isolated points and discret metric

Question

Let $(X, d)$ be a compact metric space, such that all its elements they are isolated points. Prove that $X$ is a finite set and that there exists a homeomorphism between $(X, d)$ and a discrete metric space.

Hello, I am trying to solve this problem, so I can prove that $X$ is a finite set by contradiction, my problem is homeomorphism, in that part I was thinking if $f(x)$ can put my point in the ball with $r$ less than $1/2$ x is equal to the center of the ball, but I don't know what function can do it specifically, thanks for the ideas folks.

Answer 1

For the first part, you can argue directly. Since every point of $X$ is isolated, the singleton $\{x\}$ is an open set for all points $x \in X$. Writing $X = \bigcup_{x \in X} \{x\}$, since $X$ is compact, $X$ can be written as a finite union of singletons, so $X$ is finite.

For the next part, if $(X,d)$ is to be homeomorphic to a discrete metric space, a natural choice is to consider the discrete metric $\rho$ on $X$, and to show that $(X,d)$ and $(X,\rho)$ are homeomorphic. What is the "simplest" bijection $\varphi: X \to X$?

If you are familiar with topology, then you can just note that $d$ and $\rho$ induce the same topology on $X$, so the map $\varphi: X \to X$ is immediately a homeomorphism. Otherwise, you can argue using the definition of continuity of a map between metric spaces to show that $\varphi: (X,d) \to (X,\rho)$ and $\varphi^{-1}: (X,\rho) \to (X,d)$ are both continuous.