I been trying to find a total ordering on $\displaystyle [0,1]^{[0,1]}$, the set of all functions from $\displaystyle [0,1]\mapsto [0,1]$. This cardinality of this set is $\displaystyle 2^{ 2^{\aleph_0}} > 2^{\aleph_0}$. I tried a few ideas and they all failed. I happen to know a total ordering exists by the axiom of choice (this is correct, right?). But I am trying to find anexplicit orderingon this set for a counterexample.