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Re: Existence of a function
Posted:
Aug 15, 2013 4:14 PM
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> Does there exist a function f: R --> R such that
> f(f(x)) =/= x for all x in R and
>
> for every a in R there exists a sequence {x_n} such
> that x_n --> f(a) and f(x_n) --> a as n-->oo ?
>
>
> Message was edited by: Pastamak
Probably yes, but I don't have a construction.
Obviously f cannot be continuous.
I will work with (0,1) rather than R.
We would like a set P of points in (0,1)x(0,1)
{(X,Y)} = {(X,f(X)} such that
P is dense in (0,1)x(0,1),
each X corresponds to exactly one f(X),
and f(f(X)) =/= X.
The existence of space-filling curves tells us it won't
be hard to make P dense, and I suspect a "random" function
f would work, after the occasional violation f(f(x))=x
is taken care of.
It would be better to construct a concrete f,
but I don't see it offhand.
Once we have P, for a given A in (0,1),
find a sequence of points
(x_n, f(x_n)) in P converging to (f(A),A).
Don Coppersmith
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