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I need to figure out what is the power of the group of functions from R to R that for each x that is not from Q, it's f(x) belongs to {x,x+1}

I can make a mapping for each function to be Q-->R * (R/Q)--> {x,x+1}
Does it ok?


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Hint: Do you know the cardinality(power) of the set(group) of functions from $\mathbb{R}$ to $\mathbb{R}$ with no restriction? Can you see how to argue that this must be the same? Alternately, you should be able to follow through the argument of the power of the unrestricted functions and make it work the same.

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Thank you for the comment! – user6163 Feb 11 '11 at 14:22
@Nir: I updated the terminology as suggested by Arturo Magidin in another post of yours. – Ross Millikan Feb 11 '11 at 17:47

For each $x$ in $\mathbb{R}-\mathbb{Q}$, you have two possible values, so the cardinality is at least $2^{|\mathbb{R}-\mathbb{Q}|}$, which is the same as $2^{|\mathbb{R}|}$, also known as $\beth_2$, since $|\mathbb{R}-\mathbb{Q}|=|\mathbb{R}|$. On the other hand, the cardinality is also at most $\beth_2$, since that is the cardinality of the set of all functions $\mathbb{R}\to\mathbb{R}$.

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Thank you for the comment! – user6163 Feb 11 '11 at 14:21

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