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I would like someone to verify my answers to the below questions are correct.

Exercise 3.2.8 from Understanding Analysis by Stephen Abbott.

Assume that $A$ is an open set, and $B$ is a closed set. Determine if the following sets are definitely open, definitely closed, both, or neither.

(a) $\overline{A \cup B}$

(b) $A-B = \{x \in A: x \notin B\}$

(c) $(A^C \cup B)^C$

(d) $(A \cap B) \cup (A^C \cap B)$

(e) $(\overline{A})^C \cap \overline{A^C}$

My Attempt.

(a) $\overline{A}$ is definitely closed for any set $A$. So, $\overline{A \cup B}$ is definitely closed.

(b) $A - B$ is definitely open. For all $x$ remaining in $A - B$, there exists an $\epsilon$-neighbourhood, that is contained entirely in $A - B$.

Consider $A = (0,1)$, $B=\{1/n:n \in \mathbf{N}\} \cup \{0\}$. $A-B$ is still open, because for the difference set is \begin{align*} \bigcup_{n=1}^{\infty}\left(\frac{1}{n},\frac{1}{n+1}\right) \end{align*} which is a countable union of open sets.

(c) $(A^C \cup B)^C$ is definitely open. $A^C$ is closed, so $A^C \cup B$ is closed and therefore $(A^C \cup B)^C$ is open.

(d) $(A \cap B) \cup (A^C \cap B) = (A \cup A^C) \cap B = \mathbb{R} \cap B = B$. This set is definitely closed.

(e) $(\overline{A})^C \cap \overline{A^C}$. I can't tell this one for sure, because its the intersection of an open set with a closed set.

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    $\begingroup$ I think your b) needs more explanation (note $A\setminus B=A\cap B^C$). $\endgroup$ – David Mitra Jan 7 at 10:08
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    $\begingroup$ @DavidMitra, can I argue: $A$ is open, $B^C$ is open, so finite intersection is open? $\endgroup$ – Quasar Jan 7 at 10:10
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    $\begingroup$ Yes.${}{}{}{}{}$ $\endgroup$ – David Mitra Jan 7 at 10:11
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Your answers for a)-d) are correct.

For e): Since $A$ is open its complement is closed. So $(\overline {A^{c})}=A^{c}$. Now $(\overline A)^{c} \cap A^{c}=(\overline A)^{c}$ which is open.

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    $\begingroup$ How is $(\overline{A})^C \cap A^C = (\overline{A})^C$? Perhaps, I'm missing something here. $\endgroup$ – Quasar Jan 7 at 10:12
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    $\begingroup$ @Quasar Since $A$ is subset of its closure we see that $(\overline A)^{c}$ is a subset of $A^{c}$. So when you intersect these two we get $(\overline A)^{c}$ $\endgroup$ – Kavi Rama Murthy Jan 7 at 10:13
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    $\begingroup$ @Quasar and Kavi Rama Murthy, relevant? proofwiki.org/wiki/… proofwiki.org/wiki/… proofwiki.org/wiki/… $\endgroup$ – BCLC Jan 7 at 10:19

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