How is a real number a proper subset of $ℚ$! (dedekind cut)? I just started studying some set theory using "Classic Set Theory: For Guided Independent Study" and i've been stuck at the 8th and 9th page for 2 days lol. It says:
A Dedekind left set is a subset of $r$ of $ℚ$ with the following properties:

*

*$r$ is a proper, non-empty subset of  $ℚ$

*if $q∈r$ and $p<q$, then $p∈r$

*$r$ has no greatest element

A real number is a Dedekind left set and $ℝ$ is the set of all such real numbers.
Let $q∈ℚ$. Then the real number corresponding to $q$ is $Q=\{p∈ℚ:p<q\}$
Everything is clear except "a real number is a Dedekind left set", how?? How is $\sqrt2$ a non empty subset of $ℚ$? How does it even make sense?
Any help please?! Thank you
 A: Well, $\sqrt2$ is the set of all rational numbers $x$ which are negative or that $x^2<2$. This set is certainly not empty, it contains all the negative rationals, etc.
But wait, you might say, this is somehow circular. How do you know to define it by $x^2<2$? Well, $\sqrt\cdot$ is not an integral part of our language. Instead we have $2$, which is a rational number, and then we say that $\sqrt2$ is the unique positive solution to $x^2-2=0$, so we define the Dedekind cut as above, and we can then show in the field $\Bbb R$, given by these Dedekind cuts, $\sqrt2$ is in fact the cut we defined, i.e. the positive solution for $x^2-2=0$.
A: "a real number is a Dedekind left set", how??  - from now and in the future you'll know, that numbers are sets. All operations, all properties between numbers now will be corresponding operations, properties between sets. Welcome to real life.
A: This is Dedekind cut construction of real number from $\Bbb{Q}$. In this construction real numbers are defined as subsets of $\Bbb{Q}$, satisfing the conditions mentioned in the question.
So here the subset $$A=\{x\in\Bbb{Q}:x>0,x^2<2\}\cup \{x\in\mathbb {Q} :x\leq 0\} $$ is a Dedekind cut and which is denoted as $\sqrt2$.
So it is not, how $\sqrt2$ can be a subset of $\Bbb{Q}$, but this set $A$ is defined as $\sqrt2$. And if we further proceed we can see, the set $B=\{x\in\Bbb{Q}:x<2\}$ is a Dedekind cut, defined as $2$ and product is defined such that $A^2=B$, that is, $(\sqrt2)^2=2$.
