Suppose $\mathcal F$ and $\mathcal G$ are families of sets and $\mathcal F \cap \mathcal G ≠ \emptyset$. Prove that $\bigcap \mathcal F \subseteq \mathcal \bigcup G$

My attempt:

$\mathcal F \cap \mathcal G ≠ \emptyset$ implies that there is at least one set, call it $A$, such that $A \in \mathcal F$ and $A \in \mathcal G$

If $\cap \mathcal F = \emptyset$ then $\cap \mathcal F \subseteq \cup \mathcal G$ because empty set is a subset of any set.

If $\cap \mathcal F ≠ \emptyset$, then it implies that there is at least one element, call it $x$, such that $x$ is in every subset of $\mathcal F$. Because $A \in \mathcal F$, it follows that $x \in \cap \mathcal F \implies x \in A$.

$\cup \mathcal G$ is the collection of all elements of all subsets in $\mathcal G$. If $A \in \mathcal G$ then $A \subseteq \cup \mathcal G$, or in other words, $\forall x(x \in A \implies x \in \mathcal \cup \mathcal G$)

Combining everything said above, we have $x \in \cap \mathcal F \implies x \in A \implies x \in \mathcal \cup \mathcal G$. Hence $\cap \mathcal F \subseteq \mathcal \cup \mathcal G$

Is it correct?

  • 1
    $\begingroup$ "because .. then" is not correct english $\endgroup$ – mathworker21 Jul 24 at 19:33
  • $\begingroup$ @mathworker21 corrected $\endgroup$ – Nelver Jul 24 at 19:34
  • $\begingroup$ perfect solution :) $\endgroup$ – mathworker21 Jul 24 at 19:35

It looks fine.

You really don't need to worry about the case when $\bigcap \mathcal F=\emptyset.$ You just need to prove that if $a\in \bigcap \mathcal F$ then $a\in \bigcup \mathcal G.$

More briefly, you've shown two things:

  1. If $A\in \mathcal F$ then $\bigcap \mathcal F\subseteq A.$

  2. If $A\in\mathcal G$ then $A\subseteq \bigcup \mathcal G.$

So given a common set $A\in \mathcal F\cap \mathcal G\dots?$


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