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I've been having a bit of trouble on the last exercise found in Enderton's Element's of Set Theory about the section on functions. Even after examining a proof I found online, I don't fully understand the proof, and I feel utterly lost in how I would construct the proof myself. As such, I would greatly appreciate any help in allowing me to understand this problem.

The exercise states:


Assume that $F: \mathscr{P} A \rightarrow \mathscr{P} A$ and that $F$ has the monotonicity property: $$X \subseteq Y \subseteq A \Leftrightarrow F(X) \subseteq F(Y).$$ Define $$B = \bigcap \{X \subseteq A \mid F(X) \subseteq X \}$$ $$C = \bigcup \{ X \subseteq A \mid X \subseteq F(X) \}$$

(a) Show that $F(B) = B$ and $F(C) = C$.

(b) Show that if $F(X) = X$, then $B \subseteq X \subseteq C$.


In the proof I found online, it stated that for $X \subseteq A$, $$F(B) = F \left( \bigcap\limits_{F(x) \ \subseteq \ X } X\right) \subseteq \bigcap\limits_{F(x) \ \subseteq \ X } F(X) \subseteq \bigcap\limits_{F(x) \ \subseteq \ X } X = B.$$ This step makes sense to me because it fits with a theorem found earlier in the book. It is the next step that confuses me. The author continues by saying that because $F(B) \subseteq B$, by the monotonicity property, $F(F(B)) \subseteq F(B)$ and $B \subseteq F(B)$. It is primarily this very last step that confuses me. Why is it that $F(F(B)) \subseteq F(B)$ implies $B \subseteq F(B)$?

For showing $F(C) = C$, the author uses a very similar argument with indexed sets, but with unions and with concluding that $C \subseteq F(C)$ The problem is that she makes the same conclusion that $F(C) \subseteq C$ follows from $F(C) \subseteq F(F(C))$.

This is my main problem that I need to solve. Why is this the case? How might I prove it? I also would like some advice on how to improve my proof-writing skills. I am able to do almost all the exercises and proofs in the book on my own, but these more tricky and nuanced proofs still seem difficult and impossible without help. My proofs are also a bit messy and convoluted. If you have any more general advice about that, I would also greatly appreciate it. I'm self-studying this material as my first advanced mathematics course, and so I'm still a novice.Please forgive me if my questions are trivial.

If you would like to see the full proof, the problem is exercise 3.30 in this solution's manual.

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By definition, $B$ is the intersection of all subsets $X$ such that $F(X)\subseteq X$. The intersection of a collection of sets is contained in each member of that collection, namely is the largest set which is contained in each member of the collection. Since $F(B)$ is in the collection, because $F(F(B))\subseteq F(B)$, then $B$ must be a subset of $F(B)$.

Similarly for the union. The union of a collection of sets contains each element of the collection, namely is the smallest set with that property. If the collection is given by all sets with the property that their image under $F$ contains them, then $F(C)$ enjoys that property, hence $F(C)$ must be contained in the union of all sets enjoying that property, and that union is defined to be $C$.

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  • $\begingroup$ Thank you! I figured it out later thanks to you and a few other resources :) $\endgroup$
    – mijucik
    May 25, 2020 at 8:06

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