Closed Real Interval is Closed Set

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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.

$\blacksquare$


Sources

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