Commonplace sets I recently started reading about sets of numbers, set builder notation, and operations on sets of numbers. To practice using different symbols (e.g., $\wedge$) and different set "layouts," I decided to delineate some common categories/sets of numbers. Some of them are named with single symbols while others are not. In addition, some are defined using more than one set or a union of sets. However, most equivalent sets that can be achieved through basic algebraic manipulation (e.g., substituting $\sqrt{-k}$ for $k\sqrt{-1}$) have been excluded.

Natural: $\Bbb{N}=\{ 1,2,3,4,\ldots\}$
Whole: $\Bbb{W}=\Bbb{N}\cup\{0\}=\{ 0,1,2,3,4,\ldots\}$
Integers: $\Bbb{Z}=\{ z\mid\left| z\right|\in\Bbb{W}\}=\{\ldots,-4,-3,-2,-1,0,1,2,3,4,\ldots\}$
Integers excluding zero: $\Bbb{Z_{0}}=\Bbb{Z}\setminus\{0\}$
Even: $\{n\mid n/2\in\Bbb{Z}\}=\{\ldots,-4,-2,0,2,4,\ldots\}$
Odd: $\{n\mid (n+1)/2\in\Bbb{Z}\}=\{\ldots,-3,-1,1,3,\ldots\}$
Prime: $\{p\mid p/k\notin\Bbb{Z}\wedge p\in\Bbb{Z}\wedge k\neq 0\wedge k\neq 1\wedge k\neq p\}=\{2,3,5,7,11,\ldots\}$
Composite: $\{p\mid p=hk\wedge h\in\Bbb{Z}\setminus\{p\}\wedge k\in\Bbb{Z}\setminus\{p\}\}$
Units: $\{u\mid 1/u\in\Bbb{W}\}=\{1\}$
Zero divisors: $\{d\mid dh=0\wedge h\neq 0\}=\{0\}$
Rational: $\Bbb{Q}=\{ q\mid q=a/b\wedge a\in\Bbb{Z}\wedge b\in\Bbb{Z_{0}}\}=\{a/b\mid a\in\Bbb{Z}\wedge b\in\Bbb{Z_{0}}\}$
Irrational: $\Bbb{R}\setminus\Bbb{Q}$
Real: $\Bbb{R}=\{r\mid -\infty <r<\infty\}=\{r\mid r\in (-\infty,\infty)\}$
Real excluding zero: $\Bbb{R}_{0}=\Bbb{R}\setminus\{0\}$
Non-real: $\Bbb{I}\cup\Bbb{C}$
Imaginary: $\Bbb{I}=\{j\mid j^{2}<0\}=\{j\mid j=k\sqrt{-1}\wedge k\in\Bbb{R_0}\}=\{k\sqrt{-1}\mid k\in\Bbb{R_0}\}$
Complex: $\Bbb{C}=\{c\mid c=a+b\sqrt{-1}\wedge a\in\Bbb{R_{0}}\wedge b\in\Bbb{R_{0}}\}=\{a+b\sqrt{-1}\mid a\in\Bbb{R_{0}}\wedge b\in\Bbb{R_{0}}\}=\{a+b\mid a\in\Bbb{R_0}\wedge b\in\Bbb{I}\}$

In the above image, the use of question marks indicates that the sub-categories are not mutually exclusive (e.g., zero is both whole and even).

Unlike some of the other sets, $\Bbb{N_0}$ and $\Bbb{R_0}$ have been equated to a single symbol not because those symbols are conventional but rather because doing so made the delineation of the sets of rational, irrational, composite, imaginary, and complex numbers easier and clearer.
Also, I was not completely sure if $\Bbb{R}=(-\infty,\infty)$, and I was very unsure if it is correct to say $\Bbb{Z}=\pm\Bbb{W}$.
So, here's my question:

*

*Have a made any 'grammatical' or 'punctuational' errors? Did I use $\wedge$ too much?


