Closure of Union of Adjacent Open Intervals

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Theorem

Let $a, b, c \in R$ where $a < b < c$.

Let $A$ be the union of the two adjacent open intervals:

$A := \left({a \,.\,.\, b}\right) \cup \left({b \,.\,.\, c}\right)$

Then:

$A^- = \left[{a \,.\,.\, c}\right]$

where:

$A^-$ is the closure of $A$.


Proof

\(\displaystyle A^-\) \(=\) \(\displaystyle \left({\left({a \,.\,.\, b}\right) \cup \left({b \,.\,.\, c}\right)}\right)^-\)
\(\displaystyle \) \(=\) \(\displaystyle \left({a \,.\,.\, b}\right)^- \cup \left({b \,.\,.\, c}\right)^-\) Closure of Finite Union equals Union of Closures
\(\displaystyle \) \(=\) \(\displaystyle \left[{a \,.\,.\, b}\right] \cup \left[{b \,.\,.\, c}\right]\) Closure of Open Ball in Metric Space
\(\displaystyle \) \(=\) \(\displaystyle \left[{a \,.\,.\, c}\right]\) Definition of Closed Real Interval

$\blacksquare$