# prove - If F ⊆ G then ∩G ⊆ ∩F

I'm self studying proofs (reading 'How to Prove it') and am stuck on how to prove this one. I'm hoping someone can tell me how they would go about.

A little more background. This problem is in a chapter on Quantifiers. I often get stuck when the 'Givens' in a proof all have universal quantifiers and no proven elements that I can plug in. As I have read, you can only do universal instantiation if you have a particular value you can plug in.

Anyway, any help is MUCH appreciated :)

If F ⊆ G then ∩G ⊆ ∩F

• What do you mean by $\cap G$? Are $F$ and $G$ just sets, or families of sets? You need more than one set to do an intersection. – Nick Feb 25 '16 at 16:27
• @Nick Families of sets. It's standard notation. – BrianO Feb 25 '16 at 16:36
• It's bad standard notation. One should really write something like $\bigcap_i G_i$. – D_S Feb 25 '16 at 16:46
• It's great, classic, standard notation. There's no index set, none is needed. If you want to be silly, go ahead and write$$\bigcup_{X\in G} X$$ $\bigcap, \bigcup$ don't only apply to functions. In ZF(C), the Union axiom isn't stated in terms of functions (indexed families); it states that for every set $A$, $\bigcup A$ exists, where $\bigcup A := \{x\mid (\exists a\in A)\,x\in a \}$. – BrianO Feb 25 '16 at 19:19

This is a very simple definition chase. Suppose $x\in \bigcap G$. Then $(\forall X\in G)\, x\in X$. Now for any $X\in F$, we have $X\in G$ because $F\subseteq G$, so $x\in X$. That is, $(\forall X\in F)\,x\in X$. Thus, by definition, $x\in \bigcap F$.
An intuitive explanation first. Do you agree that $$E_1 \cap E_2 \cap E_3 \subseteq E_1 \cap E_2 \subseteq E_1\ ?$$ General case. If $F \subseteq G$, we get $$\bigcap_{i \in F} E_i = \{ x \in E \mid \text{for all i \in F, x \in E_i}\} \text{ and } \bigcap_{i \in G} E_i = \{ x \in E \mid \text{for all i \in G, x \in E_i}\}$$ Thus if $x \in \bigcap_{i \in G} E_i$, one has $x \in E_i$ for all $i \in G$ and in particular, for all $i \in F$. The inclusion $\bigcap_{i \in G} E_i \subseteq \bigcap_{i \in F} E_i$ follows.
• @maybedave The notation in this answer isn't exactly the same as in your question. Your $F,G$ are just sets of sets, but here they're treated as index sets, and the sets whose intersection is taken are these $E_i$ things that don't figure in your question at all. For starters, think $X$ not $i$, and $E_i = E_X = X$. Thus $\bigcap_{i\in G}E_i$ is just $\bigcap_{X\in G}X = \bigcap G$. – BrianO Feb 25 '16 at 19:30