Interior of Closed Real Interval is Open Real Interval

Theorem
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology. Let $\closedint a b$ be a closed interval of $\R$.

Then the interior of $\closedint a b$ is the open interval $\left({a, b}\right)$.

Proof
By definition, the interior of $\closedint a b$ is the largest open set contained in $\closedint a b$.

From Open Sets in Real Number Line it follows that $\openint a b$ is an open set of $\R$.

By definition of open interval, $\openint a b$ is contained in $\closedint a b$.

Suppose $U$ is an open set of $\R$ which is contained in $\closedint a b$.

The only way $U$ could be bigger than $\openint a b$ is if either $a \in U$ or $b \in U$ or both.

Suppose $a \in U$.

Then by Open Sets in Real Number Line it follows that $a \in \openint p q$ for some $p, q \in \R$ such that $\openint p q \subseteq U$.

From Real Numbers are Close Packed:
 * $\exists r \in \R: p < r < a$

and so:
 * $\exists r \in U: r < a$

which means:
 * $r \notin \openint a b$

That is, $a$ is not in an open set of $\struct {\R, \tau_d}$ which is contained in $\closedint a b$.

Thus $a$ is not in the interior of $\closedint a b$.

By a similar argument it is shown that neither is $b$ in the interior of $\closedint a b$.

Hence the result.