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I'm studying for my exam of Logic later this week and I came across this question:

Suppose that $\mathscr{F} \subseteq\mathcal P(X)$ and $\mathscr{G} \subseteq \mathcal P(X)$ so that every element of $\mathscr{F}$ a subset is of every element of $\mathscr{G}$. Then applies $\bigcup\mathscr{F} \subseteq \bigcap \mathscr{G}$.

I think that this is a very obvious theorem but I can't find any easy and clear why of proofing it. Is there anyone who can help me?

Note: $\bigcup\mathscr{F}$ and $\bigcap\mathscr{F}$ are in our course defined as:

$$\bigcup\mathscr{F} = \{x\in X\mid\exists A\in\mathscr{F}\text{ s.t. } x \in A\}$$

$$\bigcap\mathscr{F} = \{x\in X\mid\forall A\in\mathscr{F}\text{ s.t. } x \in A\}$$

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  • $\begingroup$ It is enough to show that every $Y\in\mathcal F$ is a subset of $\cap\,\mathcal G$. This is true by hypothesis, and so $\cap\,\mathcal G$ contains all the sets $Y\in\mathcal F$, hence their union, which is $\cup\,\mathcal F$. $\endgroup$ – Olivier Bégassat Jan 11 '15 at 18:41
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Let $x \in \cup \mathscr{F}$. Then there exists some $A \in \mathscr{F}$ such that $x \in A$.

Let $B$ be any set from $\mathscr{G}$. By assumption, $A \subseteq B$. So $x \in B$. As $B \in \mathscr{G}$ was arbitrary, $x \in \cap \mathscr{G}$.

As $x$ was arbitrary, $\cup \mathscr{F} \subseteq \cap \mathscr{G}$.

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  • $\begingroup$ Thank you very much! This is indeed a very clear proof. $\endgroup$ – Pieter Verschaffelt Jan 12 '15 at 7:55
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HINT: Show that $\bigcup\scr F$ is a subset of every element of $\scr G$, then conclude it is a subset of $\bigcap\scr G$. Remember that taking a union (or intersections) of subsets of some $A$, will result in a subset of $A$.

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