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Kim Lee
science forum beginner
Joined: 09 May 2006
Posts: 3
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 Posted: Tue May 09, 2006 1:50 am Post subject:
Countable set question
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Hello everyone, I am preparing for a final and would like some help.
Suppose:
A_0 >= A_1 >= A_2 >= ... >= A_alpha >= ... (alpha < w_1) is a sequence
of disjoint sets. (A >= B reads as "B is a subset of A").
How does one prove or disprove that there must be an alpha < w_1 with
A_alpha a countable set? |
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Robert B. Israel
science forum Guru
Joined: 24 Mar 2005
Posts: 2151
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| Posted: Tue May 09, 2006 2:03 am Post subject:
Re: Countable set question
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In article <ktCdnXQSm832aMLZnZ2dnUVZ_vqdnZ2d@comcast.com>,
Kim Lee <kimmy4life@comcast.net> wrote:
| Quote: | Hello everyone, I am preparing for a final and would like some help.
Suppose:
A_0 >= A_1 >= A_2 >= ... >= A_alpha >= ... (alpha < w_1) is a sequence
of disjoint sets. (A >= B reads as "B is a subset of A").
How does one prove or disprove that there must be an alpha < w_1 with
A_alpha a countable set?
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???
If A_0 >= A_1 and they are disjoint, then A_1 must be empty...
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada |
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Kim Lee
science forum beginner
Joined: 09 May 2006
Posts: 3
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| Posted: Tue May 09, 2006 2:09 am Post subject:
Re: Countable set question
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Robert Israel wrote:
| Quote: | In article <ktCdnXQSm832aMLZnZ2dnUVZ_vqdnZ2d@comcast.com>,
Kim Lee <kimmy4life@comcast.net> wrote:
Hello everyone, I am preparing for a final and would like some help.
Suppose:
A_0 >= A_1 >= A_2 >= ... >= A_alpha >= ... (alpha < w_1) is a sequence
of disjoint sets. (A >= B reads as "B is a subset of A").
How does one prove or disprove that there must be an alpha < w_1 with
A_alpha a countable set?
???
If A_0 >= A_1 and they are disjoint, then A_1 must be empty...
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
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what if it read:
Suppose A_0 >= A_1 >= A_2 >= ... >= A_alpha >= ... (alpha < w_1) is a
sequence of sets, with Intersection A_alpha (for alpha < w_1) = null set. |
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Ryan Reich
science forum Guru Wannabe
Joined: 21 May 2005
Posts: 120
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| Posted: Tue May 09, 2006 3:35 am Post subject:
Re: Countable set question
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Kimmy wrote:
| Quote: | Robert Israel wrote:
In article <ktCdnXQSm832aMLZnZ2dnUVZ_vqdnZ2d@comcast.com>,
Kim Lee <kimmy4life@comcast.net> wrote:
Hello everyone, I am preparing for a final and would like some help.
Suppose:
A_0 >= A_1 >= A_2 >= ... >= A_alpha >= ... (alpha < w_1) is a sequence
of disjoint sets. (A >= B reads as "B is a subset of A").
How does one prove or disprove that there must be an alpha < w_1 with
A_alpha a countable set?
???
If A_0 >= A_1 and they are disjoint, then A_1 must be empty...
Robert Israel israel@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
what if it read:
Suppose A_0 >= A_1 >= A_2 >= ... >= A_alpha >= ... (alpha < w_1) is a
sequence of sets, with Intersection A_alpha (for alpha < w_1) = null set.
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Well then it's false: take the sequence where for each i, A_i is w_1
with the ordinals less than i removed. Any tail of w_1 is uncountable,
so none of the A_i is countable (indeed, at each stage only countably
many elements have yet been removed), yet their intersection is empty.
The problem got a lot easier when I realized that it was a "prove or
disprove", not a "prove". :)
--
Ryan Reich
ryan.reich@gmail.com |
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