# F and G are families of sets and $\mathcal F \cap \mathcal G ≠ \emptyset$, prove that $\bigcap \mathcal F \subseteq \mathcal \bigcup G$

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?

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

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 $$a\in \bigcup \mathcal G.$$
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 $$A\in \mathcal F\cap \mathcal G\dots?$$