*Other than taking away one set from the setof real numbers (e.g., real take away composite equals prime), are there any quick or easy ways of defining the sets that I missed?
Sorry for the novice question; I just finished pre-calculus. Much thanks in advance! —C.T.
 A: I hope you don't consider this an overly harsh critique but:
--Natural: N={1,2,3,4,…}
Fine.
--Whole: W=N∪{0}={0,1,2,3,4,…}
Fine.
--Integers: Z={z∣|z|∈W}={…,−4,−3,−2,−1,0,1,2,3,4,…}
Now you have your first real problem.  You have defined what universal set $z$ is in, and what the definition of $|z|$ is.  Consider $z = 1/\sqrt{2} + \sqrt{-1}*1/\sqrt{2}$.  Then $|z| = 1$ but clearly $z$ is not an integer.
Basically you have a constructivist choice to make.  You can start from the basic $\mathbb U = \{1\}$ and build yourself up, or start for the universal top $\mathbb C$ and restrict yourself down.
Better to do $\mathbb Z = \{x| x $ may be a solution to $a + x = b|a,b \in \mathbb W\}$.
In fact if you want to be real thorough you should do:
$\mathbb U = \{1\}$
$\mathbb N = $ the most basic set where $1 \in \mathbb N$ and whenever $n \in \mathbb N \implies n + 1 \in \mathbb N$.
$\mathbb W = \mathbb N \cup $ {$0| n + 0 = n; n \in \mathbb N$ has solution}
$\mathbb Z = \mathbb W \cup \{z| w + z = 0; w \in \mathbb W\}$
--Integers excluding zero: Z0=Z∖{0}
Fine.
---Even: {n∣n/2∈Z}={…,−4,−2,0,2,4,…}
---Odd: {n∣(n+1)/2∈Z}={…,−3,−1,1,3,…}
Fine.
Prime: {p∣p/k∉Z∧p∈Z∧k≠0∧k≠1∧k≠p}={2,3,5,7,11,…}
Composite: {p∣p/k∈Z0∧k≠0∧k≠1∧k≠p}
---Units: {u∣1/u∈W}={1}
okay
---Zero divisors: {d∣dh=0∧h≠0}={0}
okay
---Rational: Q={q∣q=a/b∧a∈Z∧b∈Z0}={a/b∣a∈Z∧b∈Z0}
Fine
---Irrational: R∖Q
What the heck is $\mathbb R$?  You can't refer to $\mathbb R$ if you haven't defined it first.
---Real: R={r∣−∞

What is your "universe"? You are assuming that your "universe" is $\mathbb R$ without defining what it is.
This is really kind of tough because you have probably never learned it.  Usually you are taught that $\sqrt{2} \not \in \mathbb Q$ and so you conclude that "$\mathbb Q$ has holes in it" so you say "$\mathbb R$ is $\mathbb Q$ with the holes filled in" but never worry about what the means or what you are using to fill the holes in with and where those numbers are coming from.
In actuality $\mathbb R$ is the most basic set where $\mathbb Q \subset \mathbb R$ and for all sets $A \subset R$ where all $x \in A$ are such that $x \le a$ for some $a \in R$ then there exist some number $b \in R$ so that all $x \in A$ are such that $x \le b$ and for all $y < x$ that there is some $w \in A$ such that $y < w$.  
This is that $\mathbb R$ has the "least upper bound property".  That any set $A$ that is bounded (all $x$ in $A$ are bounded in size) there is some number that is an upper bound but is the smallest upper bound.  
$\mathbb Q$ does not have the least upper bound property.  Let $A = \{q < \sqrt{2}|q \in \mathbb Q\}$.  Then all the $x \in A$ have $x < \sqrt{2}$ but there is no smallest rational number that is bigger or as big as all $x \in A$.  The smallest such number is $\sqrt{2}$ and it isn't rational.
---Real excluding zero: R0=R∖{0}
Fine.
---Non-real: I∪C
What is $\mathbb I$?  What is $\mathbb C$?  And $\mathbb I \subset \mathbb C$ so $\mathbb I \cup \mathbb C = \mathbb C$.  And $\mathbb R \subset \mathbb C$ so.... $\mathbb R \subset $ the non-reals???
This is just wrong.  You mean Non-real = $\mathbb C /\mathbb R$.  But you have to define $\mathbb C$ first.
---Imaginary: I={j∣j2<0}={j∣j=k−1−−−√∧k∈R0}={k−1−−−√∣k∈R0}
What you universe does $j$ live in?  Before you were assuming your universe was $\mathbb R$.  Now you are assuming it is something..."bigger".
Need to do $i = $ solution to $i^2 + 1 = 0$.  $\mathbb I = \{i*x|x \in \mathbb R\}$.  $\mathbb C = \{a + bi| a,b \in \mathbb R\}$.
--Complex: C={c∣c=a+b−1−−−√∧a∈R0∧b∈R0}={a+b−1−−−√∣a∈R0∧b∈R0}={a+b∣a∈R0∧b∈I}
$a,b$ can be 0.
