I'm reading this proof that says that a non-trivial discrete space is not connected. I understood that the proof works because it separated the discrete set into a singleton ${x}$ and its complementar. Since they're both open, their intersection is empty and their union is the entire space, this is a separation that is not trivial, therefore the space is not connected. But why a finite set of points is open? I remember that I proved that this set is closed, since I just have to pick a ball in the complementary, with radius such that its the minimum of the distances to those points. I know that if a set is closed it doesn't mean it's not open, but how to prove it?
Update: what's the simples proof that does not involve topology, only metrics?
Best Answer
You're a little muddled about what open and closed mean. In a metric space, open and closed sets are defined using the concept of a ball. This proof does not deal with metric spaces, but with topological spaces, which are more general, so there is no such thing as a ball here. Here, to know whether the two halves of the separation are open, you just need to know whether they're in the topology $\tau$, and that's trivially true because any subset of the space is in $\tau$ by the definition of $\tau$.