Interior may not equal Exterior of Exterior/Proof 2
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be a subset of the underlying set $S$ of $T$.
Let $A^e$ be the exterior of $A$.
Let $A^\circ$ be the interior of $A$.
Then it is not necessarily the case that:
- $A^{ee} = A^\circ$
Proof
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $T = \struct {S, \tau}$ be the Sorgenfrey line on $\struct {S, \preccurlyeq}$.
Let $A \subseteq S$ denote the subset of $S$ defined as:
- $A = \openint a {+\infty}$
By Exterior of Exterior in Sorgenfrey Line is not necessarily Interior:
- $A^{ee} = \hointr a {+\infty}$
while:
- $A^\circ = \openint a {+\infty}$
The result is apparent.
$\blacksquare$
Sources
- 1978: Lynn Arthur Steen and J. Arthur Seebach, Jr.: Counterexamples in Topology (2nd ed.) ... (previous) ... (next): Part $\text I$: Basic Definitions: Section $1$: General Introduction: Closures and Interiors