Although basic set theory was already discussed in module 0280, that was discussed without the use of quantifiers. Quantifiers are useful tools for a variety of math topics, set theory is a part of those topics. By reexamining basic set theory, we reinforce the understanding of both set theory as well as quantifiers.
Given two items, \(x\) and \(y\), how do we know they are the same? In order words, how do we evaluate \(x=y\)?
First of all, if one is a set and other is not, then \(x=y\) is false.
Given that \(x\) and \(y\) are not sets, then the “normal” comparison applies.
Given that \(x\) and \(y\) are sets, then we can say the following:
\((x=y) \Leftrightarrow ((\forall e \in x(e \in y)) \wedge (\forall e \in y(e \in x)))\)
There is a more concise statement that states the same:
\((x=y) \Leftrightarrow (\forall e((e \in x) \Leftrightarrow (e \in y)))\)
Recall the notation \(E=\{x|P(x)\}\) describes a set \(E\) where all its elements make the predicate \(P\) true. We mentioned how this means \(x \in E \Leftrightarrow P(x)\) in module 0280. The more correct statement is as follows:
\(\forall x(x \in E \Leftrightarrow P(x))\)
This makes it clear where we get (how we bind) the value of the variable to some “thing.” \(x\) can be anything in the universe! Without the quantifier \(\forall \), one can ask the question where do we get a value for \(x\) to be evaluated.
How do we know that a set \(S\) is empty?
\(\neg \exists x(x \in S)\)
This literally says “there does not exist a thing \(x\) such that \(x\) is in \(S\).”
\(\forall x((x \in (A \cap B)) \Leftrightarrow (x \in A \wedge x \in B))\)
and
\(\forall x((x \in (A \cup B)) \Leftrightarrow (x \in A \vee x \in B))\)
\(\forall x \in A(\forall y \in B((x,y) \in (A \times B))) \wedge \forall (x,y) \in A \times B(x \in A \wedge y \in B)\)
\(A \subseteq B \Leftrightarrow \forall e \in A(e \in B)\)
\(A \subset B \Leftrightarrow (A \subseteq B \wedge \exists x \in B(\neg x \in A))\)