Union of Real Intervals is not necessarily Real Interval

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Theorem

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


Then $I_1 \cup I_2$ is not necessarily a real interval.


Proof

Proof by Counterexample:

Consider the real intervals:

$I_1 = \left({0 \,.\,.\, 2}\right)$
$I_2 = \left({4 \,.\,.\, 6}\right)$

Then we have that:

$1 < 3 < 5$

where:

$1 \in I_1 \cup I_2$
$5 \in I_1 \cup I_2$

but:

$3 \notin I_1 \cup I_2$

Thus $I_1 \cup I_2$ is not a real interval.

$\blacksquare$


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