MHB Proving a Set Theory Statement Regarding Families of Sets

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The discussion focuses on proving the statement that if F is a subset of the intersection of families G1 and G2, then the union of their intersections is a subset of the intersection of F. The proof begins by assuming an arbitrary element from F and demonstrating its membership in G1 and G2. The correct structure for the proof involves fixing arbitrary elements from the intersections of G1 and G2 and showing that they belong to the intersection of F. A suggestion is made to properly nest the reasoning within the proof to enhance clarity. Overall, the proof is validated with recommendations for improvement in its logical structure.
logan3
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I was wondering if anyone could please check my work and reasoning for this problem. Thank-you! (Also, would this be considered a direct proof? How might a contradiction and IFF proof look like and compare?)

Problem: Suppose F, G1 and G2 are nonempty families of sets. Prove that if FG1G2, then ∩ G1 ∪ ∩ G2 ⊆ ∩ F.

Solution: Suppose F ⊆ G1 ∩ G2. Let A be an arbitrary element of F. Then since F ⊆ G1 ∩ G2 and A ∈ F, A ∈ G1 ∩ G2.

Let x be an arbitrary element of ∩ G1 and y be an arbitrary element of ∩ G2, which are defined since G1 and G2 are nonempty. Then by definition 2.3.5. (see below), ∀A (A ∈ G1 → x ∈ A) and ∀A (A ∈ G2 → y ∈ A). Thus, x, y ∈ A. Since A is an arbitrary element of F and x, y ∈ A, then x, y ∈ ∩ F, which is defined since F is nonempty. But x and y are arbitrary elements of ∩ G1 and ∩ G2, respectively, therefore ∩ G1 ∪ ∩ G2 ⊆ ∩ F.

Definition 2.3.5. Suppose F is a family of sets. Then the intersection is the set ∩ F and defined as: ∩ F = {x | ∀AF (xA)} = {x | ∀A (AFxA)} (Velleman, 2006, p. 77).
 
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Your proof is correct, but I would change one subtle point. Proofs of statements of the form $\forall x\,P(x)$ often have the following shape: fix some $x$; prove $P(x)$; since $x$ was arbitrary conclude $\forall x\,P(x)$. If you have several such lines of reasoning within one proof, they should be properly nested. For example:

Code:
fix some x
  fix some y
    prove Q(y)
  conclude ∀y Q(y)
  use ∀y Q(y) to prove P(x)
conclude ∀x P(x)

In your case you are proving $\bigcap G_1\cup\bigcap G_2\subseteq\bigcap F$, so you should start by fixing an arbitrary $x\in\bigcap G_1\cup\bigcap G_2$ and considering two cases: $x\in\bigcap G_1$ and $x\in\bigcap G_2$. (You are considering these cases simultaneously, which is fine.) Your next task is to prove $\forall A\;(A\in F\to x\in A)$, so you fix an arbitrary $A$ and assume $A\in F$. This implies $A\in G_1$ and $A\in G_2$, so $x\in A$. This concludes a subproof of $\forall A\;(A\in F\to x\in A)$, i.e., $x\in \bigcap F$. At this point $A$ does not exist because the subproof that considered a specific $A$ is closed, but $x$ still does. Finally, you conclude $\forall x\;(x\in \bigcap G_1\cup\bigcap G_2\to x\in \bigcap F)$ and close the scope of $x$.
 
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