Variables a and b assume values of two number sets as part of a grid search. a assumes the first 21 multiples of 1440, b assumes values from 0 to 1 in steps of $\dfrac{1}{100}$. To clarify ... $$a = [1440 \cdot 0,~1440 \cdot 1,~...,~1440 \cdot 21]$$ $$b = [0,~0.01,~...,~1]$$
I formulated this as seen below, which I am very unsure about. $$a \in \{i \in \mathbb{Z} \cap [1,21] | 1440\}$$ $$b \cdot 10^{2} \in \mathbb{Z} \cap \frac{1}{100}\cdot [0,100]$$
Is this correct?
Notation is convention, and depending on your background and your target audience, some conventions maybe different.
Usually in mathematics, we list set elements in between curly braces $$\{\text{set,elements,here,between,commas,}\cdots \text{,lots,of,elements}\}$$ and we reserve the bracket notation $[x,y]$ to denote all real numbers between $x$ and $y$: $$[x,y] = \{ r \in \mathbb{R} \mid x \leq r \leq y\}$$
So let's look at this set: $\mathbb{Z} \cap \dfrac{1}{100} \cdot [0,100]$. Considering $[0,100]$ as the real number interval, we have that $$\frac{1}{100}\cdot [0,100] = \left\{\frac{r}{100}\mid r \in [0,100]\right\} = [0,1]$$
So that set is the interval $[0,1]$. Intersecting it with the integers would yield the binary set $\{0,1\}$. So saying that $b\cdot 10^2$ belongs to this set states that $b\cdot 10^2$ equals $0$ or $1$ (i.e., $b = 0$ or $b = 0.01$).
Let's say that we want to refer to all values of $b$ ranging from $0$ to $1$ in steps of $\dfrac{1}{100}$. One way of doing this in the notation we established is simply by stating: $$b \in \frac{1}{100} I_{100}\,.$$
Another way, this time using the set-builder notation:
$$b \in \left\{\frac{x}{100} \mid 0\leq x \leq 100\right\}$$
More compactly: $b = \dfrac{x}{100}$ for $0 \leq x \leq 100$ an integer. I leave the analysis of the multiples of $1440$ for you. Also, even if $[0,100]$ meant $I_{100}$, intersecting it with $\mathbb{Z}$ would also yield you a set with only $2$ elements (why?).