1. ## Metric spaces

Let A not be the empty set, and be a finite subset of R^2 with the Euclidean metric, d. Prove that A with the relative metric d|A is equivalent to (A, d') where d' is the discrete metric.

I am not even sure what this is asking me to do. Any help would be great! Thanks!

2. I guess that by "equivalent" you meant homeomorphic. in the discrete space every point is an open set, meaning $\forall x \in A, \; \{ x \}$ is an open set.
now, for $A \subset \mathbb{R}^2$ where A is finite (nonempty) subspace, define $m = min \{ d(x,y) | x,y \in A, x \neq y \}$, or m=1 if |A|=1. because A is finite, m is well defined. now for every $x \in A$ the ball of radius m/2 around x contain only x, so {x} is an open set - therefore this topology is homeomorphic to the discrete topology

3. That all makes sense to me, but the only thing I'm confused about is the "relative metric denoted d|A. When it says prove A with the relative metric d|A is equivalent to (A,d') where d' is the discrete metric. I'm not sure what the relative metric means?

4. the metric d is defined over all R^2. to make A into a metric space you need a metric on A, so d|A is probably the notation for the metric d reduced to the points in A

5. A function f is homeomorphic if :
- f is bijective (and continuous)
- $f^{-1}$ is continuous

2 distances d (in A) and d' (in A) are equivalent if there exist $M,N > 0$ such that :
$\forall x,y \in A ~,~ M d(x,y) \leqslant d'(x,y) \leqslant N d(x,y)$

is it ?