + realization that it can!

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#18 mmiguel

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Posted 21 October 2012 - 04:15 AM

So the cube has other points, but not more points.



I don't think we can say the the same about the edge with respect to the cube:
The edge does not have other points than those contained in the cube.

Do you agree?

The subset operator would still show asymetry here.
The less-than operator would not show asymmetry between the cardinalities of the two sets.

Our intuition tells us, if A is a strict subset of B, then |A| < |B|.
Our conclusion from this discussion is that this transformation, from a statement of sets, to a statement of cardinalities, does not hold when sets are infinite.

Is there some other commonly accepted mathematical construct, e.g. denoted for set A by [A], that evaluates to a number, such that if A is a strict subset of B, then [A] < [B] even for infinite sets?

Edited by mmiguel, 21 October 2012 - 04:20 AM.

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#19 mmiguel

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Posted 21 October 2012 - 04:31 AM

I don't think we can say the the same about the edge with respect to the cube:
The edge does not have other points than those contained in the cube.

Do you agree?

The subset operator would still show asymetry here.
The less-than operator would not show asymmetry between the cardinalities of the two sets.

Our intuition tells us, if A is a strict subset of B, then |A| < |B|.
Our conclusion from this discussion is that this transformation, from a statement of sets, to a statement of cardinalities, does not hold when sets are infinite.

Is there some other commonly accepted mathematical construct, e.g. denoted for set A by [A], that evaluates to a number, such that if A is a strict subset of B, then [A] < [B] even for infinite sets?


One more requirement for [A]: for finite sets, [A] = |A|
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#20 mmiguel

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Posted 21 October 2012 - 04:48 AM

One more requirement for [A]: for finite sets, [A] = |A|


One more requirement:

If A is the real interval [0,1], and B = [0,0.5) U (0.5,1] U {2}
Then [A] = [B]

B is the same as A, except that 0.5 is removed, and 2 is added.
B is no longer a subset of A, but from an intuitive perspective should have the same "quantity of points".
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