Assume $K \subset \mathbb{R}$ is non-empty and compact, and $F \subset \mathbb{R}$ is non-empty and closed. Decide if the following sets are definitely compact, definitely closed, both, or neither.
(1) $K \cap F$
My attempt: Since $(K \cap F) \subseteq K$, all open covers of $(K \cap F)$ must have a finite subcover which means that $(K \cap F)$ is definitely compact.
(2) $\overline{F^c \cup K^c}$
My attempt: Since $\overline{F^c \cup K^c}$ is the closure of $F^c \cup K^c$, it must (definitely) be closed. Now, since $K$ is bounded, $K^c$ must be unbounded which means that $\overline{F^c \cup K^c}$ is not compact.
(3) $K\setminus F = K \cap F^c$
My attempt: Since ($K \cap F^c) \subseteq K$, it is definitely compact (same reasoning as in (1)).
(4) $\overline{K \cap F^c}$
My attempt: It is definitely closed since $\overline{K \cap F^c}$ is the closure of $K \cap F^c$. Since $K$ is bounded, and since $(K \cap F^c) \subseteq K$, $\exists [a, b]$ s.t. $(K \cap F^c) \subseteq [a, b]$. But, since $\overline{K \cap F^c}$ is the smallest set closed containing $K \cap F^c$ there must exist a larger closed set which contains $\overline{K \cap F^c}$ which makes $\overline{K \cap F^c}$ bounded. So, $\overline{K \cap F^c}$ is definitely compact.
Can someone please verify these answers? Also, can someone explain what the question means by "both"? If I say that a set is definitely compact, then it must definitely be closed. Then, how can a set be "both" closed and compact? Thanks!
Note that only closed subset of a compact set is compact .Your $1^{st}$ and $2^{nd}$ looks ok. In third attempt how can you sure $K∩F^c$ is compact while given $F$ is closed. $F^c$ is open . Since compact set is closed in $\mathbb{R}$ . Intersection of closed set and open set need not be closed. Therefore $K∩F^c\subseteq K$ is not compact.