Dedekind cut with $-3$ Im studying some set theory, and my book gives me this definition:
"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 $Q=\{p∈ℚ:p<q\}$"

Then it gives me this exercise:
"Write down a description, in terms of rational numbers and operations of $ℚ$, of the Dedekind left sets corresponding to the following real numbers: -3, (and some more)"
the result of this exercise is:
$\{p∈ℚ:p<-3\}$

Now finally my question:
Does this mean that $-3=\{p∈ℚ:p<-3\}$? Isn't this also saying that "the real number corresponding to $-3$ is $-3$? by the definition above?
 A: Let me quote the following comment from your previous question:

Another issue could be that the common view  $\mathbb{Q} \subset\mathbb {R} $ is incompatible with elements of $\mathbb {R} $ being subsets of $\mathbb {Q} $. If that's bothering you, then understand that in reality we don't have $\mathbb{Q} \subset\mathbb {R} $ but rather $\mathbb{R} $ has a subset (let's denote it by $\mathbb{Q}^{*} $) of rational real numbers which is isomorphic to $\mathbb {Q} $. The same happens with all set inclusions in the chain $\mathbb{N} \subset\mathbb {Z} \subset\mathbb {Q} \subset \mathbb {R} \subset \mathbb {C} $. – Paramanand Singh
, posted on 22 June, 2020.

The point here is that $-3$ the rational number is not $-3$ the real number, and neither is $-3$ the natural number.
Your task is to understand how do the rational numbers embed into the real numbers, as defined here by Dedekind left sets. So when we write $$-3 = \{q\in\Bbb Q\mid q<-3\},$$ the left $-3$ is a real number, it is a set of rational numbers, whereas the right $-3$ is a rational number (which is presumably a set of ordered pairs of integers, which are sets of ordered pairs of natural numbers, which are presumably the von Neumann ordinals; but even if the rational numbers are just "given", it does not change anything here).
But after we have done all that, and maybe a bit more (i.e. defining $\Bbb C$ as well), we can decide that it is easier to work under the assumption that the rational numbers are real numbers, and the above tells you exactly what are the "new rational numbers".
A: $-3$ may now be considered as a rational number (perhaps defined as an element of the equivalence class containing the ordered pair $(-3,1)$, not sure how that book constructs the rationals) or as a real number, in which case it is the set of all rationals less than $-3$. They are different things set theoretically, but the rationals considered as a subset of the reals of course 'obey the same behavior' (with respect to addition, multiplication, order), which means there is an isomorphism between the previously constructed rationals and the elements of $\mathbb{R}$ which are considered rational; if $r$ is a rational in our old construction,
the corresponding 'rational' in $\mathbb{R}$ would be $\{ p \in \mathbb{Q} \, | p < r\}$.
A: It would be better to clarify what the symbol $-3$ actually means in this case, because we can't write, in a formal manner $-3:=\{p\in \mathbb Q\mid p<-3\}$ otherwise we have self reference.
It might be better to write $D(-3):=\{p\in \mathbb Q\mid p<-3\}$, where $D(-3)\in \mathbb R$, and $-3\in \mathbb Q$.
A: Yes, it exactly does mean that.
The problem is, before going to construct real number (by any method, Dedekind cut, Cauchy complition) first you have to forget what you have assumed to be a real number in high school. Otherwise such questions will arise frequently. We also used to wonder a lot when we studied this first.
Starting with, till now we only know rational number as equivalence classes of a relation defined on the set of integers. Then we are constructing a set $R$ which contains subsets of $\Bbb{Q}$ satisfying the three conditions. After defining the operations we prove that the $R$ is a complete ordered field.
Then we embed $\Bbb{Q}$ inside $R$ by the map $q\mapsto \{x\in\Bbb{Q}:x<q\}$.
Once we show this map is monomorphism then we see $\Bbb{Q}$ as a subset $R$ and write rationals as you have written $q=\{x\in\Bbb{Q}:x<q\}$. Then we denote $R$ as $\Bbb{R}$, sets of real numbers.
I hope it is clear now.
