Let $(M,d)$ be a metric space. Prove that $A \subset M$ is closed iff whenever every disk about $x$ meets $A$ then $x \in A.$

Question

Let $(M,d)$ be a metric space. Prove that $A \subset M$ is closed iff whenever every disk about $x$ meets $A$ then $x \in A.$

My attempt: I am trying to create a monotone sequence of $x_n \in U(x, \epsilon) \cap A$ such that $x_n \to x.$ So, if $x_1 \in U(x, \epsilon) \cap A,$ then there exists $\delta > 0$ such that $U(x_1, \delta) \subset U(x, \epsilon) \cap A.$ How do I define the open ball about $x_2 < x_1 ? $ Could anyone advise me on that? Hints will suffice, thank you.

Answer 1

Hint For $\Rightarrow$, pick $\epsilon =\frac{1}{n}$.

For $\Leftarrow$, try to prove that the complement of $A$ is open.

Don't focus on monotone sequence, as this concept doesn't make sense in metric space. Ignore monotony.