Proposition: For any bijection $f:I\to B\subset \mathbb{R}$ on an
interval
i 
We understand this as a common notation for open and closed intervals. In the open case we also allow the values $\infty$ for the right and $\infty$ for the left boundery. Thus $\mathbb{R}$ and e.g. $\mathbb{R}}^{>0$ are regarded as intervals as well.

we have:
f continuous$\text{\hspace{1em}}\Rightarrow \text{\hspace{1em}}{f}^{1}$ continuous

[6.3.11] 
Proof: We start with a closed interval I and proceed indirectly. Suppose that there is an $a\in B$ and a sequence $({a}_{n})$ in B such that ${a}_{n}\to a$ but with $({f}^{1}({a}_{n}))$ not converging to ${f}^{1}(a)$. Then we find an $\epsilon >0$ such that the special $\epsilon$neighbourhood $]$ lacks almost all of the sequence members. Thus we can create the recursion
$\begin{array}{l}{k}_{1}>1\text{\hspace{1em}}\wedge \text{\hspace{1em}}{f}^{1}({a}_{{k}_{1}})\notin ]{f}^{1}(a)\epsilon ,{f}^{1}(a)+\epsilon [\\ {k}_{n+1}>{k}_{n}\text{\hspace{1em}}\wedge \text{\hspace{1em}}{f}^{1}({a}_{{k}_{n+1}})\notin ]{f}^{1}(a)\epsilon ,{f}^{1}(a)+\epsilon [\end{array}$
to get a subsequence $({f}^{1}({a}_{{k}_{n}}))$ in I with all it's members being outside $]{f}^{1}(a)\epsilon ,{f}^{1}(a)+\epsilon [$ [1]. As I is closed $({f}^{1}({a}_{{k}_{n}}))$ is a bounded sequence and thus has an accumulation point b according to the BolzanoWeierstrass theorem [5.8.6]. Due to [5.8.5] there is now a further subsequence $({f}^{1}({a}_{{k}_{{j}_{n}}}))$ such that
${f}^{1}({a}_{{k}_{{j}_{n}}})\to b$
With I being closed we have $b\in I$ and as f is continuous at b we see ${a}_{{k}_{{j}_{n}}}\to f(b)$. On the other hand we know that ${a}_{{k}_{{j}_{n}}}\to a$. Therefor we have $f(b)=a$ i.e. $b={f}^{1}(a)$. But this actually means ${f}^{1}({a}_{{k}_{{j}_{n}}})\to {f}^{1}(a)$ and thus contradicts [1].
Now let I be any interval and $a\in B$ be arbitrary. We find a closed interval $J\subset I$ with ${f}^{1}(a)\in J$. The restriction $fJ$ is continuous on J and using a result from part 6 ([6.6.6]) we find that $f(J)$ is an interval. Thus $f}^{1$ and $(fJ)}^{1$ locally coincide at a. So $f}^{1$ is continuous at a because we know from the first part that $(fJ)}^{1$ is continuous at a.
