This is exercise $3.6.11.$ from the book How to Prove it by Velleman $($$2^{nd}$ edition$)$:
Suppose $\mathcal F$ is a family of sets that has the property that for every $\mathcal G\subseteq \mathcal F$, $\bigcup\mathcal G\in\mathcal F$. Prove that there is a unique set $A$ such that $A\in\mathcal F$ and $\forall B\in\mathcal F(B\subseteq A)$.
I almost spent my whole day on solving the above exercise and after I gained a sufficient amount of disappointment I turned into the partial solution manual at the end of the book. As it turned out, I was not able to solve the problem because I had forgotten that in exercise $3.3.8$ I had proved "$A\in\mathcal F$ implies $A\subseteq \bigcup\mathcal F$" and I should have used it in solving the above exercise.
I am not done with chapter $3$ of Velleman's textbook and yet until now I have solved almost $120$-$130$ exercises most of which are such statements like "$A\in\mathcal F$ implies $A\subseteq \bigcup\mathcal F$." So my question is how could I recognize which of these statements I should remember such that in the future it might help me solve another problem?
Thanks for your attention.
For statements like these I’d say that it’s more a matter of absorbing them at the intuitive level than of remembering them explicitly. You want to reach a point at which it’s obvious that if $A\in\mathscr{F}$, then $A\subseteq\bigcup\mathscr{F}$: after all, that’s really just the definition of union. I can, however, offer some retrospective advice about Exercise $3.6.11$.
You’re asked to show that there is an $A\in\mathscr{F}$ that contains every member of $\mathscr{F}$. In other words, you’re asked to show that $\mathscr{F}$ has a maximal element with respect to the order $\subseteq$. Now relate that to the hypothesis that $\bigcup\mathscr{G}\in\mathscr{F}$ for each $\mathscr{G}\subseteq\mathscr{F}$. Bigger subcollections of $\mathscr{F}$ will have bigger unions, and $\mathscr{F}$ itself is the biggest possible subcollection of $\mathscr{F}$, so we should probably look at $\bigcup\mathscr{F}$ as a candidate for that maximal element $A$; after all, the hypothesis does guarantee that $\bigcup\mathscr{F}$ is an element of $\mathscr{F}$. So let’s set $A=\bigcup\mathscr{F}$ and try to show that if $B$ is any element of $\mathscr{F}$, then $B\subseteq A$.
All that we really know about $A$ is that it’s $\bigcup\mathscr{F}$, so we probably have to use that fact. And $B\in\mathscr{F}$, so $B$ is one of the sets whose union we took in order to form $A$. Oh, of course: that means that $B$ must be a subset of $A$! And now it’s just a matter of cleaning up the argument, perhaps something like this:
Of course this still leaves the uniqueness of $A$ to be proved, but uniqueness is generally easier to prove than existence, because there’s a natural approach to such proofs: assume that two objects both have the property in question, and either show directly that they must be the same object, or assume that they’re distinct and derive a contradiction.
So suppose that $A'\in\mathscr{F}$ has the property that $B\subseteq A'$ for all $B\in\mathscr{F}$. Then $A\subseteq A'$, and since $A$ is also maximal, $A'\subseteq A$, so $A=A'$, and $A$ is indeed the unique maximal element of $\mathscr{F}$.