Tightenning Interval Mapping I am having problems formalizing a proof for the following statement contained in another proof (regarding existance of periodic point for continuous mappings of the real line):
Consider a continuous mapping $f \colon R \to R$ where $R$  is  the real line. Consider two compact intervals $I_1 = [a_1,b_1]$ and $I_2 = [a_2, b_2]$. Supose that $I_1 \subseteq f(I_2)$. Prove that there exist two points $x,y \in I_2$ such that $f([x,y]) = I_1$.
The result seems intuitive in a drawing and the book I am reading makes it seem pretty straightforward. Is it really straightforward? Any ideas for getting a real proof? Any help would be appreciated.
 A: This is an example of an intuitively obvious property, whose proof
is not that simple. I don’t think you can avoid using infima and
suprema of sets in the proof.
You can proceed as follows, by using those two lemmas :
Lemma 1 (making the extremities fit). Suppose that $f$ is continous, that $J=[a,b]$ and $J'=[a',b']$ are two compact intervals such that $J \subseteq f(J')$. Then there
is a third interval $J''=[a'',b''] \subseteq J'$ such that
$J \subseteq f(J'')$ and either $f(a'')=a$ and $f(b'')=b$, or
$f(a'')=b$ and $f(b'')=a$.
Lemma 2 (dichotomy). Suppose that $f$ is continous, that $J=[a,b]$ and $J'=[a',b']$ are two compact intervals such that $J \subseteq f(J')$. If $f(J')\neq J$, so that
there is a $c'\in J'$ such that $f(c')\not\in J$, then there is an interval
$J''$ which is one of $[a',c']$ or $[c',b']$ satisfying $J \subseteq f(J'')$.
Proof of lemma 1 By the intermediate value theorem, (ITV)
there is a $x\in J'$ such that $f(x)=a$, and 
a $y\in J'$ such that $f(y)=b$.  Then the interval 
$J''=[{\sf min}(x,y),{\sf max}(x,y)]$ connecting $x$ 
to $y$ does the job, because of ITV again.
Proof of lemma 2 Applying lemma 1, we find $z_1,t_1$ such that
$a'\leq z_1 <t_1 \leq b'$, $J\subseteq f([z_1,t_1])$ and $f$ sends the 
extremities of $[z_1,t_1]$ to the extremities of $[a,b]$. If $c'\not\in[z_1,t_1]$, then one of $[a',c']$ or $[c',b']$ contains $[z_1,t_1]$ and we are done. So we can assume that $c'\in[z_1,t_1]$.  The value 
$f(c')$ is either $<a$ or $>b$ : two possibilities.
Also, $f(z_1)=a$ and $f(t_1)=b$, or $f(z_1)=b$ and $f(t_1)=a$ : two possibilities
again. This makes a total of $2\times 2=4$ cases. I explain one case at random
here, all the others are similar. Suppose for example that $f(c')>b$ and
$f(a')=b$ and $f(b')=a$. By ITV, there is an $x\in [c',b']$ such that $f(x)=b$,
and we may take $J''=[c',b']$.
Proof (of your result) Applying lemma 1, we find $z_1,t_1$ such that
$a_2\leq z_1 <t_1 \leq b_2$, $I_1\subseteq f([z_1,t_1])$ and the extremities
of $I_1$ and $[z_1,t_1]$ fit. Now, consider the set
$$
A=\lbrace a\in [z_1,t_1] | I_1\subseteq f([a,t_1]) \rbrace \tag{1}
$$
Note that $A$ is nonempty because $z_1\in A$. So that 
$\alpha={\sf sup}(A)$ is well-defined and in $[z_1,t_1]$. Let $y\in I_1$.
For any integer $n\geq 1$, there is an $a_n\in A$ such that
$\alpha-\frac{1}{n} < a_n< \alpha$ (by definition of sup). Then
$y\in f([a_n,t_1])$, so that there is a $u_n\in[a_n,t_1]$ such that
$f(u_n)=y$. The sequence $(u_n)_{n\geq 1}$ is bounded  as it stays inside
$[z_1,t_1]$, so by the Bolzanno-Weierstrass property it must have a convergent
subsequence. If $u$ is the limit of that subsequence, we must have $f(u)=y$.
Since this works for any $y\in I_1$, we have just shown that
$$
I_1 \subseteq f([\alpha,t_1]) \tag{2}
$$
Let us now repeat on the right what we just did on the left : put
$$
B=\lbrace b\in [\alpha,t_1] | I_1\subseteq f([a,b]) \rbrace \tag{3}
$$
As above, $B$ will have an inf $\beta$ satisfying $I_1 \subseteq f([\alpha,\beta])$.
Using lemma 1 again,we find numbers $z_2,t_2$
satisfying $\alpha\leq z_2 <t_2 \leq \beta$, $\lbrace f(z_2),f(t_2) \rbrace=
\lbrace a_1,b_1 \rbrace$ and $I_1 \subseteq f([z_2,t_2])$. Because of the definition
of $\alpha$ and $\beta$, we must have $z_2=\alpha$ and $t_2=\beta$. Finally, if
$f([\alpha,\beta]) \not\subseteq I_1$, using lemma 2 we would contradict
the definition of $\alpha$ or $\beta$. In the end $f([\alpha,\beta])=I_1$ by double inclusion.
