Problem on understanding the definition of "Real Numbers"

Question

Assume that a cut is defined like this :

A cut of a set like $M$ is a subset of $M$ like $A$ such that :
1. $A \neq \emptyset $ and $A\neq M$
2. $\forall a,b \in M \space \space a\in A \space\land b<a \implies b \in A$
3. $A$ doesn't have a maximum.

Then, we can define real numbers like this :
$R=\{A\subseteq \Bbb Q :A \text{ is a cut}\}$

Now, I have three problems.

1. Isn't $\Bbb R$ supposed to be larger than $\Bbb Q$? I've learned that $\Bbb Q\subset \Bbb R$ . But with this definition, $\Bbb R$ is the union of many subsets of $\Bbb Q$. So again, it's a subset of $\Bbb Q$ .

2. What's the reason of adding the 3rd statement?

3. From second statement, Can we conclude that $A$ doesn't have minimum?

Note 1 : By $\Bbb Q$, I mean Rational Numbers.

Note 2 : This is the definition of one my professors who wanted to explain it briefly. I'm not sure if it's complete or wrong. That's why I asked it here.

Thanks in advance.

Answer 1

1) $\mathbb{R}$ is not the union of many subsets of $\mathbb{Q}$, it qualifies as a so called limit union of subsets of $\mathbb{Q}$. For example, the set of all rational numbers such that $x^2 < 2$ has no equivalent rational representation. This is the union of all sets of the form $x^2 < q$ where $q < 2$, which are all represented by rational numbers, but then the cut itself is not a rational number.

2) I don't know, to be fair, but the answer somewhat stems from the fact that if both open and closed intervals were allowed, then each real number would have identity crisis, because we would not know whether to associate it to the open or closed interval.

3) Yes, we can conclude that $A$ does not have a minimum whenever $M$ doesn't have a minimum, because then for every $a$ in $A$ we have some $b \in M$ smaller than $a$, then $b \in A$, so $A$ has no minimum. However, if $M$ has a minimal element, then $A$ can consist solely of that minimal element, for example.

Answer 2

1. The elements of $\Bbb R$, defined like this, are (infinite) subsets of $\Bbb Q$. Hence no element of $\Bbb R$ is an element of $\Bbb Q$. For now, $\Bbb R$ and $\Bbb Q$ are disjoint. However, we can identify $\Bbb Q$ with a subset of $\Bbb R$ in a natural way, by identifying the rational number $q$ with the cut $\{\,x\in \Bbb Q\mid x<q\,\}$.

2. If we allow maxima, the sets $\{\,x\in\Bbb Q\mid x\le 0\,\}$ and $\{\,x\in\Bbb Q\mid x< 0\,\}$ would both be cuts. This is not what we want (it just doesn't work out when proceding).

3. No, see above.