# If every subset in a topological space is closed, prove it is a discrete space

This is question from Topology without tears by S. Morris. It reads;

Let $(X, \tau)$ be a topological space with the property that every subset is closed. Prove that it is a discrete space.

I actually think it's not true though, my counter example:

$$let\quad X = \{x, y\}\quad and \quad \tau = \{\emptyset, X\}$$

Cleary $\tau$ is a topology and all it's subsets are closed, but it's not a discrete space. What am I missing?

• $\{x\}$ is not closed under your topology. Its complementary, $\{y\}$, is indeed not open. Jun 21, 2018 at 9:29
• To answer your question, because every subset is closed, every subset is also open. In particular, the singletons are open. This is the very definition of the discrete topology. Jun 21, 2018 at 9:30
• Ahh I see, sorry that seems so obvious now. Jun 21, 2018 at 9:31
• I feel it's answered already. What do I do with the question now, just keep it up? Jun 21, 2018 at 9:35
• @MatticeVerhoeven: You should accept the answer. Jun 21, 2018 at 10:17

If every subset of $X$ is closed, then every subset of $X$ is open! Hence $\tau$ is the discrete topology on $X$. In your example not every subset of $X$ is closed (e.g. $\{x\}$).