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.
Best Answer
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
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 $a\in\{q\in\mathbb{Q}: q<1\}$.