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?

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    $\begingroup$ $\{x\}$ is not closed under your topology. Its complementary, $\{y\}$, is indeed not open. $\endgroup$ – Suzet Jun 21 '18 at 9:29
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    $\begingroup$ 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. $\endgroup$ – Suzet Jun 21 '18 at 9:30
  • $\begingroup$ Ahh I see, sorry that seems so obvious now. $\endgroup$ – Mattice Verhoeven Jun 21 '18 at 9:31
  • $\begingroup$ I feel it's answered already. What do I do with the question now, just keep it up? $\endgroup$ – Mattice Verhoeven Jun 21 '18 at 9:35
  • $\begingroup$ @MatticeVerhoeven: You should accept the answer. $\endgroup$ – tomasz Jun 21 '18 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. $\{x\}$).


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