Friday 25 April 2014

elementary set theory - Cardinality of set of real continuous functions



The set of all $\mathbb{R\to R}$ continuous functions is $\mathfrak c$. How to show that? Is there any bijection between $\mathbb R^n$ and the set of continuous functions?


Answer



The cardinality is at least that of the continuum because every real number corresponds to a constant function. The cardinality is at most that of the continuum because the set of real continuous functions injects into the sequence space $R^{N}$ by mapping each continuous function to its values on all the rational points. Since the rational points are dense, this determines the function.



The Schroeder-Bernstein theorem now implies the cardinality is precisely that of the continuum.




Note that then the set of sequences of reals is also of the same cardinality as the reals. This is because if we have a sequence of binary representations $.a_1a_2..., .b_1b_2..., .c_1c_2...$, we can splice them together via $.a_1 b_1 a_2 c_1 b_2 a_3...$ so that a sequence of reals can be encoded by one real number.


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