0
$\begingroup$

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?

$\endgroup$
6
  • 1
    $\begingroup$ $\{x\}$ is not closed under your topology. Its complementary, $\{y\}$, is indeed not open. $\endgroup$
    – Suzet
    Jun 21, 2018 at 9:29
  • 2
    $\begingroup$ 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, 2018 at 9:30
  • $\begingroup$ Ahh I see, sorry that seems so obvious now. $\endgroup$
    – Mattice Verhoeven
    Jun 21, 2018 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, 2018 at 9:35
  • $\begingroup$ @MatticeVerhoeven: You should accept the answer. $\endgroup$
    – tomasz
    Jun 21, 2018 at 10:17

1 Answer 1

Reset to default
5
$\begingroup$

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. $\{x\}$).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .