MathDB

Is is true that if the perfect set

F\subseteq [0,1]

is of zero Lebesgue measure then those functions in

C^1[0,1]

which are one-to-one on

F

form a dense subset of

C^1[0,1]

? (We use the metric

d(f,g)=\sup_{x\in[0,1]} |f(x)-g(x)| + \sup_{x\in[0,1]} |f'(x)-g'(x)|

to define the topology in the space

C^1[0,1]

of continuously differentiable real functions on

[0,1]

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