TITLE: A Basic question on intuition of rational cut set in the construction of real numbers
QUESTION [6 upvotes]: The intuition for cuts presumably comes from the standard experience of approximation by terminating decimals. For example, we can approximate $\sqrt{2}$ by the sequence $1,1.4,1.41,1.414,1.4142, \dots$. Now, why a cut set for $\sqrt{2}$ requires "all" rationals less than $\sqrt{2}$. Intuitively, only these rationals $1,1.4,1.41,1.414,1.4142, \dots$ uniquely identifies $\sqrt{2}$. Are not they sufficient ?
Related question : Why the addition of two cut set is defined as the set of all possible summation of elements of the two cutsets ?

REPLY [3 votes]: Now, why a cut set for $\sqrt{2}$ requires "all" rationals less than $\sqrt{2}$.

This is Dedekind's idea (these cut sets are often called "Dedekind cuts", after him), and it works. Simple as that. Your idea works too - there's more than one way to skin a cat.
But as soon as you write down the statement "$\mathbb{R} = \{\text{Dedekind cuts of }\mathbb{Q}\}$", a few things are very obvious:

you get $\mathbb{Q}$ sitting inside $\mathbb{R}$
you can define addition (just add all the elements in two cut sets, and you get another cut set) and multiplication (similarly)
you can define <, and $\mathbb{R}$ is ordered
$\mathbb{R}$ (by definition) has the least-upper-bound property: any non-empty subset bounded above has a supremum
etc.

You can check all of this with your definition too, and it comes down to the same thing. It's just a little messier.
Notice that I haven't asked you to write down a Dedekind cut for, say, $\pi$ explicitly. (How would you do this?) That's because we don't really want to know that $\pi$ exists at this stage - all we want to know right now is that the number system $\mathbb{R}$ exists and has all the nice properties we think it does. And, of course, it does. Once we've done that, we can worry about all the individual numbers that we need.