Show that any subset of the discrete metric space is open. Also show that any subset of the discrete metric space is closed. I have been struggling with the definition of open/closed and how to visualize them. I understand that a set being open means that if I take any point in that set, I can find a ball around it for some r>0 such that the ball is still inside the set. And closed just means the compliment of the set is open.
I know that for a discrete metric: d: X x X -> R with
d(x,y) = 0 if x = y and d(x,y) = 1 if x ≠ y.
What does it mean for B(x,r) = {x} exactly?
I am having trouble visualizing what is going on and why the set is both open and closed.
Thank you!
 A: I think the open sets are called open because they do not contain their boundary. The same can be said about closed sets, they are the ones that contain their boundary. I believe this all comes from the intervals in the real line though.
When it comes to metric spaces I prefer to think of open sets as sets such that each element of the set has a neighbourhood that is contained in the set. In the case of the discrete metric it is easy to see this, because every point is a neighbourhood for itself.
As you become more and more comfortable with topology and analysis and metric spaces you will see there are a lot of different equivalent phrasings for different definitions, and you will be able to choose the one that makes the thing you want to prove the most obvious depending on the circumstance.
A: First, note that point sets are open since the open ball $B(x,0.5)$ is equal to the set $\{x\}$ for any point $x\in X$.
From there we can use the property that unions of open sets are open to deduce that every set is open.
Then it follows that every set must also be closed.
Note $B(x,r) = \{x\}$ means that the only point in $X$ whose distance from $x\in X$ is less than $r$ is $x$ itself.
