Interior of Closed Real Interval is Open Real Interval

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Theorem

Let $\left({\R, \tau_d}\right)$ be the real number line under the usual (Euclidean) topology.

Let $\left[{a \,.\,.\, b}\right]$ be a closed interval of $\R$.


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


Proof

By definition, the interior of $\left[{a \,.\,.\, b}\right]$ is the largest open set contained in $\left[{a \,.\,.\, b}\right]$.

From Open Sets in Real Number Line it follows that $\left({a \,.\,.\, b}\right)$ is an open set of $\R$.

By definition of open interval, $\left({a \,.\,.\, b}\right)$ is contained in $\left[{a \,.\,.\, b}\right]$.


Suppose $U$ is an open set of $\R$ which is contained in $\left[{a \,.\,.\, b}\right]$.

The only way $U$ could be bigger than $\left({a \,.\,.\, b}\right)$ 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 $a \in \left({p \,.\,.\, q}\right)$ for some $p, q \in \R$ such that $\left({p \,.\,.\, q}\right) \subseteq U$.

From Real Numbers are Close Packed:

$\exists r \in \R: p < r < a$

and so:

$\exists r \in U: r < a$

which means:

$r \notin \left({a \,.\,.\, b}\right)$

That is, $a$ is not in an open set of $\R$ which is contained in $\left[{a \,.\,.\, b}\right]$.

Thus $a$ is not in the interior of $\left[{a \,.\,.\, b}\right]$.


By a similar argument it is shown that neither is $b$ in the interior of $\left[{a \,.\,.\, b}\right]$.


Hence the result.

$\blacksquare$