a) Prove: for any $X,Y \subset \mathbb R$ we have $int(X) \cup int(Y) \subset int(X \cup Y)$...

Question

a) Prove: for any $X,Y \subset \mathbb R$ we have $int(X) \cup int(Y) \subset int(X \cup Y)$.

b) Give an example of sets for which $int(X) \cup int(Y) \ne int(X \cup Y)$.

a) Let $x \in int A$ then exist $ \epsilon \gt 0$ such that $(x - \epsilon, x + \epsilon) \subset A$ then $(x - \epsilon, x + \epsilon) \subset A \cup B$ and $(x - \epsilon, x + \epsilon) \subset int(A \cup B)$ the same for $B$, so $intA \cup intB \subset int(A \cup B )$.

b) A=(0, 1] and B=[1, 2)

How is my answer?

Thanks.

Answer 1

Note that $A \subset B$ implies that $\mbox{int}(A) \subset \mbox{int}(B)$.

We know that $$ X \subset X \cup Y $$ which implies that $$ \mbox{int}(X) \subset \mbox{int}(X \cup Y) \tag{1} $$

We also know that $$ Y \subset X \cup Y $$ which implies that $$ \mbox{int}(Y) \subset \mbox{int}(X \cup Y) \tag{2} $$

From (1) and (2), we conclude that $$ \mbox{int}(X) \cup \mbox{int}(Y) \subset \mbox{int}(X \cup Y) \tag{2} $$

To show that the set inequality can be strict, we give the following standard example with $X = \mathbf{R}$, the usual topology.

Take $A = (0, 1]$ and $B = [1, 2)$. Then $$ \mbox{int}(A) = (0, 1), \ \mbox{int}(B) = (1, 2) $$

Hence, $$ \mbox{int}(A) \cup \mbox{int}(B) = (0, 1) \cup (1, 2) $$

But $$ A \cup B = (0, 2) $$ which is an open set in $\mathbf{R}$.

Hence, $$ \mbox{int}(A \cup B) = (0, 2). $$

For this example, we have $$ \mbox{int}(A) \cup \mbox{int}(B) = (0, 1) \cup (1, 2) \neq \mbox{int}(A \cup B) $$