Metric on $[-\infty,\infty]$ by homeomorphism I want to precise a topology to use on $[-\infty,\infty]$. This space is homeomorphic to  $[0,1]$ by taking $$f(x) = \tan (\pi(x-1/2))$$ as map. Taking $d(x,y) = |x-y|$, $x,y\in[0,1]$ can I "map" the metric using $f$ to get a metric on $[-\infty,\infty]$? Maybe $\overline{d}(x,y) = d(f^{-1}(x),f^{-1}(y))$ would work?
 A: Yes, that works, and is a standard way of getting a metric on some space. We have
Lemma: Let $(X,\,d)$ a metric space, and $\varphi \colon Y \to X$ a bijection. Then $\delta \colon Y \times Y \to \mathbb{R};\; \delta(y,\,z) = d(\varphi(y),\,\varphi(z))$ is a metric on $Y$, and $\varphi$ is an isometry (in particular a homeomorphism) between $(Y,\,\delta)$ and $(X,\,d)$.
Proof: We have $\delta(y,\,z) \geqslant 0$, since $d$ takes only non-negative values, and
$$\delta(y,\,z) = 0 \iff d(\varphi(y),\,\varphi(z)) = 0 \iff \varphi(y) = \varphi(z) \iff y = z.$$
We have $\delta(y,\,z) = d(\varphi(y),\,\varphi(z)) = d(\varphi(z),\,\varphi(y)) = \delta(z,\,y)$, so $\delta$ is symmetric.
We have $\delta(x,\,z) = d(\varphi(x),\,\varphi(z)) \leqslant d(\varphi(x),\,\varphi(y)) + d(\varphi(y),\,\varphi(z)) = \delta(x,\,y) + \delta(y,\,z)$ for all $x,\,y,\,z \in Y$, hence $\delta$ satisfies the triangle inequality, hence is a metric.
By the very definition of $\delta$, $\varphi$ is an isometry $Y \to X$, hence also a homeomorphism for the topologies induced by $d$ resp. $\delta$.
Furthermore, we also have the
Lemma: Let $(Y,\tau)$ a topological space, and $(X,\,d)$ a metric space, and $\varphi \colon Y \to X$ a homeomorphism. Then the metric $\delta$ on $Y$ constructed as above induces the topology $\tau$ on $Y$.
Proof: Let $y \in Y$ and $V$ a $\tau$-neighbourhood of $y$. Since $\varphi$ is open (because it's a homeomorphism), $\varphi(V)$ is a neighbourhood of $\varphi(y)$, hence there is an $\varepsilon > 0$ with $d(x,\,\varphi(y)) < \varepsilon \Rightarrow x \in \varphi(V)$. But that means that $B^\delta_\varepsilon(y) = \{ z \in Y : \delta(z,\,y) < \varepsilon\} \subset V$, i.e. the topology induced by $\delta$ is finer than $\tau$.
Conversely, let $y \in Y$ and $\varepsilon > 0$. Then $\varphi(B^\delta_\varepsilon(y)) = B^d_\varepsilon(\varphi(y))$ is a neighbourhood of $\varphi(y)$, and by continuity of $\varphi$, there is a $\tau$-neighbourhood $V$ of $y$ with $\varphi(V) \subset B^d_\varepsilon(\varphi(y)) \iff V \subset B^\delta_\varepsilon(y)$, which means that $\tau$ is finer than the topology induced by $\delta$.
