# Question about relationship between metrics

1. Jun 15, 2010

### mnb96

Hello,
Let X and Y be two spaces equipped respectively with two metrics $d_1$ and $d_2$, and let's consider a mapping $f:(X,d_1)\rightarrow(Y,d_2)$

How can I formalize the fact that for any point x in X, all the points that are "very close" to x (respect to the metric $d_1$) must be mapped into points in Y that are "very close" to f(x) (with respect to the metric $d_2$).

Thanks!

2. Jun 15, 2010

You're talking about a continuous mapping f between X and Y. You can express this very similar as (for example) in real analysis (epsilon-delta), only you have to invoke the different metrics into your notation. So, for any given ε > 0 there exists some δ > 0 such that d1(p, x) < δ implies d2(f(p), f(x)) < ε - this means that f is continuous at the point p from X. If f is continuous at any point of X, then it is continuous on X.

3. Jun 15, 2010

### mnb96

Ah...I see.
and if I add the requirement that f is bijective I obtain a homeomorphism between X and Y? is it correct?

Thanks a lot.

4. Jun 15, 2010

Yes, along with the condition that the inverse of f is continuous too.

5. Jun 20, 2010

### Bacle

Now here is a point I am still confused with: when is a homeomorphism between
metric spaces as isometry.? I know every isometry is an isomorphism, but not
the other way around ( I guess the homeo. between (0,1) and R is maybe the
clearest counterexample.). Do we need some sort of Lipschitz condition.?
Anyone know.?

6. Jun 21, 2010

### zhentil

Bacle, it's precisely when the pullback metric (if it's a homomorphism, this becomes a metric) agrees with the given metric. It's an interesting result in Riemannian geometry that a homeomorphism which preserves the distance given by the Riemannian metric is actually an isometry (in the Riemannian sense) - in particular, it's smooth.