Closure of Union of Adjacent Open Intervals

Theorem

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

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

$A := \openint a b \cup \openint b c$

Then:

$A^- = \closedint a c$

where:

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

Proof

\(\ds A^-\)

\(=\)

\(\ds \paren {\openint a b \cup \openint b c}^-\)

\(\ds \)

\(=\)

\(\ds \openint a b^- \cup \openint b c^-\)

Closure of Finite Union equals Union of Closures

\(\ds \)

\(=\)

\(\ds \closedint a b \cup \closedint b c\)

Closure of Open Ball in Metric Space

\(\ds \)

\(=\)

\(\ds \closedint a c\)

Definition of Closed Real Interval

$\blacksquare$