I believe that the set of all R→R continuous functions is c, the cardinality of the continuum. However, I read in the book “Metric spaces” by Ó Searcóid that set of all [0,1]→R continuous functions is

greater thanc:“It is demonstrated in many textbooks that Q

is countable, that 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 R, and that there are sets of greater and greater cardinality.”I understand that (via composition with the continuous function tan or arctan) these sets of continuous functions have the same cardinality. Therefore, which claim is correct, and how do I prove this?

**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 RN 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 .a1a2...,.b1b2...,.c1c2..., we can splice them together via .a1b1a2c1b2a3... so that a sequence of reals can be encoded by one real number.

**Attribution***Source : Link , Question Author : kennytm , Answer Author : Michael Hardy*