Question:
Prove that a finite subset in a metric space is closed.
My proof-sketch: Let $A$ be finite set. Then $A=\{x_1, x_2,\dots, x_n\}.$ We know that $A$ has no limits points. What's next?
Definition: Set $E$ is called closed set if $E$ contains all his limits points.
Context: Principles of Mathematical Analysis, Rudin
Best Answer
Thinking $A$ as a subset of a metric space $M$. An easy approach will be to use that the single points $\{x_j\}\subset M$ are closed (you know why?), then of course $$ A=\bigcup_{j=1}^n \{ x_j \} $$ Since $A$ is a finite union of closed sets, it is itself closed.