Open Real Interval is Subset of Closed Real Interval

Theorem

Let $a, b \in \R$ be real numbers.

Then:

$\openint a b \subseteq \closedint a b$

where:

$\openint a b$ is the open interval between $a$ and $b$

$\closedint a b$ is the closed interval between $a$ and $b$.

Proof

Let $x \in \openint a b$.

Then by definition of open interval:

$a < x < b$

Thus:

$a \le x \le b$

and so by definition of closed interval:

$x \in \closedint a b$

$\blacksquare$

Sources

• 1975: Bert Mendelson: Introduction to Topology (3rd ed.) ... (previous) ... (next): Chapter $1$: Theory of Sets: $\S 2$: Sets and Subsets