Countability subset of the reals proof

hlin818
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Homework Statement



Let (a,b)=XUY, X,Y arbitrary sets where (a,b) is an arbitrary interval. Prove that either X or Y has the same cardinality as that of (a,b).

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The Attempt at a Solution



Really lost.
 
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first show (a,b) is uncountable this shows XUY must be uncountable, now assume X & Y are both countable and look for a contradiction
 
lanedance said:
first show (a,b) is uncountable this shows XUY must be uncountable, now assume X & Y are both countable and look for a contradiction

If we do this by contradiction wouldn't the negation be to assume that X and Y do not have cardinality equal to (a,b)?
 
no, you wnat a ssume that neither has cardinality equivalent to an interval and find a contradiction that says it can't be so
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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