My Math Forum Dichotomy paradox reverse engineering

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September 11th, 2012, 08:12 PM   #11
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Quote:
 Originally Posted by Maschke As I indicated in my first post, you would have to tell us. Because your process is not defined at t = 1.
If set theory tells us, that the union of [0, 0.5], [0.5, 0.75], [0.75, 0.875], ... is [0, 1), isn't this the same as defining the limit case?

 October 2nd, 2012, 12:31 PM #12 Senior Member   Joined: Aug 2010 From: Germany Posts: 132 Thanks: 1 Re: Dichotomy paradox reverse engineering If we allow only finitely many copies of a line segment s in [0, 1], where the size of s is an element of (0, 0.5], how can we allow infinitely many line segments { [0, 0.5], [0.5, 0.75], [0.75, 0.875], … } in [0, 1], where the size of each line segment is an element of (0, 0.5]? We wouldn’t allow an infinite number of copies of any element of { [0, 0.5], [0.5, 0.75], [0.75, 0.875], … } in [0, 1], right?
 October 2nd, 2012, 04:07 PM #13 Global Moderator     Joined: Nov 2006 From: UTC -5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms Re: Dichotomy paradox reverse engineering Huh?
October 2nd, 2012, 10:44 PM   #14
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Quote:
 Originally Posted by netzweltler If we allow only finitely many copies of a line segment s in [0, 1], where the size of s is an element of (0, 0.5], how can we allow infinitely many line segments { [0, 0.5], [0.5, 0.75], [0.75, 0.875], … } in [0, 1], where the size of each line segment is an element of (0, 0.5]? We wouldn’t allow an infinite number of copies of any element of { [0, 0.5], [0.5, 0.75], [0.75, 0.875], … } in [0, 1], right?
I guess number of copies is ambiguous. For example, we allow four copies of the segment [0.5, 0.75] in [0, 1], which are [0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1]. So, for each element of the set { [0, 0.5], [0.5, 0.75], [0.75, 0.875], … } is true, that the number of copies in [0, 1] is finite, right?

October 3rd, 2012, 11:33 AM   #15
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Quote:
 Originally Posted by netzweltler I guess number of copies is ambiguous. For example, we allow four copies of the segment [0.5, 0.75] in [0, 1], which are [0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1]. So, for each element of the set { [0, 0.5], [0.5, 0.75], [0.75, 0.875], … } is true, that the number of copies in [0, 1] is finite, right?
So when you say that "we allow n copies of A in B" you seem to mean that n|A| <= |B| < (n+1)|A|, where |[a, b]| = b - a. In that case you're asking if each element of {2, 4, 8, 16, ...} is finite (it is).

October 5th, 2012, 01:08 PM   #16
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Quote:
Originally Posted by CRGreathouse
Quote:
 Originally Posted by netzweltler I guess number of copies is ambiguous. For example, we allow four copies of the segment [0.5, 0.75] in [0, 1], which are [0, 0.25], [0.25, 0.5], [0.5, 0.75], [0.75, 1]. So, for each element of the set { [0, 0.5], [0.5, 0.75], [0.75, 0.875], … } is true, that the number of copies in [0, 1] is finite, right?
So when you say that "we allow n copies of A in B" you seem to mean that n|A| <= |B| < (n+1)|A|, where |[a, b]| = b - a. In that case you're asking if each element of {2, 4, 8, 16, ...} is finite (it is).
Does it make some kind of sense, that there is enough space for infinitely many segments { [0, 0.5], [0.5, 0.75], [0.75, 0.875], … } in [0, 1] even if there is no space for infinitely many copies of one of these segments of this set?

October 5th, 2012, 01:30 PM   #17
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Quote:
 Originally Posted by netzweltler Does it make some kind of sense, that there is enough space for infinitely many segments { [0, 0.5], [0.5, 0.75], [0.75, 0.875], … } in [0, 1] even if there is no space for infinitely many copies of one of these segments of this set?

Yes it makes perfect sense. The infinitely many intervals keep getting smaller and smaller. The sum of their lengths converges.

You're asking if it makes sense that

1/2 + 1/4 + 1/8 + /16 + ... = 1

but

1 + 1 + 1 + 1 + ... is infinite.

Well yes, it makes perfect sense. Doesn't it?

In fact it's a necessary condition of a convergent infinite series that its terms go to zero. All you've done is illustrate that fact, which is proved in freshman calculus.

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