If $f:(M,d )\to (M,p)$ is a homeomorphism, are $d,p$ equivalent? Let $M$ be a set and $\delta$, $\rho$ metrics on $M$. If $f:(M,\delta)\to(M,\rho)$ is a homeomorphism, are $\delta$, $\rho$ equivalent metrics?
Not necessarly $f=\operatorname{id}_M$ (since result is obvious).
 A: The answer is no. Let $X=\{0\}\cup\{1/n:n\in\Bbb Z^+\}$, and let $\rho$ be the usual metric inherited from $\Bbb R$. Define $$f:X\to X:n\mapsto\begin{cases}1,&\text{if }n=0\\0,&\text{if }n=1\\
n,&\text{otherwise}\;.\end{cases}$$ Define a new metric $\delta$ on $X$ by $\delta(m,n)=\rho\big(f(m),f(n)\big)$. It’s easy to check that $\delta$ really is a metric on $X$ and that $f$ is not just a homeomorphism, but an isometry between $\langle X,\delta\rangle$ and $\langle X,\rho\rangle$. Both spaces are a simple sequence together with its limit point. But $0$ is isolated in $\langle X,\delta\rangle$ but not in $\langle X,\rho\rangle$, so the two metrics do not generate the same topology.
The sequence $\langle 1/n:n\ge 2\rangle$ converges to $0$ with respect to $\rho$ and to $1$ with respect to $\delta$.
A: Let $X=\mathbb{R}\times\{0\}\dot\cup\mathbb{R}\times\{2\}$, and define $\delta$ such that:
$$
\left\{\begin{array}{cccc}
\delta((x,0),(y,2))&=&1,&\forall x,y\in\mathbb{R}\\
\delta|_{(\mathbb{R}\times\{0\})\times(\mathbb{R}\times\{0\})}&=&\mbox{discrete metric}&\\
\delta|_{(\mathbb{R}\times\{2\})\times(\mathbb{R}\times\{2\})}&=&\min\{\mbox{usual metric},1\}&
\end{array}\right.
$$
That is, $X$ is a disjoint union of two $\mathbb{R}$ such that we have the discrete metric on one and the usual on other.
Define $\rho$ equivalently:
$$
\left\{\begin{array}{cccc}
\rho((x,0),(y,2))&=&1,&\forall x,y\in\mathbb{R}\\
\rho|_{(\mathbb{R}\times\{0\})\times(\mathbb{R}\times\{0\})}&=&\min\{\mbox{usual},1\}&\\
\rho|_{(\mathbb{R}\times\{2\})\times(\mathbb{R}\times\{2\})}&=&\mbox{discrete}&
\end{array}\right.
$$
If $f:X\longrightarrow X$ sends $\mathbb{R}\times\{0\}$ to $\mathbb{R}\times\{2\}$ and vice-versa, then $f^{-1}=f$ and $f:(X,\delta)\longrightarrow(X,\rho),f:(X,\rho)\longrightarrow(X,\delta)$ are continuous, so, $f$ is a homeomorphism, but $\delta,\rho$ aren't equivalents.
In general, let $(X,d)$ a metric space and define a new metric $d'$ such that $d'(x,y)=\min\{d(x,y),1\}$. Then $d'$ (as you can check) is a new metric and $d'(x,y)\le1,\forall x,y\in X$. In particular, this shows that $\rho(x,y)=\min\{d(x,y),1\}$, for $x,y\in\mathbb{R}$, is a metric in $\mathbb{R}$, where $d$ is the usual metric.
Let $(X_1,d_1),(X_2,d_2)$ metric spaces, with $d_i(x,y)\le1,\forall x,y\in X_i,i=1,2$. Define $Z=X\dot\cup Y$ and $d:Z\times Z\longrightarrow\mathbb{R}$, such that:

*$d(x,y)=d(y,x)=1,\forall x\in X_1,\forall y\in X_2$.

*$d|_{X_i\times X_i}=d_i,i=1,2$
To see the triangular inequality, let $x\in X_1,y\in X_2$ and $z\in Z$. Then, $d(x,y)= 1$ and $d(x,z)+d(z,y)\ge 1$ since $z\in X_1$ or $z\in X_2$, so, in this case holds. Now, let $x,y\in X_1$ and $z\in Z$. If $z\in X_1$, then we have $d(x,y)\le d(x,z)+d(z,y)$, since in this case $d=d_1$ and $d_1$ is a metric. If $z\in X_2$, then, since $d(x,y)\le 1$, and $d(x,z)=d(z,y)=1$, we have $d(x,y)\le d(x,z)+d(z,y)$. The same argument holds if $x,y\in X_2$.
This shows the triangular inequality for $d$, then $(Z,d)$ is a metric space. In particular, $\rho,\delta$ are indeed metric.
