For $n_1,n_2,\ldots\in\N$ we put \begin{eqnarray*} I_{n_1}&\colon=&(\la n_1+1\ra,\la n_1\ra),\\ I_{n_1,n_2}&\colon=&(\la n_1,n_2\ra,\la n_1,n_2+1\ra),\\ I_{n_1,n_2,n_3}&\colon=&(\la n_1,n_2,n_3+1\ra,\la n_1,n_2,n_3\ra, \quad\mbox{etc.} \end{eqnarray*} Then the following holds:
  1. $\theta$ is a homeomorphism from $I_{n_1,\ldots,n_k}$ onto $I_{n_2,\ldots,n_k}$.
  2. $I_{n_1,\ldots,n_k,n_{k+1}}\sbe I_{n_1,\ldots,n_k}$.
  3. For all $j\leq k$: $a_j|I_{n_1,\ldots,n_k}=n_j$.
  4. The open interval $I_{n_1,\ldots,n_k}$ has length at most $2^{-k}$. Thus if $a_j(x)=a_j(y)=n_j$ for all $j\leq k$, then $|x-y|\leq2^{-k}$.
1. is a consequence of exam 2.
2. Suppose $x\in I_{n_1,\ldots,n_{k+1}}$, i.e. $x=\la n_1,\ldots,n_{k+1}+t\ra$. We show by induction that for $n$ even and $n$ odd, respectively: \begin{eqnarray*} &&\la n_1,\ldots,n_k+1\ra<\la n_1,\ldots,n_{k+1}+t\ra<\la n_1,\ldots,n_k\ra \quad\mbox{and}\\ &&\la n_1,\ldots,n_k+1\ra>\la n_1,\ldots,n_{k+1}+t\ra>\la n_1,\ldots,n_k\ra \end{eqnarray*} For $k=1$ this follows from $n_1+1 > n_1+\la n_2+t\ra > n_1$. For odd $k$ this is equivalent to: $$ \la n_2,\ldots,n_k+1\ra > \la n_2,\ldots,n_{k+1}+t\ra > \la n_2,\ldots,n_k\ra $$ but that's just the assertion for $k$ even.
3. For all $x\in I_{n_1,\ldots,n_k}$ there is $t\in(0,1)$ - in fact $t=\theta^k(x)$, such that $x=\la n_1,\ldots,n_k+t\ra$. It follows that $a_1(x)=[1/x]=[n_1+\la n_2,\ldots,n_k+t\ra]=n_1$, $a_2(x)=[1/\theta(x)]=[1/\la n_2,\ldots,n_k+t\ra]=n_2$, etc.
4. Let $x,y\in I_{n_1,\ldots,n_k}$, then there are $a,b\in I_{n_3,\ldots,n_k}$ such that $x=1/(n_1+1/(n_2+a))$ and $y=1/(n_1+1/(n_2+b))$. Hence $$ y-x =\frac{n_2+a}{n_1(n_2+a)+1}-\frac{n_2+b}{n_1(n_2+b)+1}\\ =\frac{a-b}{(n_1(n_2+b)+1)(n_1(n_2+a)+1)} $$ i.e.: $|y-x|\leq|a-b|/4$.