Intersection of Real Intervals is Real Interval
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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 $\text{B} \ 7$