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let $f$:R->N such that given any $x_1$,$x_2$ belong to R such that $x_1 < x_2$ then there exists a $x_3$ belongs to $(x_1,x_2)$ such that $f(x_3)> max(f(x_1),f(x_2))$. How many such mappings are possible ?

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There are $\aleph_0^\aleph$ such mappings. The upper bound is trivial. For the lower bound, pick any function $f\colon \mathbb{R} \longrightarrow \mathbb{N}$ such that $f(\pm \frac{p}{q}) = q$.

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    Please let me know in wordings as i am not able to follow the notation.2010-11-08
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    what is $\aleph_0$ ?2010-11-08
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    Aleph-null (i.e. countable infinity). See: http://en.wikipedia.org/wiki/Aleph_number2010-11-08
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    @Rajesh D: I believe that Yuval means that the cardinality of the set of such functions ("how many" there are) is equal to the cardinality of the set of functions from $\mathbb{R}$ to $\mathbb{N}$, because the map can send the irrationals anywhere in the examples given, and the irrationals have the same cardinality as $\mathbb{R}$.2010-11-08
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    the answer is cool! So the cardinality of the set of mappings is even greater than that of Real numbers ?(please let me know whether i got it correct.2010-11-08
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    @Rajesh: That is correct.2010-11-08
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    I think your exponent should be c, the power of the continuum. Aleph without a subscript is unclear.2010-11-08