Period of 3 implies chaos Let $f(x)$ be a continuous function from $\mathbb{R}\rightarrow\mathbb{R}$.
Let's denote $k$-times repeated application of the function, $f(f(f(...f(x)...)))$ as $f^{(k)}(x)$.
Let's call any $x$ a periodic point with period $n$ if $f^{(n)}(x)=x$.
Is it true that if a point with period 3 exists, then points with all possible periods exist?
In other words is it true that
$$\exists x:f^{(3)}(x)=x\Rightarrow \forall n>0 \exists y:f^{(n)}(y)=y$$
and if so, why?
 A: See the Monthly paper of that exact title Period three implies chaos - which made the JSTOR All-Stars list as the third most frequently accessed Monthly article. See also A simple proof of Sharkovsky's theorem and this paper for some history on Sharkovsky's theorem.
A: I found Chaos: An Introduction to Dynamical Systems by Kathleen T. Alligood, Tim D. Sauer, and James A. Yorke, to be a sensible text.  A small section of the text contains a brief explanation of Sharkovsky's Theorem.  
In short, Sharkovsky uses this unique ordering of the natural numbers in such a way as to ensure that for each natural number n, the existence of a period-n point implies the existence of periods containing all the periods of orders higher than n. In my opinion, the text provides a reasonable simplification of this ordering and gently walks one through a decipherable step by step proof.
A: This is (a special case of) a famous theorem, and of course there are many published proofs. However, knowing the statement, you might be able to work out a proof yourself.
The simple key fact for the proof is:
Lemma. Suppose that $f$ is a continuous function on the compact interval $I=[a,b]$ such that the image of $f$ contains $I$. Then $f$ has a fixed point.
(Hint: intermediate value theorem.)
Now suppose you have a period 3 orbit. First of all think about how the three points $x_1<x_2<x_3$ of the orbit can be permuted by the function - you will notice that you can assume without loss of generality that $f(x_1)=x_2$, $f(x_2)=x_3$ and $f(x_3)=x_1$. Consider the two intervals $I_1=[x_1,x_2]$ and $I_2=[x_2,x_3]$. What do we know about the image of these intervals under $f$? Already you should see that $f$ must have a fixed point. With a little bit more work, you can see that $f$ has periodic points of every period $n$. (Show that there is a subinterval $J$ of $I_2$ such that $f^{n-1}(J)= I_1$ and $f^j(J)\subset I_2$ for $1\leq j<n-1$.)
A: I believe you want Sharkovsky's Theorem.
A reasonable explanation is in:
Kaplan, Harvey. "A cartoon-assisted proof of Sarkowskiĭ's theorem."
Amer. J. Phys. 55 (1987), no. 11, 1023–1032.
MR917121
DOI:10.1119/1.14928"
