How to define sets using set-builder notation where the elements of the set are in AP?

Question

I am trying to define sets such as: $$A=\{...,-8,-3,2,7,12,17,...\}$$ $$B=\{...,-\frac{3}{2},-\frac{3}{4},0,\frac{3}{4},\frac{3}{2},\frac{9}{4},3,\frac{15}{4},\frac{9}{2},...\}$$ using set-builder notation and I'm not sure how to proceed. The obvious observation for sets such as these is that the elements of the sets are in Arithmetic Progression (AP) with an associated common difference; however, my problem is that since there is no element which can be considered as the initial value, I'm not sure how to come up with an equation to define the AP that also caters to the definition of the set.

Answer 1

If it's an infinite sequence in each direction, first choose any "initial" term $a$. (It doesn't matter which is "initial" in this case.) Then, for common difference $d$, we could write

$$S = \{ x \in \mathbb{R} \mid x = a+nd \text{ for some } n \in \mathbb{Z} \}$$

or, less clunkily,

$$S = \{a+nd \mid n \in \mathbb{Z} \}= \{a+nd\}_{n \in \mathbb{Z}}$$

For instance, I would write $B$ with $a=0$ (it's convenient) and $d=3/4$, to get

$$B = \left\{ \frac 3 4 n \right\}_{n \in \mathbb{Z}}$$

or $A$ with $a=2$ and $d=5$ to get

$$A = \{2+5n\}_{n \in \mathbb{Z}}$$

In essence, the jump to be making here is that, for "typical" sequences (extending in only one direction), we would normally be using $\mathbb{N}$ or a subset thereof, instead of $\mathbb{Z}$.