Prove every subset of in the discrete metric is clopen

Hey fellow math enthusiasts! I am reading in ”Introduction to Topology” by Gameline and Greene and I got stuck on an exercise in the first chapter, and I’d love some help on understanding their solution. The problem is as follows:

”Given a set $$X$$ and metric $$d(x, y) = 1$$ if $$x \neq y$$ and $$d(x, y) = 0$$ if $$x = y$$ then we want to prove that every subset of the resulting metric space $$(X, d)$$ is both open and closed.”.

And the solution is as follows:

”Since each ball $$B(x; \frac{1}{2})$$ reduces to the singleton set $${x}$$, every subset is a union of open balls, hence every subset is open.”.

My interpretation of the solution is that they are just providing the way to reason. They only showed that each subset is open but not closed.

In the book there is a theorem that states that a subset of $$X$$ is open if and only if it is a union of open balls in $$X$$, and is being used in the proof.

I get that in $$X$$ each subset is a singleton set $${x}$$ or a collection of singletons and since each singleton can be rewritten in X as an open ball $$B(x; \frac{1}{2})$$ then each collection of singletons can be written as a union of these open balls and thus each subset of $$X$$ is open.

But how do we get that each subset is closed? My idea is that we look at the complements of the sets we considered above. Since each of these complement-sets also obey the same structure (so in practice would not be distinguishable from the sets above) they too, using the ball-argument, can be showed to be open sets. Then using the argument (theorem in book) that if a subset is open, it’s complement is closed. So both the subsets of individual singletons or collections of singletons are both open and closed.

How do I formalize? Any feedback is greatly appreciated. Thanks in advance.

/Isak

• Possible duplicate of discrete metric, both open and closed. Apr 14 '19 at 10:07
• You essentially have the entire formal argument here. It may be that the original text considered the part you added to be easy enough for the reader to figure out themselves. Apr 14 '19 at 13:33
• Why is no answer accepted? Nov 3 '21 at 18:21

Another way to reason: suppose $$x \in \overline{A}$$ (the closure of $$A$$) for some arbitary subset of $$A$$. Then every ball around $$x$$ intersects $$A$$, in particular $$B(x,\frac{1}{2})=\{x\}$$ must intersect $$A$$, which means $$x \in A$$.
So for all $$A \subseteq X$$, $$\overline{A} \subseteq A (\subseteq \overline{A})$$ so $$A = \overline{A}$$ and every subset $$A$$ is closed.