# Closed Real Interval is Closed Set

## Theorem

Let $\R$ be the real number line considered as an Euclidean space.

Let $\closedint a b \subset \R$ be a closed interval of $\R$.

Then $\closedint a b$ is a closed set of $\R$.

## Proof

 $\ds \closedint a b$ $=$ $\ds \set {x \in \R: x \ge a \land x \le b}$ Definition of Closed Real Interval $\ds \leadsto \ \$ $\ds \R \setminus \closedint a b$ $=$ $\ds \R \setminus \set {x \in \R: x \ge a \land x \le b}$ $\ds$ $=$ $\ds \set {x \in \R: x < a \lor x > b}$ De Morgan's Laws: Disjunction of Negations $\ds$ $=$ $\ds \openint {-\infty} a \cup \openint b \infty$ Definition of Open Real Interval

From the corollary to Open Real Interval is Open Set, both $\openint {-\infty} a$ and $\openint b \infty$ are open sets in $M$.

From Union of Open Sets of Metric Space is Open it follows that $\openint {-\infty} a \cup \openint b \infty$ is open in $\R$.

But $\openint {-\infty} a \cup \openint b \infty$ is the relative complement of $\closedint a b$ in $\R$.

The result follows by definition of closed set.

$\blacksquare$