Let $A=\operatorname{Axis}(\phi)$ and let $p$ be the nearest-point projection of $x$ to $A$. If the translation length is $\tau>0$, the unique path from $x$ to $\phi x$ is the concatenation
$$
[x,p]\cup[p,\phi p]\cup[\phi p,\phi x].
$$
The first and last pieces both have length $d(x,A)$, while the middle one has length $\tau$. Therefore
$$
d(x,\phi x)=\tau+2d(x,A).
$$
The displacement is minimized exactly when $x\in A$, proving that the axis is the minimum set of the <displacement function>.

On $A$, the power $\phi^r$ translates through $r\tau$ when $r>0$ and in the opposite direction through $|r|\tau$ when $r<0$. The same formula applied to $\phi^r$ shows that its minimum set is $A$. Hence
$$
\operatorname{Axis}(\phi^r)=\operatorname{Axis}(\phi)
\qquad(r\ne0).
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 133 / 2 / b / Solution

For $x,y\in X$, the <triangle inequality> and the fact that $\phi$ is an <isometry> give
$$
\begin{aligned}
|d_\phi(x)-d_\phi(y)|
&=|d(x,\phi x)-d(y,\phi y)|\\
&\leq d(x,y)+d(\phi x,\phi y)\\
&=2d(x,y).
\end{aligned}
$$
Thus the <displacement function> is $2$-Lipschitz, hence <continuous function>[continuous].

Solved by gpt-5.6-sol high.

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 133 / 4 / a / Solution

{title2=$d_g(x)$}

For an isometry $g$ of a metric space, its displacement function is $d_g(x)=d(x,gx)$. It is $2$-Lipschitz.


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