There are a few possibilities, but here is the one approach. Even the starting point—the set of natural numbers $\mathbb{N}$—can be defined in several ways, but the standard definition takes $\mathbb{N}$ to be the set of finite von Neumann ordinals. Let us assume that we do have a set $\mathbb{N}$, a constant $0$, a unary operation $s$, and binary operations $+$ and $\cdot$ satisfying the axioms of second-order Peano arithmetic.
First, we need to construct the set of integers $\mathbb{Z}$. This we can do canonically as follows: we define $\mathbb{Z}$ to be the quotient of $\mathbb{N} \times \mathbb{N}$ by the equivalence relation
$$\langle a, b \rangle \sim \langle c, d \rangle \text{ if and only if } a + d = b + c$$
The intended interpretation is that the equivalence class of $\langle a, b \rangle$ represents the integer $a - b$. Arithmetic operations can be defined on $\mathbb{Z}$ in the obvious fashion:
$$\langle a, b \rangle + \langle c, d \rangle = \langle a + c, b + d \rangle$$
$$\langle a, b \rangle \cdot \langle c, d \rangle = \langle a c + b d, a d + b c \rangle$$
(Check that these respect the equivalence relation.)
Again, this is not the only way to construct $\mathbb{Z}$; we can give a second-order axiomatisation of the integers which is categorical (i.e. any two models are isomorphic). For example, we may replace the set $\mathbb{Z}$ by $\mathbb{N}$, since the two sets are in bijection; the only thing we have to be careful about is to distinguish between the arithmetic operations for $\mathbb{Z}$ and for $\mathbb{N}$. (In other words, $\mathbb{Z}$ is more than just the set of its elements; it is also equipped with operations making it into a ring.)
Next, we need to construct the set of rational numbers $\mathbb{Q}$. This we may do using equivalence relations as well: we can define $\mathbb{Q}$ to be the quotient of $\mathbb{Z} \times (\mathbb{Z} \setminus \{ 0 \})$ by the equivalence relation
$$\langle a, b \rangle \sim \langle c, d \rangle \text{ if and only if } a d = b c$$
The intended interpretation is that the equivalence class of $\langle a, b \rangle$ represents the fraction $a / b$. Arithmetic operations are defined by
$$\langle a, b \rangle + \langle c, d \rangle = \langle a d + b c, b d \rangle$$
$$\langle a, b \rangle \cdot \langle c, d \rangle = \langle a c, b d \rangle$$
And as before, we can give an axiomatisation of the rational numbers which is categorical.
Now we can construct the set of real numbers $\mathbb{R}$. I describe the construction of Dedekind cuts, which is probably the simplest. A Dedekind cut is a pair of sets of rational numbers $\langle L, R \rangle$, satisfying the following axioms:
- If $x < y$, and $y \in L$, then $x \in L$. ($L$ is a lower set.)
- If $x < y$, and $x \in R$, then $y \in R$. ($R$ is an upper set.)
- If $x \in L$, then there is a $y$ in $L$ greater than $x$. ($L$ is open above.)
- If $y \in R$, then there is an $x$ in $R$ less than $y$. ($R$ is open below.)
- If $x < y$, then either $x \in L$ or $y \in R$. (The pair $\langle L, R \rangle$ is located.)
- For all $x$, we do not have both $x \in L$ and $x \in R$. ($L$ and $R$ are disjoint.)
- Neither $L$ nor $R$ are empty. (So $L$ is bounded above by everything in $R$ and $R$ is bounded below by everything in $L$.)
The intended interpretation is that $\langle L, R \rangle$ is the real number $z$ such that $L = \{ x \in \mathbb{Q} : x < z \}$ and $R = \{ y \in \mathbb{Q} : z < y \}$. The set of real numbers is defined to be the set of all Dedekind cuts. (No quotients by equivalence relations!) Arithmetic operations are defined as follows:
- If $\langle L, R \rangle$ and $\langle L', R' \rangle$ are Dedekind cuts, their sum is defined to be $\langle L + L', R + R' \rangle$, where $L + L' = \{ x + x' : x \in L, x' \in L' \}$ and similarly for $R + R'$.
- The negative of $\langle L, R \rangle$ is defined to be $\langle -R, -L \rangle$, where $-L = \{ -x : x \in L \}$ and similarly for $-R$.
- If $\langle L, R \rangle$ and $\langle L', R' \rangle$ are Dedekind cuts, and $0 \notin R$ and $0 \notin R'$ (i.e. they both represent positive numbers), then their product is $\langle L \cdot L' , R \cdot R' \rangle$, where $L \cdot L' = \{ x \cdot x' : x \in L, x' \in L', x \ge 0, x' \ge 0 \} \cup \{ x \in \mathbb{Q} : x < 0 \}$ and $R \cdot R' = \{ y \cdot y' : y \in R, y \in R' \}$. We extend this to negative numbers by the usual laws: $(-z) \cdot z' = -(z \cdot z') = z \cdot -z'$ and $z \cdot z' = (-z) \cdot -z'$.
John Conway gives an alternative approach generalising the Dedekind cuts described above in his book On Numbers and Games. This eventually yields Conway's surreal numbers.
The set $\{x| x\in \mathbb N\land x<1000\}$ is finite and it can easily be presented in the set builder notation...
All integers greater than $0$ but less than $6$:
$$\{n| n\in\mathbb N\land n>0 \land n<6\}$$
All positive even integers less than $11$:
$$\{n| n\in\mathbb N\land n<11\land (\exists k\in\mathbb N: n=2k)\}$$
or, alternatively,
$$\{2k| k\in\mathbb N\land 2k<11\}$$
All perfect squares less than $101$:
$$\{n| n\in\mathbb N\land n<101\land (\exists k\in\mathbb N: n=k^2)\}$$
or, again, alternatively as
$$\{k^2| k\in\mathbb N\land k^2<101\}$$
Best Answer
You have a few errors in that. Try (and yes, comma instead of $\land$ is not uncommon) $$A=\{b^2+2k: k\in\mathbb Z, b\in\mathbb N, b>3\}\cap M $$