I know that a discrete topological space is where all singletons are open.
For example, $\mathbb{N}$ with the subspace topology inherited from $(\mathbb{R}, \mathfrak{T}_{usual})$. This is the case because we can find $\{n\} = (a,b) \cap \mathbb{N}$ which is open. Hence all singletons are open.
But are all sets are clopen? Closed?
My thoughts: Suppose we take a singleton $\{x\}$ in a discrete space $X$, we know singleton is open, hence $\{x\}^c$ is closed. But it is the arbitrary union of singletons, so it is also open, so all sets are clopen.
Best Answer
Let $F$ be an arbitrary subset of $X$ where $X$ is equipped with discrete topology.
As you said: in a discrete topological space all singletons are open.
As you said: arbitrary unions of singletons are open so $F^c=\bigcup_{x\in F^c}\{x\}$ is an open set.
(You don't even need this subroute: in a discrete space all sets are open by definition)
Then its complement $F$ is a closed set.