We discuss the notions of set builder. It is another way to describe a set.

In **set-builder form**, a set #\blue A# is described by means of a property #P# for an element #a# to belong to #\blue A#.

This means that an object #a# is an element of #\blue A# if and only if property #P# holds for #a#.

The corresponding notation is #\blue A = \{a \mid P \}#. This can be read as "The set #\blue A# consists of all elements #a# that satisfy property #P#".

We use the following two methods for formulating the set builder more succinctly. Here #P# is a statement, #S# is a set, and #f(x)# is an expression involving #x#.

- #\{a\in S \mid P \} =\{a \mid P \land \left(a\in S\right)\}#
- #\{f(x) \mid P \} =\{a \mid P \land a=f(x)\}#

**Example**

\[\begin{array}{rcl}\blue{\mathbb{Z}}&=&\{\dots, -2, -1, 0, 1, 2, \dots\}\\&& \\ \blue {\mathbb{Z}}&=&\{x \mid x ~\text{is an integer} \} \end{array}\]

The set of even integers is

\[\left\{x\,\left|\, x\in\mathbb{Z} \text{ and } 2 \text{ divides } x\right.\right\}\]

It can also be given in one of the following forms.

- #\left\{a\in\mathbb{Z}\,\left|\, 2 \text{ divides } a\right.\right\}#
- #\left\{2x\,\left|\,x\in\mathbb{Z}\right.\right\}#

If we have \[ P = a\in \mathbb{Z} \text{ and } a \lt 10\text{ and } a\gt 0\] then #\left\{a\mid P\right\}# is the set of all natural numbers smaller than #10#.

If we instead have \[ P = \left(\rv{x,y}\text{ lies on the unit circle around the origin}\right)\] then #\left\{ \rv{x,y} \mid P \right\}# describes the set of points in the plane on the unit circle.

If #P# is always false, then #\blue A=\emptyset#.

If #P # is always true, then #\blue A# is the "universe"; each element we can think of belongs to #A#. This set is too big to be of use for us, so it hardly ever appears in mathematics and in our course.

The property #P# in the set-builder #\blue A = \{a\mid P\}# is a statement about #a# that can take the values true or false. Such statements are known as *propositions* in logic.

Verifying whether #P# is satisfied (that is, takes the value true) for a particular object #a# amounts to deciding whether #a# belongs to the set. For a good definition of a set, some requirements on the property #P# are needed.

We will not go into details, but give an impossible property in the Paradox appearing further on.

We also note that caution is needed in the choice of *variables (also called proposition letters)* occurring in #P#. For instance, for the definition #\blue A = \{a\mid P\}# to make sense, we cannot have #\blue A# occurring in #P# (imagine

#\blue A = \{a\mid a\not\in A\}#) and the possibilities are scarce if the variable #a# does not occur in #P# (think of the comment *Two extremes*).

The variable #a# is regarded as a **bound variable**. This relates to the following two important properties.

- The name of the variable has no other meaning than to relate one occurrence to another inside #\{a\mid P\}#. So if #\blue A = \{a\mid 3a\in \mathbb{N}\}#, then we regard the description as being the same as #\blue A = \{b\mid 3b\in \mathbb{N}\}#. Of course we need to avoid using a variable name that is already taken elsewhere in #P#. For example, in the description #\blue A = \left\{a\,\left|\, {a}\in \mathbb{N}\land \frac{a}{m}\in \mathbb{N}\right.\right\}#, where #m# is a natural number, we cannot replace #a# by #m#. We would get #\blue A = \left\{m\,\left|\, {m}\in \mathbb{N}\land \frac{m}{m}\in \mathbb{N}\right.\right\} = \{m\}#, where #m# is a natural number. Which is not the same.
- The
**scope** of the name #a# is just the expression #\{a\mid P\}#. If #a# occurs outside this expression, it does not stand for the same internal variable #a#. So if #\blue A = \{a\mid 3a\in \mathbb{N}\}#, the expression #a\not\in \blue A # still makes sense; it just means #3a\not\in \mathbb{N}#.

A classical paradox making clear that requirements on #P# are needed arises from the property

\[P= \left(b \text{ is a barber who shaves all persons, and those only, who do not shave themselves}\right)\]

Now consider \(B=\left\{b\mid P\right\}\). If #b\in B#, then #b# is a barber who shaves all persons who do not shave themselves and none other. This implies that #b# is a barber. But does #b# shave himself? If he doesn't, then #b\in B# implies that #b# does shave himself, a contradiction. So #b# must shave himself. But #b\in B# implies that #b# shaves none other than those that do not shave themselves, so #b# does not shave himself, again a contradiction.

Logic can be set up so as to avoid such conflicts. Here, the paradox serves as a warning that propositions used for defining sets must be reasonable (pun intended).

The *method of describing sets by enumerating its elements* does not work for sets, such as #\mathbb{R}#, that are not enumerable. A typical example where the set-builder form works and enumeration does not is an interval of real numbers.

For instance, the interval of all real numbers #x# lying between #-\pi# and #\pi#, is given by

\[\ivcc{-\pi}{\pi} = \left\{x\mid x\in\mathbb{R}\ \land x\ge -\pi \land x\le \pi\right\}\]

This set is not enumerable.

Determine the value of the pair #\rv{a,b}# of integers for which the set

\[\left\{ x\,\left|\, x=a\cdot m+b\text{ for some natural number }m\right.\right\} \]

is equal to

\[\left\{9,15,21,\ldots\right\}\]

#\rv{a,b}=# #\rv{6,3}#

The expression #a\cdot m+b# increases for #m=1,2,\ldots#, so the first element , # 9 #, in the explicit description of the set must be equal to #a\cdot 1+ b# and the second to #a\cdot 2 + b#. This gives the system of equations

\[\eqs{ a+b &=& 9\\2a+b &=& 15}\]

The only solution of this system of two linear equations with two unknowns is #\rv{a,b}=\rv{6,3}#. Since #a# and #b# are integers, the answer is #\rv{a,b}= \rv{6,3}#.