Why we cannot extend real numbers with dedekind cuts?

You can perform the Dedekind cut construction on $mathbb{R}$, but you don't get anything new.

Specifically, here's the abstract setting. Suppose we have a linear order $L$ with no first or last point - for example, $mathbb{Q}$ or $mathbb{R}$ (ordered as usual). We can define a new linear order $Ded(L)$ as follows:

• An element of $Ded(L)$ is a set $Xsubseteq L$ such that $X$ is nonempty, is downwards-closed (with respect to the ordering of $L$), is not all of $L$, and has no greatest element.

• Elements of $Ded(L)$ are ordered by inclusion: $Xle_{Ded(L)}Y$ iff $Xsubseteq Y$.

There is a natural embedding $e$ of $L$ into $Ded(L)$, given by $$xmapsto {y: y< x}.$$ In this sense $hat{L}$ "extends" $L$.

---

Now given a particular $L$, we can ask whether $Ded(L)$ properly extends $L$ in the above sense; are there in fact elements of $Ded(L)$ which are not in the range of the embedding $e$, that is, not of the form ${y: y<x}$ for some $xin L$?

The answer depends on the $L$ in question. For example, if we take $L=mathbb{Q}$ (ordered as usual) then we do indeed get new elements: the set $${y: y^2<2}$$ is an element of $Ded(mathbb{Q})$ but is not of the form ${y: y<x}$ for any $xinmathbb{Q}$.

However, for other choices of $L$ we do not get anything new, that is, the corresponding embedding $e$ is actually a bijection (or more fancily, an isomorphism). And $mathbb{R}$ is indeed such an example: $Ded(mathbb{R})$ is "the same thing as" $mathbb{R}$, in the sense that every element of $Ded(mathbb{R})$ has the form ${y: y<x}$ for some $xinmathbb{R}$. This is a good exercise, and relies crucially on the least upper bound principle for $mathbb{R}$:

We can now ask, "Where did the least upper bound principle for $mathbb{R}$ come from?" Of course the answer is: from the Dedekind cut construction itself! In fact, what's really going on is this:

For any appropriate $L$, we have $Ded(L)cong Ded(Ded(L))$ (although possibly $Lnotcong Ded(L)$).