# If $d_1,d_2$ are not equivalent metrics, is it true $(X,d_1)$ is not homeomorphic to $(X,d_2)$?

Consider the statement:

If $(X,d_1)$ and $(X,d_2)$ are metric spaces and $d_1,d_2$ are not equivalent metrics, then $(X,d_1)$ is not homeomorphic to $(X,d_2)$.

I think this is true, however I can't seem to prove it. Since the metrics are not equivalent, they induce different topologies on X but is this enough to say that the spaces are not homeomorphic?

$d_1$ and $d_2$ are equivalent metrics if and only if the identity map is a homeomorphism $(X,d_1)\to(X,d_1)$.

However, it is possible for $(X,d_1)$ and $(X,d_2)$ to be homeomorphic through a non-identity map even though $d_1$ and $d_2$ are not equivalent.

For example, take $X=\mathbb R$ and define $$f(x) = \begin{cases} -x & \text{when }|x|<1 \\ x & \text{otherwise} \end{cases}$$ and then $$d_1(x,y) = |x-y| \qquad d_2(x,y) = d_1(f(x),f(y))$$

Then, plainly, $f$ is a homeomorphism $(\mathbb R,d_1)\to(\mathbb R,d_2)$, but the metrics are not equivalent, because the set of positive reals is open under $d_1$ but not under $d_2$. (More precisely: $1$ is positive, but every open ball centered on $1$ under $d_2$ contains a non-positive number).

This statement is false.

To take a concrete example: let $X = \mathbb{R}$ and let $d_0$ be the post-office metric w.r.t. $0$ and let $d_1$ be the post-office metric w.r.t. $1$.

Here the post-office metric with respect to some fixed $p$ means that $d_p(x,y) = |x - p| + |y - p|$: the distance is the Euclidean distance a letter from $x$ to $y$ takes when letters always have to go to and from the post-office point $p$.

One easily checks this is a metric, for any $p$ (we could start with any metric on any space as the basis in fact, in stead of the Euclidean one, but I want a concrete example).

All points in $(X,d_p)$ are isolated except $p$, which has the usual Euclidean neighbourhoods: $d_p(x,p) = |x-p|$ and for $q \neq p$, the ball with radius $|q-p|$ around $q$ only contains $q$ (you cannot even reach the post-office..), so $\{q\}$ is open.

The latter means that $d_0$ and $d_1$ induce different topologies on $\mathbb{R}$, as they have different open sets: $\{1\}$ is open in $(X,d_0)$ but not in $(X,d_1)$, and vice versa for $\{0\}$.

But they are homeomorphic to each other: just move one post-office to the other: $h(x) = x+1$ is a homeomorphism from $(X,d_0)$ to $(X,d_1)$.

There are many examples of nonequivalent distances generating the same topology and thus for which the corresponding topological spaces are homeomorphic (via de identity map).

As a prototype, consider the space of sequences $X^{\mathbb N}$, where $X$ is a finite set with the discrete topology, equipped with the product topology. Given $\beta>1$, the distance $$d_\beta(x_1x_2\cdots,y_1y_2\cdots)=\sum_{n=1}^\infty\beta^{-n}|x_i-y_i|$$ generates the product topology, but $d_{\beta_1}$ is not equivalent to $d_{\beta_2}$ for $\beta_1\ne\beta_2$.

• This is a different meaning of "equivalent metrics". – zyx Apr 24 '16 at 14:21
• @zyx Actually it is exactly as I wrote, you should use the canonical notion. – John B Apr 24 '16 at 15:17

Suppose $X = Y_1 \cup Y_2$ is a disjoint union of two copies of $Y$ with two inequivalent bounded metrics $d_1$ and $d_2$ on the copies. (Define the distance to be some large constant $c$ between pairs of points on different copies, so that $d_i < c$.)

Then $(d_1,d_2)$ and $(d_2, d_1)$ are inequivalent metrics on $X$ but there is a self-homeomorphism of $X$ that interchanges $Y_1$ and $Y_2$ and exchanges the two metrics on $X$.

Equivalent (in the topological sense) metrics are homeomorphic by the identity map. This does not rule out the possibility of a non-identity homemorphism of $X$ that identifies two inequivalent metrics on $X$.

• If $(d_1,d_2)$ is a metric on $X$, then how do you define $(d_1,d_2)(y_1,y_2)$ where $y_i\in Y_i$? – Henning Makholm Apr 24 '16 at 13:23
• @HenningMakholm, the two Y's should be at "infinite" distance from each other to not interfere with the triangle inequality. This can be realized in examples by taking bounded $d_i$ and placing the Y's further apart than max($d_1, d_2$). – zyx Apr 24 '16 at 13:30