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In Section 1.5 in the book "Daniel W. Cunningham, Set Theory: A First Course., Cambridge", the author noted that:

While reading these axioms, keep in mind that in set theory everything is a set, including elements of a set.

If so, given $\mathbb{R}$ is the universe, what are the subsets of $\sqrt{2}$? Besides, how to interpret the set $\sqrt{2}$?

Tran Khanh
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    It depends on how you define real numbers in your theory. If you use a Dedekind cut of rationals then each real number has two elements (each of which is a set of rationals) and so four subsets: two with single elements (remembering each of those elements is a countably infinite set), one with no elements, and one with both elements. – Henry Oct 20 '23 at 10:40
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    It's better to start with the question, how do we represent any natural number as a set. Representing Reals is not hard, but it is quite a bit involved, unfortunately. Mostly any Intro Set Theory Book will likely cover the details for you, all that really matters is that however you define 2, the naturals follow the Peano Axioms, and that Howver You define the Reals, you satisfy that you are a superset of the Rationals satisfying the completeness Axiom. The particular representation isn't really that important – Michael Carey Oct 20 '23 at 18:01

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There are too many ways to "Define" Integers/rations/real/ETC. Each way will have a new way to represent $\sqrt{2}$ & hence each way will give various Sub-Sets.

Here is one such way to "Define" real numbers with nothing other than Sets :

Let $\phi=\{\}$ be the only thing we have. It is the NULL SET.
I will call it $0$.

Now I will form the Set $\{\phi\}=\{0\}$ , which is having 1 element.
I will call that Set $1$.

Then I will form the Set $\{0,1\}$ which I will call $2$.
Naturally , I will form the Set $\{0,1,2\}$ which I will call $3$.

We can see that $n+1=\{0,1,2,\cdots,n\}$ in my System.
Now I got all Natural Numbers.
[[ I am including $0$ & Positive Numbers here. I am not including Negative Integers to keep is easy to visualize at the high-level ]]

In this way , we can see that :
OPERATION $MIN(C,D)=C \cap D$
OPERATION $MAX(C,D)=C \cup D$

[[ OPERATION ADDITION & OPERATION ADDITION & ETC are out of scope of this longish Answer. ]]

Then I will make new Sets $\{A,\{A,B\}\}$ which I will call $A/B$
[[ The Single Element is the Numerator. The Double Element will contain the Denominator. This is necessary to make some order , otherwise $\{A,B\}=\{B,A\}$ , hence there will be no way to know which is Numerator & which is Denominator. ]]

Example : $\{1,\{1,2\}\}$ , $\{1,\{1,2\}\}$ , $\{1,\{1,2\}\}$ , $\{1,\{1,2\}\}$ , ETC will represent rational number $0.5$
[[ Making Unique representation can be done , but out of scope of this longish Answer ]]

Now I got all rational numbers.

Lastly , Irrational Numbers are not so easy.
With rational number , I can "Define" real numbers , including Irrational Numbers , in various ways : One very common way is "Dedekind Cuts" , which will include the Set representing 2 collections of 2 rational numbers.

Thus , we see that real numbers are Sets which are actually having 2 Sets representing 2 rational numbers which have Sets representing 2 Integers , which in turn have Sets representing smaller Integers , which will eventually represent NULL SET & nothing else !

Example : $X=\sqrt{2}=\{Y,Z\}$ where $Y$ contains all rational number less than $X$ & $Z$ contains all rational numbers more than $X$.
Sub-Sets of $X=\sqrt{2}=\{Y,Z\}$ are : $\{\}$ , $\{Y\}$ , $\{Z\}$ , $\{Y,Z\}$

[[ Complex Numbers & Vectors & