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So this may be a well-trodden question -- I've reviewed the previous answers and similar questions (this one was pretty close), but I think my confusion is more specific.

I get the definition of Dedekind cuts, especially the lower cut.

$$\forall q \in \Bbb Q, L_q:=\{s \in \Bbb Q: s < q\}$$

What I don't get is how we know we can define an appropriate right hand side of the inequality for every irrational number (i.e., for when $q \notin \Bbb Q$, which is the whole point of doing this)

For the $q \in \Bbb Q$ the cut $q:= \{s \in \Bbb Q: s < q\}$ makes total sense to me and utilizes the (already established) ordering of $\Bbb Q$.

What I don't get is that if we have some irrational number $k \notin \Bbb Q$, we can't simply say:

$$k:= \{s \in \Bbb Q: s < k\}$$

Since $s<k$ is not defined outside of $\Bbb Q$, and $k \not \in \Bbb Q$.

I've seen somewhat complex versions for algebraic irrationals such as :

$$\sqrt{2} := \{s \in \Bbb Q: s\geq 0, s^2 < 2\} \cup \{s \in \Bbb Q: s < 0\}$$

And I think we see others for $e,\pi$ and other known transcendentals.

My question is, how do we know we can define enough Dedekind cuts so that we get a complete, ordered field (using the operations for $+,\times, x^{-1}, -$ given for Dedekind cuts (plus set inclusion for ordering).

As an example, when I say something like $\gamma < e^{\sqrt{\pi}}$ (where $\gamma$ is the Euler constant) how do I know there are Dedekind cuts corresponding to $\gamma$, $e^{\sqrt{\pi}}$ ? Obviously if they did exist then the inequality would be nicely defined per set inclusion, but I am struggling to see what the rational (!) "RHS" of the inequality would be for such cuts.

Any help on the general question or my example would be most appreciated. Thanks!


EDIT (9/26/2022):

So I think I see what my issue was (many thanks to the commenters and people with answers). Here's where my thinking is -- would appreciate feedback if I've course-corrected properly.

  1. We know that $\Bbb Q$ is incomplete (even without the concept of "real" number)

  2. We expect that there should be quantities/numbers for these gaps, so we don't have irrational (pardon the pun) results like "Right triangles only exist if all edges are of rational length".

  3. Therefore, we define a complete ordered field which, by definition, has enough elements to allow every triangle to exist (for example), not just the ones with rational edges. More simply, we could also lean on the intuition that space is not "quantized" and so if I start at a point 1 ft away from an object and move to a point 2 ft away, then my distance from the object took on every value in between at least once (I didn't mysteriously hop over any distances).

  4. Armed with our "axiomatic/algebraic" solution to our problem, we now want to show that such things can actually exist in terms of things we already know exist (vs just being defined). Enter Dedekind Cuts, the sole purpose of which is to demonstrate that there is a set of objects, defined only using the established properties of $\Bbb Q$, that satisfy the complete ordered field axioms.

  5. We then can formally place the usual symbols for numbers (e.g., $e,\pi,2,\sqrt{3},...$) that we intuitively know how to use in 1:1 correspondence with elements of the set of all Dedekind Cuts, knowing that the formal/algebraic statement $\exists c \in \Bbb R: c\cdot c=\sqrt{\pi}$ has an equivalent (up to isomorphism) statement in terms of an element $c^*$ in the set of Dedekind Cuts $D_{\Bbb Q}$ of $\Bbb Q$: $\exists c^* \in D_{\Bbb Q}: c^* \times c^*=\sqrt{\pi}$.

Take Away

The question I had about "How do we know that there is a cut corresponding to $\sqrt{\pi}$?" was misguided, as I was implicitly injecting my intuitive understanding of $\pi$ and $\sqrt{x}$ into the mix.

As stated in the comments, if I want to identify the cut for, say $\sqrt{\pi}$ I actually need to define that number so I have criteria to form the specific cut. However, the key here is that we don't require every real number to be definable, just that they are formalized as members of a complete ordered field, and hence exist outside of any concrete definition.

For example, let $d \in \Bbb R: \sqrt{2} < d < \sqrt{3}$. From the properties of Dedekind cuts, we know such a $d$ exists and there is a unique Dedekind cut for every choice of $d$.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Sep 26, 2022 at 20:06

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It sounds like you're expecting that every Dedekind cut is definable by a set-builder expression of the form $\{q\in\Bbb Q:\phi(q)\}$ (where $\phi$ is in some restricted language of rational arithmetic, say). But the definition of Dedekind cuts doesn't require this—we just literally define the set of Dedekind cuts as the set of all downward-closed subsets of $\Bbb Q$, i.e. the sets $L\subseteq\Bbb Q$ with the property that for all $a,b\in\Bbb Q$, if $a<b$ and $b\in L$ then $a\in L$. When we proceed to define operations on this set and show that it's a complete ordered field, we're using this abstract definition, so we're not assuming (or even developing the metamathematical tools that would allow us to formally talk about whether) every real number is individually definable.

If you're curious about a particular real number, you should ask how that number is defined (presumably by algebraic operations, limits, etc. operating on other numbers, e.g. $e=\sum_{n=0}^\infty\frac1{n!}$) and check that those operations all boil down to ones that have been defined on Dedekind cuts.

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  • $\begingroup$ Thank you! So the "practical" point of even needing the reals is that we lack quantities in the rationals that can serve as solutions to some mathematical expressions (the quantity is always just out of reach of any sequence of rationals). But, for example, how do I know that $e < \pi$ is true in the reals using dedekind cuts, how would we show that the cut for $e$ is a proper subset of the cut for $\pi$? $\endgroup$
    – Annika
    Sep 26, 2022 at 19:23
  • $\begingroup$ Yeah, the constructions of the reals are essentially just proofs that a complete ordered field exists - they "take the completion of the rationals" by cleverly characterizing the "holes". Compare this to how the algebraic numbers are defined as solutions to polynomial equations: this can only yield countably many new elements (since there are only countably many equations to solve), whereas the reals are uncountable. $\endgroup$
    – Karl
    Sep 26, 2022 at 20:40

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