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This question arose while reading Velleman's How to Prove It in section 2.3. For context, the definition of ∩F is { x |∀A ∈F(x ∈ A)}={ x |∀A(A ∈F → x ∈ A)} and ∪F is { x |∃A ∈F(x ∈ A)}={ x |∃A(A ∈F∧x ∈ A)}. F is a family of sets.

The question, however, doesn't have much to do with these definitions than it does with the title and a more generalized issue of these two statements when used with respect to sets. Why is the definition for ∩F defined as ∀A(A ∈F → x ∈ A) but not ∀A(A ∈F∧x ∈ A)? Aren't they both saying checking for the same conditions: that A belongs to F and x belongs to A?

I understand why they aren't the same by transforming the definition for ∩F into a logical statement: ∀A(A does not belong to F or x belongs to A), which is obviously different than ∀A(A ∈F∧x ∈ A).

I suppose I just need some convincing in some way other than a logical statement.

Andrés E. Caicedo
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user2793618
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2 Answers2

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$\forall A(A\in F\land x\in A)$ can only be true if everything is an element of $F$, which is not the case. So that definition would give you the same as $\{ x\ \mid \text{false}\}$, which is the empty set no matter what $F$ is.

hmakholm left over Monica
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  • That raises two questions. One: why does ∀A(A does not belong to F or x belongs to A) not mean not every A does not belong to F? Two: why doesn't the ∪F use the if then format? – user2793618 Aug 15 '18 at 21:41
  • @user2793618: Because it is possible for $A\notin F \lor x\in A$ to be true for all $A$, namely if it so happens that $x$ is an element of every element of $F$. – hmakholm left over Monica Aug 15 '18 at 21:43
  • That's right silly me. It's an or statement. How about question two? But also realistically for A∉F∨x∈A, A∉F will not be true right? – user2793618 Aug 15 '18 at 21:45
  • @user2793618: Because $\exists A(A\in F\to x\in A)$ is true no matter what $F$ and $x$ are -- just choose $A$ to be something that is not in $F$. – hmakholm left over Monica Aug 15 '18 at 21:46
  • $\forall A(A\in F\wedge x\in A)$ is always false because: By putting $A=F$, we have $F\in F\wedge x\in F$. However, by some regularity axiom, if $F$ is a set, it is false that $F\in F$. – Danny Pak-Keung Chan Aug 15 '18 at 21:47
  • @user2793618: $A\notin F$ will certainly be true for a great many different $A$s -- in fact that vast majority of things in the universe are not elements of your $F$. – hmakholm left over Monica Aug 15 '18 at 21:48
  • Thanks guys! I appreciate it. – user2793618 Aug 15 '18 at 21:48
  • @HenningMakholm will A∉F be true for those X's that pass the element hood test? – user2793618 Aug 15 '18 at 21:50
  • @user2793618: $A\notin F$ does not depend on $x$ at all. – hmakholm left over Monica Aug 15 '18 at 21:51
  • Sorry my response was unclear. If we have this statement, { x |∀A(A∉F∨x∈A)}, and for a particular x that passes the element hood test, A∉F would be false right? – user2793618 Aug 15 '18 at 21:54
  • @user2793618: It would be false for some $A$s and true for others. This is the case no matter what $x$ is, because $A\notin F$ does not depend on what $x$ is. – hmakholm left over Monica Aug 15 '18 at 21:55
  • In what case would that happen? It seems weird because the if statement asserts that A must belong to F in order for x to belong to A – user2793618 Aug 15 '18 at 21:59
  • @user2793618: It happens in every case that $A\notin F$ does not depend on $x$. It happens in every case that $A\notin F$ is true for some $A$s and false for other $A$s. Yes, the form ${x\mid \forall A(A\in F\land x\in A)}$ asserts that every $A$ must be in $F$ in order for $x$ to be in the set being defined. That is why that form doesn't work. (But there is nothing in that formula that can reasonably or unreasonably be described as an "if statement"). – hmakholm left over Monica Aug 15 '18 at 22:01
  • I got it now. Thanks. – user2793618 Aug 15 '18 at 22:03
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(It is not a direct answer to the question.)

For most cases, we have that $\cap F\subseteq\cup F$ excepts when $F=\emptyset$. In that case, $\cap F=$ universal set while $\cup F=\emptyset$. Play attention to this special case.

You may do an experiment and try to consider the case: $F=\{A_1, A_2, A_3\}$. Then, you will observe that $\cap F=A_2\cap A_2\cap A_3$ and $\cup F = A_1 \cup A_2 \cup A_3$.

Danny Pak-Keung Chan
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