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 ?
a Mapping problem
2
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real-analysis
functions
1 Answers
2
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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0Please let me know in wordings as i am not able to follow the notation. – 2010-11-08
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0what is $\aleph_0$ ? – 2010-11-08
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0Aleph-null (i.e. countable infinity). See: http://en.wikipedia.org/wiki/Aleph_number – 2010-11-08
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2@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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0the 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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0@Rajesh: That is correct. – 2010-11-08
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2I think your exponent should be c, the power of the continuum. Aleph without a subscript is unclear. – 2010-11-08