0
$\begingroup$

This question and the given solution are from my lecture notes. I don't understand why the solution makes sense. I'm only showing the proof for the closed case.

Question: Let $(X,d)$ be a metric space, where $d$ is the discrete metric. Prove that every subset $A\subseteq X$ is both open and closed.

Solution: Since $A$ is open $\forall A$ $\implies $ $X\backslash A$ is also open $\forall A$ $\implies$ $A$ is closed.

I understand the final step, just not the initial one. I know a set can be both open and closed, but why does $A$ open imply that $X\backslash A$ is open? Is it a property of the discrete metric?

  • 1
    $A$ being open does not imply that $X \setminus A$ is open, but "$A$ is open $\forall A$" applies to all subsets of $X$, and thus in particular it applies to $X \setminus A$. – 2017-01-03
  • 1
    It should better have been "Since $B$ is open for all $B$, it follows that $X\setminus A$ is open". The occurrence of $A$ before the arrow is bound by the "for all", yet it's confusing to use the same name for different things in the same context. – 2017-01-03
  • 0
    I think the gist of the solution is "$X\setminus A$ is open, being a subset of $X$, so that its complement $A$ is closed" – 2017-01-03

2 Answers 2

0

We have $A$ is open $\forall A \subseteq X$ $(\star)$.

Let $B = X \setminus A$. Then $B \subseteq X$, so $B$ is open by $(\star)$.

0

The discrete topology, by definition, is $\tau_d = \left\{A \in 2^X\right\} = 2^X$, so, every subset of $X$ is in $\tau_d$, i.e. $A$, $X\setminus A \in \tau_d$.