Suppose for some $F\subseteq \mathcal{P}(A)$, we have $U = \{X \subseteq A: \forall S \in F . S \subseteq X\}.$ We want to show that $\bigcup F = \bigcap U$, and then subsequently generalize and show that for some $L \subseteq \mathcal{P}(A)$, $\bigcap F = \bigcup L$.

So there's essentially 2 things to prove, and one is just the opposite of the other. I'm not sure why $U$ is defined so abstractly, but it seems to me that $U$ is just a subset of $F$, where each element is a subset of A as well. Can we simplify $U = \{S\subseteq A: S \in F\}$, or just $U \subseteq F$?

I have no idea how I could begin proving the union of sets equals the intersection of sets and vice versa. If $U\subseteq F$, how could we have $\bigcup F = \bigcap U$? My thoughts:

  1. Maybe I need to show that $\bigcup F \subseteq \bigcap U$ and $\bigcup U \subseteq \bigcap F$ to show the first one. It might be able to show that for any set $X \in F$, we have $X \in F \lor X \in U$. But since $X\in U$ implies $X \in \bigcap U$, we have our relationship shown immediately.

Can I prove along this line of argument?

But the second part doesn't have $L$ related to $F$ by anything so it is even harder to prove it.


To prove this, you have to show two things: $\bigcap U \subseteq \bigcup F$ and $\bigcap U \supseteq \bigcup F$. This proves equality of sets. The idea behind proving something like "$A\subseteq B$" is to assume $a\in A$ and then show $a\in B$. This question can be approached like this, but here are a few hints.


  • To show $\bigcap U \subseteq \bigcup F$: Does $\bigcup F \in U$?
  • To show $\bigcup F \subseteq \bigcap U$: If $x\in\bigcup F$, then for each $X\in U$, is $x\in X$?

Edit: Note that $F$ is just a set of sets. $U$ is the set of sets $X$ for which any set $S\in F$ is a subset of $X$. That is, every set in $U$ contains every set in $F$.

Concerning the set $L$, you are wanting to just make an analogous set to $U$, except this time so that we have $\bigcap F = \bigcup L$. Two hints for this:

  • Try writing $L$ in a similar form to $U$. That is, in the form of: $$L = \{X\subseteq A \mid \forall S\in F.\underline{\text{blank}}\},$$ where you fill in the $\underline{\text{blank}}$.
  • If you still have difficulties, think about what $U$ was. $U$ was the set of sets containing $\bigcup F$. We took the intersection of $U$ to get only $F$. In this way, $\bigcap U$ is the "smallest set containing $\bigcup F$". In the case of $L$, because you are taking its union, you will end up with the "largest set containing $\bigcap F$".
  • $\begingroup$ Thanks for your answer. Can you clarify if the $x$ and $X$ are the same thing? Also, do you have a hint on how the second part of the proof could be solved since we don't know $F \subseteq L$ or vice versa? $\endgroup$ – oldselflearner1959 Feb 11 '18 at 16:11
  • $\begingroup$ Also, if we can show $X \in U$, how do I then show that $X \in \bigcap U$? aren't they different? $\endgroup$ – oldselflearner1959 Feb 11 '18 at 16:16
  • $\begingroup$ Here, $x$ is an element of $\bigcup F\subseteq A$, while $X\in U$ is a subset of $A$. So no they are not the same thing. I have edited the mistake I made typing up the question originally to address your last comment, and I will add another edit to address $L$ in a bit. $\endgroup$ – anakhro Feb 11 '18 at 16:19
  • $\begingroup$ For the first part, I think it's possible to get from $x \in \bigcup F \implies \forall x \in \forall X \in U$ but I cannot get $\bigcap U$ still. $\endgroup$ – oldselflearner1959 Feb 11 '18 at 16:58
  • $\begingroup$ Sorry, which inclusion are you having difficulties with? $\bigcup F\subseteq\bigcap U$ or $\bigcap U\subseteq\bigcup F$? $\endgroup$ – anakhro Feb 11 '18 at 17:01

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