Mathematics Asked by thisguy on December 20, 2020

I know we can use cuts in real numbers but it has no effect, and hence we get nothing new. I wonder why we cannot extend real numbers with cuts and which property of cuts makes it impossible to extend it. How can we show that cuts do not extend real numbers?

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 $Lnotcong Ded(L)$).

Answered by Noah Schweber on December 20, 2020

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