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## Homework Statement

Let [itex] (X,d) [/itex] be a metric space. Show that d is equivalent to the discrete metric if and only if every point in X is an isolated point

## Homework Equations

Two metrics [itex] d, d' [/itex] are said to be equivalent if they have to same convergent sequences with the same limit points

The discrete metric is [itex] d(x,y) = \begin{cases} 0 & \text{ if } x = y \\ 1 & \text{otherwise} \end{cases} [/itex]

[itex] a \in X [/itex] is an isolated point if [itex] \exists r>0 [/itex] such that [itex] B(a;r) = \{a\} [/itex].

## The Attempt at a Solution

I might just be getting hung up on technicalities but I can't really think of how to attack either end of the problem.

In the [itex] \Rightarrow [/itex] direction, my problem is that I have to show it for any metric that is equivalent to the discrete metric - so I probably need to use something about convergent sequences under the discrete metric. But I can't really think of anything useful.

In the [itex] \Leftarrow [/itex] direction, I wanted to use contradiction somehow. Since in a discrete metric space every subset is necessarily open, I want to say something about a non-discrete metric -equivalent having a closed subset, but I don't know if that's really true or not.

I've seen in some literature a definition that states that a metric space can be broken up into a set of limit points and isolated points. In which case they say that a discrete metric space has no limit points. But I find this confusing, since isn't every element a limit point of a constant sequence?

I would greatly appreciate any input.