Find all open sets in a discrete metric space.

My attempt: Let $(X,d)$ be a discrete metric space and $U$ be a nonempty subset of $X$. We want to show $U$ is open in $X$, i.e., for each $x\in U$ there exists $r>0$ such that $B_d(x,r) \subset U$. Clearly $\{x\} = B_d(x,1) \subset U$. So $U$ is open in $X$. Since $U$ is arbitrary, so all open sets in a discrete metric space are the power set of that set.

Is my answer correct?

  • $\begingroup$ $B_d(x,1)=\{x\}$ isn't correct. You need a radius strictly less than 1. $\endgroup$
    – Fabrizio G
    Commented Jun 29, 2022 at 13:42
  • $\begingroup$ @FabrizioGambelín $B_d(x,1)= \{ y \in X \mid d(x,y)<1\}$. So, if the radius is $1$ then also $B_d(x,1) = \{x\}$. I think so. $\endgroup$
    – user1234
    Commented Jun 29, 2022 at 13:53
  • $\begingroup$ Oh you're right, sorry. I was thinking the ball with $d(x,y)\leq 1$ in my head. My bad. $\endgroup$
    – Fabrizio G
    Commented Jun 29, 2022 at 17:50

1 Answer 1


Let $\varnothing\neq U\subset X$ and $x\in U$. What we need is to find a radius $\varepsilon>0$ such that $B(x,\varepsilon)\subset U$.

If $0<\varepsilon<1$, then $B(x,\varepsilon)=\{x\}$, hence every singleton is open. Recall that open sets in a topology are arbitrary union of open sets, finite intersection and of course the whole space and empty set. Then, you can caractherize this way all the open sets since any set in $\mathcal{P}(X)$ can be written as a union of singletons.


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