3
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

$d_1$ 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).

$\endgroup$
1
$\begingroup$

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 $(X,d_1)$.

$\endgroup$
0
$\begingroup$

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 $\beta_1\ne\beta_2$.

$\endgroup$
  • $\begingroup$ This is a different meaning of "equivalent metrics". $\endgroup$ – zyx Apr 24 '16 at 14:21
  • $\begingroup$ @zyx Actually it is exactly as I wrote, you should use the canonical notion. $\endgroup$ – John B Apr 24 '16 at 15:17
0
$\begingroup$

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 $X$.

$\endgroup$
  • $\begingroup$ If $(d_1,d_2)$ is a metric on $X$, then how do you define $(d_1,d_2)(y_1,y_2)$ where $y_i\in Y_i$? $\endgroup$ – Henning Makholm Apr 24 '16 at 13:23
  • $\begingroup$ @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($d_1, d_2$). $\endgroup$ – zyx Apr 24 '16 at 13:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.