Intersection of Real Intervals is Real Interval

Theorem

Let $I_1$ and $I_2$ be real intervals.

Then $I_1 \cap I_2$ is also a real interval.

Proof

Let $x, y \in I_1 \cap I_2$.

From the definition of a real interval, it suffices to show that:

for each $z \in \R$ with $x \le z \le y$ we have $z \in I_1 \cap I_2$.

Let $z$ be a real number with:

$x \le z \le y$

Since $x, y \in I_1$, we have:

$z \in I_1$

from the definition of a real interval.

Similarly, since $x, y \in I_2$, we have:

$z \in I_2$

from the definition of a real interval.

So:

$z \in I_1 \cap I_2$

as required.

$\blacksquare$

Sources

• 1964: William K. Smith: Limits and Continuity ... (previous) ... (next): $\S 2.1$: Sets: Exercise $\text{B} \ 7$