Interior Points Question

Question

Solution Attempt

As $A \subseteq A \cup B$ we have $Int(A) \subseteq A \subseteq A \cup B$

Similarly, $Int(B) \subseteq A \cup B$

$\implies Int(A) \cup int(B) \subseteq A \cup B$

Now $Int(A) \cup Int(B)$ is an open subset of $A \cup B$. As $Int(A \cup B)$ is the union of all open subsets of $A \cup B$ we have that $Int(A) \cup Int(B) \subseteq Int(A \cup B)$

Answer 1

The equation is not true in general. Consider the following counterexamples

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Counterxample #1

provided by Chinnapparaj R

Let $A=[-1,0], B=[0,1]$. In this case, $int(A) = (-1,0),\ int(B) = (0,1)$ so that \begin{align} int(A)\cup int(B)&= (-1,0)\cup(0,1) \\&=(-1,1)\setminus\{0\}\\ \end{align} However, $A\cup B = [-1,1]$ so that $int(A\cup B) = (-1,1)$, which is not a subset of $(-1,1)\setminus \{0\}$.

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Counterexample #2

provided by Ian

$\newcommand\Q{\mathbb{Q}} \newcommand\R{\mathbb{R}}$ Consider the subsets $\Q,\ \R\setminus\Q\subset\R$. Now, $int(\Q)=\emptyset$, and $int(\R\setminus\Q)=\emptyset$ so that $int(\Q)\cup int(\R\setminus\Q)=\emptyset$. However, $\Q\cup(\R\setminus\Q)=\R$, which is not a subset of $\emptyset$.