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Countable sets [0,1]
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Kim Lee
science forum beginner


Joined: 09 May 2006
Posts: 3

PostPosted: Tue May 09, 2006 7:12 am    Post subject: Countable sets [0,1] Reply with quote

So I have a few questions on countable/uncountable sets when dealing
with the closed interval [0,1].

Let < be the usual ordering on the real line.

1. Is there a countable set X subset of [0,1], ordered by <, and order
isomorphic to w^2?

2. Is there a Y subset of [0,1], Y well ordered by <, with Y
uncountable? Prove or Disprove.

I believe this is true. Is my proof here correct?
[0,1] is equinumerous to the real line and the cardinality of the real
line is 2^(aleph_0) which is uncountable.
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Denis Feldmann
science forum beginner


Joined: 23 Apr 2006
Posts: 5

PostPosted: Tue May 09, 2006 7:22 am    Post subject: Re: Countable sets [0,1] Reply with quote

Kimmy a écrit :
Quote:
So I have a few questions on countable/uncountable sets when dealing
with the closed interval [0,1].

Let < be the usual ordering on the real line.

1. Is there a countable set X subset of [0,1], ordered by <, and order
isomorphic to w^2?

Yes (use the set (1/2+1/4+...1/2^m)+(1/2^m(1/3+1/9+...1/3^p))


Quote:

2. Is there a Y subset of [0,1], Y well ordered by <, with Y
uncountable? Prove or Disprove.

No (proof : use the fact that a non-denumerable series with terms
strictly positive diverges)


Quote:

I believe this is true. Is my proof here correct?
[0,1] is equinumerous to the real line and the cardinality of the real
line is 2^(aleph_0) which is uncountable.

No : you forgot the all-important word "well-ordered"
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David C. Ullrich
science forum Guru


Joined: 28 Apr 2005
Posts: 2250

PostPosted: Tue May 09, 2006 11:43 am    Post subject: Re: Countable sets [0,1] Reply with quote

On Tue, 09 May 2006 01:12:12 -0600, Kimmy <kimmy4life@comcast.net>
wrote:

Quote:
So I have a few questions on countable/uncountable sets when dealing
with the closed interval [0,1].

Let < be the usual ordering on the real line.

1. Is there a countable set X subset of [0,1], ordered by <, and order
isomorphic to w^2?

In fact this is true with any countable ordinal in place of w^w.

Quote:
2. Is there a Y subset of [0,1], Y well ordered by <, with Y
uncountable? Prove or Disprove.

I believe this is true. Is my proof here correct?
[0,1] is equinumerous to the real line and the cardinality of the real
line is 2^(aleph_0) which is uncountable.


************************

David C. Ullrich
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Dave L. Renfro
science forum Guru


Joined: 29 Apr 2005
Posts: 570

PostPosted: Tue May 09, 2006 3:33 pm    Post subject: Re: Countable sets [0,1] Reply with quote

Kimmy wrote (in part):

Quote:
1. Is there a countable set X subset of [0,1],
ordered by <, and order isomorphic to w^2?

David C. Ullrich wrote (in part):

Quote:
In fact this is true with any countable ordinal
in place of w^w.

Or even for any countable linearly ordered set with
[0,1] replaced by the collection of rational numbers
in any specified open interval. More generally, any
nonempty [countable] linear ordering that is dense
in itself and unbounded contains order isomorphic
copies of every countable linear ordering. I think
this was proved by Cantor in the 1880's, but I don't
have my notes on linearly ordered sets with me
right now to double check this.

No ulterior motive for this post, by the way,
other than there's nothing I felt like commenting
on right now in the 11:14 5K Collatz conjecture
thread, so I thought I'd poke my nose into this
thread.

Dave L. Renfro
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The World Wide Wade
science forum Guru


Joined: 24 Mar 2005
Posts: 790

PostPosted: Tue May 09, 2006 6:00 pm    Post subject: Re: Countable sets [0,1] Reply with quote

In article <RICdnfaDqfpG3f3ZRVn-tA@comcast.com>,
Kimmy <kimmy4life@comcast.net> wrote:

Quote:
2. Is there a Y subset of [0,1], Y well ordered by <, with Y
uncountable? Prove or Disprove.

HInt: For each x in Y, there exists the "next" n(x) in Y. The
intervals (x,n(x)) are pairwise disjoint.
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