= Period three implies all periods
For a <continuous function> from an interval to itself with a three-cycle $a<b<c$, reflection if needed gives $F(a)=b,F(b)=c,F(c)=a$. Set $I=[a,b],J=[b,c]$. The <intermediate value theorem> gives $F(I)\supseteq J$ and $F(J)\supseteq I\cup J$. Successive closed-interval pullbacks realize every cyclic covering word by a point fixed under the corresponding iterate. The word $I J^{n-1}$ has a unique visit to the interior of $I$ per cycle, forcing least period $n$; the only boundary ambiguity is the original three-cycle, which realizes $n=3$ and cannot realize any other such word. A covering of $J$ by itself also gives a fixed point. Thus every positive least period occurs.
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