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SETS: RULES OF INFERENCE

The method of comprehension

The method of comprehension is the main tool for proving statements involving setbuilder notation.

Warning

Method of proof for setbuilder notation:

Let $P(x)$ be an assertion and let $A$ be the set $\{x\,|\,P(x)\}$. Then the following two statements are true:

The elements of $\{x\,|\,P(x)\}$ are exactly all those $x$ that make $P(x)$ true.

Example

For an integer $n$, the assertion \[\{n\,|\,1\lt n\lt6\}=\{2,3,4,5\}\] is correct.

Other examples

The method of comprehension and equivalence

Another way of stating the method of comprehension is that the statements "$a\in A$" and "$P(a)$" are equivalent. This means:

The method of comprehension works both ways.

Example

Let $E$ be the set of even integers. Then:

Warning

The definite article “the” has a special role when defining a set. For example “the set of even integers” automatically means the set of all even integers. There is more about this in the Glossary.

Set inclusion

Inclusion is discussed in detail in the article Subsets and Inclusion. There is a rule of inference for inclusion:

Method of proof for inclusion

To prove that $A\subseteq B$, you must prove that any element of $A$ is an element of $B$.

Examples

Thus the empty set is included in every set. But the empty set is not an element of every set. For example, it is not an element of $\{1,2,3\}$, as you can see by looking at the list of elements in $\{1,2,3\}$ -- you don't see "$\emptyset$" in that list, do you?

The notion that the empty set is an element of every set is a myth widely believed by undergraduate math students. Get over it.

Set equality

This is a basic fact about sets:

$A\subseteq B$ and $B\subseteq A$ if and only if $A$ and $B$ are the same set.

In other words, For "$A=B$" to be true, $A$ and $B$ have to have exactly the same elements.

Inequality

So how do you show that two sets are not equal? Answer: You have to show that at least one of the following statements is true:

  1. There is an element $x\in A$ for which $x\notin B$.
  2. There is an element $x\in B$ for which $x\notin A$.

Think through the three examples below. Altogether, they show you that it can happen that one of the statements above can be true and the other false, and also that both statements can be true.


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