Consider the interval $(0,1)$ on the real numbers, the supremum of this interval is the real number 1.

In the context of finding the supremum in a set of Dedekind cuts, I have seen the definition as the union of all of the Dedekind cuts in the set as a way to understand the supremum.

However, I am having difficulty conceptualizing how taking the union of the infinite number of Dedekind cuts in $(0,1)$ can be shown as equal to the Dedekind cut corresponding to $\{p \in Q: p<1\}$

This is because I do not know the correct way to conceptualize an infinite number of unions. My intuition is that an infinite union of Dedekind cuts of real numbers in $(0,1)$ will result in a set that has rational numbers increasingly closer to 1, but I have some difficulty believing these sets are equivalent because I am not sure what it means for two sets with an infinite number of elements to be equivalent.

I feel that it would need to be proved somehow that these two sets, $\{p \in Q: p<1\}$ and the union of Dedekind cuts on (0,1), have the identical elements and the same number of elements. How do I show that these two infinite sets are equivalent?

I was wondering if someone can shed some light on how to interpret this. Thanks.

  • 1
    $\begingroup$ The union will contain all rational numbers less than $1$. Clearly, any of he Dedekind cuts is contained in that set. And given any $p\in\mathbb{Q}$ with $p\lt 1$, the cut determined by $p+\frac{1-p}{2}$ will be in the union and contain $p$. $\endgroup$ Jan 2, 2020 at 20:01
  • $\begingroup$ So if I give a formula that shows I can find a larger rational number between an arbitrary rational number less than 1 and 1, this can prove that the sets are equivalent? $\endgroup$ Jan 2, 2020 at 20:10
  • $\begingroup$ You need to show that every element in each set in the union is contained in $\{p\in\mathbb{Q}\mid p\lt 1\}$, and that every element in $\{p\in\mathbb{Q}\mid p\lt 1\}$ is contained in at least one set that is in the union. Exactly how you do tthat will vary; you should also specify exactly what your definition of Dedekind cut is, since there are several ways of setting them up (whether they are pairs of sets, just the left set, just the right set, may contain a maximum, may not contain a maximum, etc.) $\endgroup$ Jan 2, 2020 at 20:15

1 Answer 1


Infinite sets (and operations on sets like infinite unions) often look more mysterious than they actually are.

Two sets are equal iff they have the same elements (depending on how formal you want to be, this is either taken for granted or explicitly given as an axiom in your system, but in either case it's immediate).

In particular, when you write

I feel that it would need to be proved somehow that these two sets, $\{p \in Q: p<1\}$ and the union of Dedekind cuts on $(0,1)$, have the identical elements and the same number of elements

you're actually making the situation look more complicated than it is - you don't need to think about the number of elements in any sense, you just need to check that they have the same elements.

This will have two directions (writing "$D$" for the union of the Dedekind cuts in $(0,1)$ for simplicity:

  • Every element of $\{p\in\mathbb{Q}: p<1\}$ is an element of $D$. Suppose $a$ is a rational less than $1$. Pick any element $x$ of $(a,1)$ and let $d_x$ be the corresponding Dedekind cut $d_x=\{q\in\mathbb{Q}: q<x\}$. Since $a$ is a rational less than $x$ we have $a\in d_x$, and by definition $d_x\subseteq D$ so $a\in D$.

  • Every element of $D$ is an element of $\{p\in\mathbb{Q}: p<1\}$. Suppose $a$ is an element of $D$. By definition, that means that for some $x\in (0,1)$ we have $a\in d_x$ (where as above $d_x=\{q\in\mathbb{Q}: q<x\}$ is the Dedekind cut corresponding to $x$). But then $a<x$, so $a<1$ since $x<1$. So $a\in\{q\in\mathbb{Q}: q<1\}$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .