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The set of all ℝ → ℝ continuous functions is c. How to show that? Is there are bijection between ℝn and the set of continuous functions?

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The result that you’ve found here is correct: there are $\mathfrak c=|\Bbb R|$ continuous real-valued functions on $[0,1]$. I find it hard to believe that Ó Searcóid made such an egregious error; could you quote exactly what he says? –  Brian M. Scott Apr 21 '13 at 0:05
    
This is from page 268 (first edition): "It is demonstrated in many textbooks that $\mathbb Q$ is countable, that $\mathbb R$ is uncountable, that every non-degenerate interval is uncountable, that the collection of continuous functions defined on $[0 , 1]$ is of a greater cardinality than $\mathbb R$, and that there are sets of greater and greater cardinality." –  Andres Caicedo Apr 21 '13 at 0:29
    
(Brian's comment and mine refer to a different version of the question, merged with this one as duplicate. The question was prompted by a claim in "Metric spaces", by Mícheál Ó Searcóid, where it is claimed that there are more continuous functions on $[0,1]$ than real numbers.) –  Andres Caicedo Apr 21 '13 at 1:43
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up vote 32 down vote accepted

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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Good answer, but your last statement does not work for infinite sequences. –  Larry Wang Jul 22 '10 at 12:27
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+1 Nice. Since the rational points are dense, this determines the function. - This is the trickiest claim in the argument, enough to count as a lacuna. It might make a good further question "Can there be two distinct, continuous functions that are equal at all rationals?" –  Charles Stewart Jul 22 '10 at 12:31
    
@Kaestur: it works for countably many reals, which I think is all that was intended. –  Charles Stewart Jul 22 '10 at 12:32
    
@Charles: Fair enough. +1 from me. –  Larry Wang Jul 22 '10 at 13:19
    
By popular demand, math.stackexchange.com/questions/505/… –  Charles Stewart Jul 22 '10 at 16:53
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It is at least $c$, since all constant functions are continuous. Now consider the fact that $\mathbb{R}$ is separable.

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