Prove that for every $x\in\mathbb{R}$, the set $\{x\}$ is a closed in $\mathbb{R}$ with respect to the usual metric.

Question

I know the set is closed if it is the complement of an open set and I know the definition of usual metric. But I don't know how to prove this question.

Answer 1

Hint: $\mathbb{R}\setminus\{x\} = (-\infty, x)\cup (x, \infty)$ is a union of open sets.

Answer 2

If the set $\{x\}$ is a closed set, then it is equivalent to prove $\mathbb{R} \setminus \{x\}$ is open. To show this, you can show that $\forall y \in \mathbb{R} \setminus \{x\}, \exists \epsilon > 0 \ni B(y, \epsilon) \cap \{x\} = \emptyset$. I'll leave the rest to you, since this shouldn't be hard to show. Thus, since $\mathbb{R} \setminus \{x\}$ is open, then $\{x\}$ is closed.