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I have been trying to understand some elementary set theory recently, and am trying to understand how the real number line can be defined using the set of rational numbers. In particular, I am trying to understand Dedekind cuts. I understand the definition of a Dedekind cut, and I have no trouble identifying if a cut is Dedekind or not. But I have troubles understanding how this definition is useful.

In a video I saw by Dr. Peyam, he claimed that $\displaystyle \sqrt[3]{2}$ can be defined the following way:

$\displaystyle \sqrt[3]{2}=\{ r\in\mathbb{Q}:r^{3}<2 \}$

I understand that this set is in fact a Dedekind cut because it has (i) no maximum, (ii) contains all rationals less than $\sqrt[3]{2}$ and (iii) is a real, nonempty subset of $\mathbb{Q}$. But I do not understand how we can define a singular number as a set containg infinitely many numbers.

Mikhail Katz
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naytte2
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    You certainly cannot define the single number $\sqrt[3]2$ as a single rational number. Remember that until you have established exactly what the definition of an arbitrary real number is, the only numbers that are available to work with are rational numbers. I don't see a generally useful way to define an arbitrary real number without using infinitely many of these numbers. – David K Feb 11 '24 at 16:35
  • Just out of curiosity, what is your formal definition of $\Bbb Q$? (This might help address (iii).) – Arthur Feb 11 '24 at 16:43
  • Other near-duplicates: https://math.stackexchange.com/q/3554031/96384, https://math.stackexchange.com/q/752220/96384, https://math.stackexchange.com/q/2753011/96384 and probably more. – Torsten Schoeneberg Feb 11 '24 at 16:43
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    Oh yes, I do in fact mean less than $\sqrt[3]{2}$. I miswrote, @TorstenSchoeneberg. Sorry for any confusion. I will update the post right away! – naytte2 Feb 11 '24 at 17:54
  • @Arthur, this might be a "naive" definition, but I imagine the definition is the set of all numbers that can be written as a quotient $p/q$, for integers $p$ and $q$. – naytte2 Feb 11 '24 at 17:57
  • Did you not previously build negative whole numbers as equivalence classes of subtractions of natural numbers, rationals as equivalence classes of certain products of integers, etc? Any number so defined (e.g., $-3$, $\frac{1}{2}$) is an infinite set by definition, and defining a new category of numbers (reals) via downsets or Dedekind cuts should at this point be sort of familiar I would think. – RobinSparrow Feb 11 '24 at 18:18
  • It turns out that the particular construction of a kind of number isnt really all that relevent. As long as we have a superset of the Rationals, which satisfies the completeness Property, it doesnt really matter how we get there. Its the properties we get out of the construction that justifies the construction in the first place. That construction of the Cube Root of 2, makes sense once you have shown that once you define what it means to exponentiate a real, if you cube that definition of root 3, indeed you get the Real Number 2. As for how we can define a number as a infinitr set of numbers. – Michael Carey Feb 11 '24 at 18:24
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    In descriptive set theory, its normal to think of reals as infinitr sequences of natural numbers. It shouldnt be all that strange, since in non set theoretic lens, we often associate individual Reals with Infinitely long decimal expansions. – Michael Carey Feb 11 '24 at 18:27

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Sometimes the best way to understand a new definition is by understanding its equivalence with an old definition that you are already familiar with. I will assume that you are familiar with the definition of a real number (say, between 0 and 1) as an unending decimal $a=0.a_1a_2a_3\ldots$ (a technical issue here is that one must also identify 1 with $0.999\ldots$ and all similar cases). Here is an observation that may help: the number $a$ splits (or "cuts") $\mathbb Q$ into two subsets: $A_L$ consisting of all the rationals less than $a$, and $A_R$ consisting of all the rationals greater than $a$ (I will ignore for the moment the issue with $a$ itself in the case when it happens to be rational). Now we have $\mathbb Q$ split into two subsets such that if $x\in A_L$ and $y\in A_R$, then always $x<y$.

Conversely, if one has a splitting of $\mathbb Q$ into a "left" set and a "right" set having the property above, then the splitting uniquely determines a real number.

This is all that is meant when one says that a real number "is" a cut on the rationals. You can actually think of it as the pair $(A_L, A_R)$, but this is merely an issue of set-theoretic formalisation that does not affect the way you actually use the real numbers.

Mikhail Katz
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  • Ah, I think I get it now. So when we think of the pair $(A_L, A_R)$, the "gap" between the sets is the real number we are looking for? – naytte2 Feb 12 '24 at 12:02
  • @naytte2, exactly. – Mikhail Katz Feb 12 '24 at 13:35
  • I see. But how can we be sure that one Dedekind cut corresponds to exactly one real number? How do we know our "gap" does not contain any other irrationals?

    (In the case of irrational numbers, of course, if the number is rational, it's simply part of one of the sets)

     – naytte2 Feb 12 '24 at 13:46
  • @naytte2, this follows from the fact that the rationals are dense. If there were two irrationals $\alpha$ and $\beta$ in between $A_L$ and $A_R$, we could find a rational number in between $\alpha$ and $\beta$, contradiction. – Mikhail Katz Feb 12 '24 at 15:04
  • Right, of course. But just to clarify, when Dr. Peyam writes that $\displaystyle \sqrt[3]{2}={ r\in\mathbb{Q}:r^{3}<2 }$, he does not actually mean that the number is the set itself, but rather that splitting the rationals this way uniqely determines the irrational number $\displaystyle \sqrt[3]{2}$. since there is only one possible set $A_R$ ? @Mikhail Katz – naytte2 Feb 12 '24 at 15:36
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    Well it's a bit of a mannerism of set theorists to insist that a Dedekind cut actually is the real number. You can safely ignore that at your level of study. @naytte2 – Mikhail Katz Feb 12 '24 at 15:39