I'm reading Charles C. Pugh's Real Mathematical Analysis, 2nd Ed. On page 12 he defines a Dedekind cut like so:

Definition. A cut in $\mathbb{Q}$ is a pair of subsets $A, B$ of $\mathbb{Q}$ such that
    (a) $A \cup B = \mathbb{Q}$, $A \neq \emptyset$, $B \neq \emptyset$, $A \cap B = \emptyset$.
    (b) If $a \in A$ and $b \in B$ then $a < b$.
    (c) $A$ contains no largest element.


Definition. A real number is a cut in $\mathbb{Q}$.

On page 17, Pugh shows that $\mathbb{R}$ has no gaps:

Next, suppose we try the same cut construction in $\mathbb{R}$ that we did in $\mathbb{Q}$. Are there gaps in $\mathbb{R}$ that can be detected by cutting $\mathbb{R}$ with scissors? The natural definition of a cut in $\mathbb{R}$ is a division $\mathcal{A}|\mathcal{B}$, where $\mathcal{A}$ and $\mathcal{B}$ are disjoint, nonempty subcollections of $\mathbb{R}$ with $\mathcal{A} \cup \mathcal{B} = \mathbb{R}$, and $a < b$ for all $a \in \mathcal{A}$ and $b \in \mathcal{B}$. Further, $\mathcal{A}$ contains no largest element. Each $b \in \mathcal{B}$ is an upper bound for $\mathcal{A}$. Therefore $y = \text{l.u.b.}(\mathcal{A})$ exists and $a \leq y \leq b$ for all $a \in \mathcal{A}$ and $b \in \mathcal{B}$. By trichotomy, $$ \mathcal{A} | \mathcal{B} = \{ x \in \mathbb{R} : x < y \} | \{ x \in \mathbb{R} : x \geq y \} . $$ In other words, $\mathbb{R}$ has no gaps. Every cut in $\mathbb{R}$ occurs exactly at a real number.

I can't convince myself that this works, since $\mathcal{A}|\mathcal{B}$ is a cut in $\mathbb{R}$ — but he's defined a real number as a cut in $\mathbb{Q}$. My mind is constructing a number system $\mathbb{S}$ where each $s \in \mathbb{S}$ is associated with some cut in $\mathbb{R}$, and I'm tempted to keep filling in the gaps of number systems ad infinitum.

What am I misunderstanding?


1 Answer 1


The point is that if you fill the gaps in $\mathbb{R}$, you get nothing new. The set of cuts in $\mathbb{R}$ is in bijection with $\mathbb{R}$, by mapping each element $y\in\mathbb{R}$ to the cut $\{x\in\mathbb{R}:x<y\}|\{x\in\mathbb{R}:x\geq y\}$ (note that this map is also how you identify $\mathbb{Q}$ as a subset of the real numbers, but in the case of $\mathbb{Q}$ it is not surjective). Moreover, this bijection preserves the order relation, so the ordered set of cuts in $\mathbb{R}$ is order-isomorphic to $\mathbb{R}$.

  • $\begingroup$ I think I understand what you're saying. My $\mathbb{S}$ does not exist because we have carefully designed $\mathbb{R}$ so that the least-upper-bound property holds, so by consequence of definition it "fills in it's own gaps". Is that a reasonable explanation? $\endgroup$ Aug 29, 2020 at 1:55
  • 1
    $\begingroup$ More or less. Another way to put it is that your $\mathbb{S}$ exists, but it is the same as $\mathbb{R}$. $\endgroup$ Aug 29, 2020 at 1:55

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