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We have a set S which is uncountably infinite. U is a subset of S and it is uncountably infinite.

What is the cardinality of the complement of U with respect to S?

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closed as too localized by François G. Dorais, Harald Hanche-Olsen, Pete L. Clark, S. Carnahan, Kevin H. Lin Apr 18 '10 at 20:35

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MO is the wrong place for this. Perhaps a homework problem? So, read the FAQ here (see the link at the top?) before posting... –  Gerald Edgar Apr 18 '10 at 13:04
    
I removed the complexity-theory tag, since this question does not involve complexity theory. –  Joel David Hamkins Apr 18 '10 at 20:03

1 Answer 1

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You cannot say anything about the cardinality of the complement. It can either be finite, countable infinite and uncountable infinite. The following examples may illustrate this.

Take $S=\mathbb{R}$. If you take $U$ to be $\mathbb{R} \setminus \{ 1 \}$. Then the complement of $U$ in $S$ has cardinality 1. If you take $U=\mathbb{R} \setminus \mathbb{Q}$ then the complement is countably infinite. If you take $U$ to be the set of numbers which are positive, then the complement is uncountably infinite.

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I understand the votes to close the question, but why was this answer downvoted? –  Qfwfq Apr 18 '10 at 19:47
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Probably because this question is readily identifiable as homework of some sort, and so should have been ignored and closed rather than answered. (I haven't voted, so am guessing at the motives of others.) –  JBL Apr 18 '10 at 20:13

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