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.

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    $\begingroup$ If every singleton is open, the every set is open, since every set is a union of singletons. Since every set is open, then every set is closed (why?) $\endgroup$ – Thomas Andrews Jul 2 '16 at 7:09
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    $\begingroup$ But your argument is wrong, because you've only shown that some sets, of the form $\{x\}^c$, are clopen. You have to show all sets are all clopen. $\endgroup$ – Thomas Andrews Jul 2 '16 at 7:10
  • $\begingroup$ @ThomasAndrews Cuz the complement of an open set is a closed set, but all sets are open, a closed set is a set, so a closed set is open, therefore all sets are clopen $\endgroup$ – Olórin Jul 2 '16 at 7:16
  • $\begingroup$ @MSEisadatingsite Your logic is still backwards. You also need the fact that all sets are complements of open sets, which you haven't mentioned. $\endgroup$ – user296602 Jul 2 '16 at 7:20
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    $\begingroup$ All sets are clopen. More specifically, any singleton is clopen and therefore a connected component. That is why it's called the discrete topology: The points just exist by themself with nothing whatsoever connecting them. $\endgroup$ – Arthur Jul 2 '16 at 8:32

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.


Hint. Use the fact that open sets are closed under arbitrary union.

  • $\begingroup$ The OP seems to already be aware of this fact and has attempted to use it ("But it is the arbitrary union of singletons, so it is also open"). They are stuck at a different point in the logic, namely justifying why all sets are complements of open sets. $\endgroup$ – user296602 Jul 2 '16 at 7:27
  • $\begingroup$ My hint was precisely intended to use this property of open sets in a different way, namely to show that any set is open. $\endgroup$ – J.-E. Pin Jul 2 '16 at 8:24

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