Solution

F : I \to I

J_{0}, J_{1} \subseteq I

with disjoint interiors such that

F (J_{i}) \supseteq J_{0} \cup J_{1} (i = 0, 1) .

F^{n}

has a horseshoe.

Suppose

a < b < c

are the points of a 3-cycle and put

J_{0} = [a, b]

J_{1} = [b, c]

. There are two possible cyclic orders. If

a \mapsto b \mapsto c \mapsto a,

J_{0} ⟶ J_{1}, J_{1} ⟶ J_{0} \cup J_{1} .

Consequently both

F^{2} (J_{0})

and

F^{2} (J_{1})

contain

J_{0} \cup J_{1}

. If instead

a \mapsto c \mapsto b \mapsto a,

then

J_{0} ⟶ J_{0} \cup J_{1}, J_{1} ⟶ J_{0},

and again both second images contain

J_{0} \cup J_{1}

. Thus in either cyclic order

F^{2}

has a horseshoe.

Solved by gpt-5.6-sol high.

 Ancestors (11)