Interior of Union is not necessarily Union of Interiors
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Theorem
Let $T = \struct {S, \tau}$ be a topological space.
Let $H_1$ and $H_2$ be subsets of $S$.
Let ${H_1}^\circ$ and ${H_2}^\circ$ denote the interiors of $H_1$ and $H_2$ respectively.
Then it is not necessarily the case that:
- $\left({H_1 \cup H_2}\right)^\circ = {H_1}^\circ \cup {H_2}^\circ$
Proof
From Union of Interiors is Subset of Interior of Union:
- $\paren {H_1 \cup H_2}^\circ \supseteq {H_1}^\circ \cup {H_2}^\circ$
It remains to be shown that it is not necessarily the case that:
- $\paren {H_1 \cup H_2}^\circ = {H_1}^\circ \cup {H_2}^\circ$
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
Let $H_1 = \closedint 0 {\dfrac 1 2}$ and $H_2 = \closedint {\dfrac 1 2} 1$.
Then:
\(\ds \paren {H_1 \cup H_2}^\circ\) | \(=\) | \(\ds \paren {\closedint 0 {\dfrac 1 2} \cup \closedint {\dfrac 1 2} 1}^\circ\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \closedint 0 1^\circ\) | by definition | |||||||||||
\(\ds \) | \(=\) | \(\ds \openint 0 1\) | Interior of Closed Real Interval is Open Real Interval |
and:
\(\ds {H_1}^\circ \cup {H_2}^\circ\) | \(=\) | \(\ds \closedint 0 {\dfrac 1 2}^\circ \cup \closedint {\dfrac 1 2} 1^\circ\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \openint 0 {\dfrac 1 2} \cup \openint {\dfrac 1 2} 1\) | Interior of Closed Real Interval is Open Real Interval | |||||||||||
\(\ds \) | \(\ne\) | \(\ds \openint 0 1\) | as $\dfrac 1 2$ is in one but not the other |
Hence the result.
$\blacksquare$
Also see
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: $6$