Closure of Interior of 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:

$A^{- \circ -} = A^{\circ -} = A^- = \closedint a c$

where:

$A^\circ$ is the interior of $A$

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

Proof

\(\ds A^{\circ -}\)

\(=\)

\(\ds A^-\)

Interior of Union of Adjacent Open Intervals: $A^\circ = A$

\(\ds \)

\(=\)

\(\ds \closedint a c\)

Closure of Union of Adjacent Open Intervals

Then:

\(\ds A^{- \circ -}\)

\(=\)

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

Interior of Closure of Interior of Union of Adjacent Open Intervals

\(\ds \)

\(=\)

\(\ds \closedint a c\)

Closure of Open Ball in Metric Space

Hence the result.

$\blacksquare$

Sources

• 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $5 \ \text{(a)}$