How to prove that the set of all bijections from the reals to the reals have cardinality c = card. of reals?

it isn't true. Sorry.





as you noted, the cardinality is actually 2^c (or larger).





but we know that FOR ANY SET S, if n is the cardinality of s, then s %26lt; 2^s. So the cardinality in question is %26gt;= 2^c %26gt;c, so can't equal c.





Or did I misunderstand your comment?





EDIT:Danny: we don't know for sure (without some other methods) that the set of bijections has the same cardinality as the set of all functions from a set to itself, since the bijections make a subset.





In fact, for finite cardinality sets, this is not true. Bijections of an n-element set have cardinality n!, while the set of all functions from an n element set to itself has cardinality n^n.

How to prove that the set of all bijections from the reals to the reals have cardinality c = card. of reals?
c = 2^(N), where N is the cardinality of natural numners.


Now Nc = c.


The cardinality of the set or all real functions of a real variable has cardinality c^c, by definition


Now c^c = (2^N)^c = 2^(Nc) = 2^c.


If we assume the continuum hypothesis, yes c%26lt;2^c





OK, I think I have a proof. Let S be the set of all bijections from real to reals. I show that the cardinality of S is at least 2^c. Since by above the cardinality is at most 2^c will imply that is equal with 2^c.





Let P the set of subsets of R.


Let f: P ---%26gt; S, defined in the following way:


If A is in P then let g be a bijection of A which doesn't have a fixed point i.e. there is no x such that g(x) = x


We define f(a) = g(a) if a is in A


f(a) = a if a is not in A.


We still have to define the image of empty set and the images of sets which have only one element. But this can be easily done (exercise)


Finally f is injective so the cardinality of S is at least cardinality of P i.e. cardinality of S is at least 2^c.
Reply:I don't know if it helps, but I saw on Wikipedia that the cardinal of continuous functions from R to R is c while the cardinal of all functions is 2^c. But I don't know where to go from there. One would need to construct a bijection from one of these sets to the set of bijectives functions.

kudzu