# 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).

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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 $\delta$.

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 Really good example. Thanks! The fact that in a set with two equivalent metrics a sequence must converges to the same value resp both metrics is a known fact but I dont know where I can find this theorem with respective proof. – Gastón Burrull May 22 '12 at 1:23 I can't find this theorem with respective proof. =( – Gastón Burrull May 22 '12 at 23:18 @Gastón: If $\delta$ and $\rho$ are equivalent metrics on a set $X$, by definition they generate the same topology $\tau$. If $\langle x_n:n\in\Bbb N\rangle$ is a sequence in $X$, and $\langle\rho(x_n,x):n\in\Bbb N\rangle\to 0$, then $\langle x_n:n\in\Bbb N\rangle$ converges to $x$ in the topology $\tau$. But $\delta$ also generates $\tau$, so $\langle\delta(x_n,x):n\in\Bbb N\rangle\to 0$. If $\delta$ and $\rho$ generate the same topology, they must have the same convergent sequences and limits, because those are determined by the topology. – Brian M. Scott May 22 '12 at 23:26 youre using more general definition of limit using open sets, in this case your answer is very clear!! Thanks. Topology view of point is sometimes easier than metric structure point of view. – Gastón Burrull May 22 '12 at 23:42

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.

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 Are you sure that they are metrics? Is so hard to prove triangular innequality when are many cases. – Gastón Burrull May 22 '12 at 18:42 @GastónBurrull: indeed! This isn't so immediate, sorry! Edited =p – Yuki May 23 '12 at 14:33 Wow.. really are metrics!, nice example! Thanks! – Gastón Burrull May 23 '12 at 15:47