Monotone Involutory Set Functions and Unions/Intersections of Families I am currently attempting to solve the following problem in elementary set theory/introductory real analysis:
Let $A$ be a nonempty set and let $f:2^{A} \to 2^{A}$ be a function satisfying:

*

*$X \subset Y \implies f(Y) \subset f(X)$ for all $X, Y \in 2^{A}$, i.e., it is monotone decreasing (in a certain sense, but do correct me if I'm wrong);


*$f(f(X)) = X$ for all $X \in 2^{A}$, i.e., $f$ is an involution.
With these properties at hand, I'm asked to show that for any family $(X_\lambda)_{\lambda \in L}$ of subsets of $A$, we have:
$$f\left(\bigcup_{\lambda \in L} X_\lambda \right) = \bigcap_{\lambda \in L} f(X_\lambda),$$
and
$$f\left(\bigcap_{\lambda \in L} X_\lambda \right) = \bigcup_{\lambda \in L} f(X_\lambda).$$
I have been able to show that
$$f\left(\bigcup_{\lambda \in L} X_\lambda \right) \subset \bigcap_{\lambda \in L} f(X_\lambda)$$
in two steps: first, note that for all fixed $\lambda_0 \in L$, we have: $X_{\lambda_0} \subset \bigcup X_\lambda$, and therefore property 1 yields:
$$f\left(\bigcup_{\lambda \in L} X_\lambda \right) \subset f(X_{\lambda_0}),$$
and since $\lambda_0$ was arbitrarily chosen, this holds for all $X_\lambda$ in the family. It therefore follows that
$$f\left(\bigcup_{\lambda \in L} X_\lambda \right) \subset \bigcap_{\lambda \in L} f(X_\lambda),$$
seeing as the intersection of the $f(X_\lambda)$ is the largest set that is contained in each member of the family $(X_\lambda)_{\lambda \in L}$ (I have already proved this statement in a previous exercise). By means of a similar argument one can show the analogous set inclusion for the second statement.
I'm stumped as to how I would go about showing the opposite set inclusion. It would obviously require that we invoke property 2, otherwise it wouldn't be there in the first place, but even so, any naïve application of it leads me to dead ends. Any hints or solutions would be much appreciated.
P.S.: This question has already been posed here, but a proper answer to it was never given. See William Elliot's comment on it.
 A: As suggested by William Elliot's comment in the post you linked, one way of thinking about the problem is to pretend $f$ is set complementation, in which case the results you're trying to show are De Morgan's laws. So if you write down a proof of De Morgan's laws and then replace complementation with $f$, this should produce a proof of your result. Go ahead and do this on your own if you want, but I've included a proof of the missing portion of your argument below for the sake of completeness.

 We wish to show $$\bigcap_\lambda f(X_\lambda)\subset f\Big(\bigcup_\lambda X_\lambda\Big).$$ Take $x$ to be an element of the left-hand side. Then $x\in f(X_\lambda)$ for each $\lambda\in L$. Equivalently, $\{x\}\subset f(X_\lambda)$ for each $\lambda\in L$, so applying $f$ and using assumptions (1) and (2) we have $f(\{x\})\supset f(f(X_\lambda)) = X_\lambda$ for each $\lambda\in L$. Thus $$f(\{x\})\supset\bigcup_{\lambda\in L}X_\lambda.$$ Applying $f$ to the above and using (1) and (2) again, we have $$\{x\} = f(f(\{x\}))\subset f\Big(\bigcup_{\lambda\in L} X_\lambda\Big).$$ Thus $x$ is an element of the right-hand side, and the desired conclusion follows. Since $$\bigcap_\lambda f(X_\lambda) = f\Big(\bigcup_\lambda X_\lambda\Big),$$ we can prove the other relation by replacing $X_\lambda$ in the above with $f(X_\lambda)$ and applying involutivity again.

