# Is $\mathbb R$ a set of sets? (Walter Rudin PMA)

I am reading Baby Rudin and worked through the Appendix of Chapter 1, on the construction of the real numbers with Dedekind cuts. I understand the construction in terms of the cuts, and describing them as the elements of an ordered field $$\mathbb{R}$$.

Now what I think I don't understand is what $$\mathbb{R}$$ itself is. At least as far as I understand, $$\mathbb{R}$$ is constructed as a set of particular sets called cuts. But are the elements of $$\mathbb{R}$$ actually cuts? I ask this because the book says if, say, $$\alpha$$ is a cut, then $$\alpha \in \mathbb{R}$$.

But aren't the elements of $$\mathbb{R}$$ just numbers? I know each cut corresponds to a number, like for example the cut: $$r^*=\{x: x

corresponds to the rational $$r$$. But the cut $$r^*$$ is a set of numbers, not the number r itself. I am just really confused with this since I just started learning real analysis. Hope this question makes sense, and thanks in advance.

• This might sound stupid, but the elements of $\mathbb{R}$ are what you define them to be. Put differently, you ask whether the elements of $\mathbb{R}$ aren't just numbers, but what is a number to begin with? – Ben Steffan Feb 10 '20 at 16:54
• But are the elements of R actually cuts?  Yes they are. and the real number is the set of all cuts on rational numbers. and yes a r* is a set of rational numbers. I did not read the book but the following might help. some text books define natural numbers as $$0= \{\emptyset\}, 1 = \{\emptyset,0\}, 2 = \{\emptyset,0,\{0\}\}, ...$$, hope you see the similarity – Aven Desta Feb 10 '20 at 17:00
• Also remember how you defined the rational numbers. a positive rational number is ordered pair of a natural number, as @Ben said, numbers are what you define them to be. the most important thing is that the definition corresponds to the way we use them and each property we already know are included there. So each real number is a cut, or a set of rational numbers satisfying a certain property. – Aven Desta Feb 10 '20 at 17:06
• @BrianSteffan I knew the definition of numbers would cause trouble. Let me say a rational number is an element of Q. The cuts are defined as sets of these elements of Q. We also Q is a subset of R, thus every element in Q is in R. So say r is in Q. Then how can r be in R if r itself is not in R? It is just the r* that corresponds to it, but clearly r=/=r* – Nick.25 Feb 10 '20 at 17:08
• @NicolásMaílloGómez, cuts define real numbers, that is rational and irrational numbers. Now we have a different definition of rational numbers. take for example r*, it is a rational number iff r is a rational number in our previous definition of rational numbers. This is the same with integers-rational numbers case. After defining rational numbers we redefine integers as a set of all ordered pairs (a,b) where b = 1 – Aven Desta Feb 10 '20 at 17:13

If you define the real numbers as cuts, then yes: each cut is a set and $$\mathbb R$$ is a set of sets. But this is not the only way to construct the real numbers. It does not really matter which definition you use, as long as your definition can be proven to be equivalent to my definition. That is, there is a one-to-one mapping $$\Phi$$ from your real numbers to my real numbers, that preserves all relevant operations: e.g. $$\Phi(x+y) = \Phi(x)+\Phi(y)$$, where the first $$+$$ is your definition of addition of real numbers and the second $$+$$ is my definition of addition.