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:

$\paren {H_1 \cup H_2}^\circ = {H_1}^\circ \cup {H_2}^\circ$


Proof by Counterexample:

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.

$\blacksquare$


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