Let $A_1,A_2 \in \mathbb R$ open sets and denote $F_1=\mathbb R−A_1$.Show that $F_1−A_2$ is closed.

Question

As is known if $A_1$ and $A_2$ are open then their complements are closed. So by definition

$F_1= \mathbb R -A_1$ is closed.

$F_1= \mathbb R -A_1$.

$F_1 - A_2 = \mathbb R - A_1 - A_2$, as the countable union of open is open, so $A_1 \cup A_2 = A_3$ is open, so $\mathbb R - A_3$ is closed. Then $F_1 - A_2$ is closed.

Are my arguments correct?

It seems too simple and when it seems too simple I've usually done something wrong.

Thanks for any help.

Answer 1

By definition, $F_1 = A_1^c$, where $S^c$ denotes the complement of $S$ in $R$.

Given that $F_2 = F_1 - A_2$.

Then it follows that

$$F_2 = F_1 \cap A_2^c = A_1^c \cap A_2^c$$

Since $A_1$ and $A_2$ are open sets, $A_1^c$ and $A_2^c$ are closed sets in $R$.

Since $F_2$ is a finite intersection of closed sets in $R$, it is closed in $R$.

This completes the proof. $\ \ \ \ \blacksquare$