Produced by Charles Wells Revised 2017-03-03 Introduction to this website website TOC website index blog Back to head of Sets chapter

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

- The name "method of comprehension" can cause semantic contamination.
- "Comprehension" to most people means
**understanding**. Ignore that meaning. In the olden days, "comprehend" meant "contain". - If I were dictator of the math language, I would call it the "Method of whatsinit".

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

- (
**ES)**If you know that $a\in A$, then you can conclude that the statement $P(a)$ is true. - (
**SE**) If $P(a)$ is a true statement then you can conclude that $a\in A$. -
**(ES)**and**(SE)**together mean that if $A=\{x\,|\,P(x)\}$, then every $x$ for which $P(x)$ is true is an element of $A$, and*nothing else is*.

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

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

- The method of comprehension allow you to state
- Since $3\in\{2,3,4,5\}$,
**ES**says $1\lt3\lt6$. - Since $1\lt3\lt6$,
**SE**says $3\in\{2,3,4,5\}$.

- Since $3\in\{2,3,4,5\}$,
- It would be wrong to say that \[\{n\,|\,1\lt n\lt6\}=\{2,3,4,6\}\] because $1\lt6\lt5$ is false, violating
**ES**. - It would be wrong to say that \[\{n\,|\,1\lt n\lt6\}=\{2,3,4\}\] because $1\lt5\lt6$ but $5\notin\{2,3,4\}$, violating
**SE**.

- $\{n\,|\,1\lt n\text{ and }1\gt n\}=\emptyset$.
- $\{n\,|\,2\gt n\text{ or }0\lt n\}=\mathbb{N}$.

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.

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

- $E=\{n\,|\,n\text{ is even.}\}$
- $42\in E$ because $42$ is even.
- $43\notin E$ because $43$ is not even.
- Suppose you know that the integer $m$ is even. Then $m\in E$.
- Suppose you know that $k$ is odd. Then $k\notin E$.

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.

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

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

- Let $A=\{1,2,3\}$ and $B=\{1,2,3,4\}$. Then $A\subseteq B$ because, going through the elements of $A$, $1\in B$, $2\in B$ and $3\in B$.
- $A\subseteq \mathbb{R}$ because $1$, $2$ and $3$ are all real numbers.
- It is not true that $\{1,2,3,\pi\}\subseteq\mathbb{Z}$, because $\pi$ is not an integer.
- For any set $A$, $\emptyset\subseteq A$ because it is vacuously true that if $x\in\emptyset$ then $x\in A$.

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.

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.

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:

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

- $A=\{1,2,3\}$, $B=\{1,2\}$.
- $A=\{1,2,3\}$, $B=\{1,2,3,4\}$.
- $A=\{1,2,3\}$, $B=\{1,2,4\}$.

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