Problem 3.3.5 from Understanding Analysis by Stephen Abott poses the following question. I'd like someone to verify if my justification is sound and rigorous.
Decide whether the following propositions are true or false. If the claim is valid, supply a short proof, and if the claim is false, provide a counterexample.
(a) The arbitrary intersection of compact sets is compact.
(b) The arbitrary union of compact sets is compact.
(c) Let $A$ be arbitrary and $K$ be compact. Then, the intersection $A \cap K$ is compact.
(d) If $F_1 \supseteq F_2 \supseteq F_3 \supseteq F_4 \supseteq \ldots$ is a nested sequence of nonempty closed sets, then the intersection $\bigcap_{n=1}^{\infty}F_n \ne \emptyset$.
My Attempt.
(a) The countably infinite intersection of closed sets is closed. If each of these sets are bounded, the intersection is also bounded. Hence, $\bigcap_{n=1}^{\infty}K_i$ is compact.
(b) The finite union of closed sets is closed. The countably infinite union of closed sets need not be closed (since the infinite intersection of open sets is not always open, for example $\bigcap_{n=1}^{\infty} \left(0,\frac{1}{n}\right) = \emptyset$, which is closed). As a result, the finite union of compact sets is compact.
(c) $A \cap K \subseteq K$. $A \cap K$ is a closed and bounded subset of $K$. $A \cap K$ is compact.
(d) This proposition is false. Let $F_n = [n,\infty)$. Then,
\begin{align*}
\bigcap_{n=1}^{\infty} F_n = \emptyset
\end{align*}
Best Answer
(a) It is correct, by I think that it's more natural to say that the intersection is a closed subset of $K_1$ and that therefore it is compact.
(b) You did not really answer the question. An arbitrary union of compacts doesn't have to be compact. For instance, $\Bbb R$, which is not compact, is equal to $\bigcup_{x\in\Bbb R}\{x\}$.
(c) This is false: take $A=(0,2)$ and $K=[1,3]$.
(d) Your answer is fine.