What is entailed by a homeomorphism with a discrete metric? I have a homework question in which I am asked to prove that several statements are equivalent. I'm confused about a particular step in this chain, namely $a)\implies b)$ below. * * * Given: Let $M$ be a metric space with metric _d_. Prove that the following are equivalent. a) $M$ is homeomorphic to $M$ equipped with the discrete metric. b) Every function $f : M \to M$ is continuous. c,d,e) etc. * * * Is the metric space in this question always the discrete one, or is the _d_ in the original question an arbitrary metric? In a) are both of the treatments of $M$ on the same metric? Assuming that _d_ from the overall premise is not the discrete metric and not both of $M$ in a) are using the discrete metric, what does the homeomorphism from $M$ to $M_{discrete}$ actually entail? Isn't continuity trivial for the discrete metric, and thus of no help in b) when the metric is just _d_ again?

Suppose a) holds. I.e. there is some homeomorphism $h$ between $(M,d)$ (where $d$ is just the metric $M$ came with, being a metric space!) and $(M,\rho)$ where $\rho$ is the discrete metric on $M$ ($\rho(x,y) = 1$, except when $x=y$ when it must be $0$).

This means that all subsets of $(M,d)$ are open in the metric topology induced by $d$, because all subsets of $M$ are open in the dicrete metric, or more formally: suppose $A \subset M$, then $h[A]$ is open in $(M,\rho)$, so $A = h^{-1}[h[A]]$ (as $h$ is a bijection!) is open in $(M,d)$ as $h$ is continuous.

Now b) is clear by using the open set definition of continuity.

So $M$ does come with some metric, i.e. for $M$ we just think $(M,d)$ and all questions on its topology are questions on the topology induced by this $d$. Being homeomorphic to the dicrete metric on $M$ then gives us all the info we need on this _topology_ on $(M,d)$, which is the only thing needed for continuity.