Given a set *A* and two metrics, *d* and *d'*, we say that *d* and *d'* are Lipschitz equivalent on *A* if there exist positive constants *h* and *k * such that

*hd'(x,y)* ≤ *d(x,y)* ≤ *kd'(x,y)*

for any *x, y* in *A*.

Why is this a useful definition, you ask? Well, the most important thing about this definition is this:

**Theorem:** *Any two Lipschitz equivalent metrics are topologically equivalent.*

**Proof:** Let *B*_{r}(x;d) denote the open ball with radius *r*, centre *x* and metric *d*, i.e. *B*_{r}(x;d) = {*y* ∈ *A* | *d(x,y) < r*}. Now observe that, for any positive *ε*,

*B*_{hε}(x;d) ⊆ *B*_{ε}(x;d') and *B*_{ε/k}(x;d') ⊆ *B*_{ε}(x;d)

From here we can easily see that any *d*-open set is a *d'*-open set and vice versa. Hence the theorem is proved. QED

Another important result which can be proved (although I'm not going to do so here) is that the metrics induced by any two norms on a finite dimensional Banach space are Lipschitz (and hence topologically) equivalent. This has the useful side-effect that any choice of norm on **R**^{n} (or **C**^{n}) gives us the same analytic properties. This allows us to streamline many proofs in analysis, simply by making an appropriate choice of norm.