Solution

every pair of nonempty open sets eventually overlap under an iterate, which is topological transitivity;
it has dense periodic points; and
nearby initial conditions can eventually separate by a fixed amount, which is sensitive dependence on initial conditions.

Write a point as a binary expansion

x = 0. b_{1} b_{2} b_{3} \dots_{2} .

Apart from the harmless choice of expansion at dyadic rationals, the binary shift representation of the doubling map is

F (x) = 0. b_{2} b_{3} b_{4} \dots_{2} .

Every open interval contains a binary cylinder specified by a finite initial word. Given cylinders

U

and

V

, choose a binary sequence beginning with the word for

U

and place the word for

V

after it. A suitable iterate shifts the second word to the front, proving topological transitivity.

Given any cylinder, repeat its defining word forever. The resulting point is periodic and lies in that cylinder, so periodic points are dense.

Finally, given

x

and any neighbourhood, choose

N

so large that changing only digits after the first

N

stays inside that neighbourhood. Choose the later tail so that after

N

shifts it is either

0

3/4

, whichever is farther from

F^{N} (x)

. The separation is at least

3/8

, so any smaller fixed constant, for example

δ = 1/4

, proves sensitivity. Hence the doubling map is chaotic in Devaney's sense.

Solved by gpt-5.6-sol high.

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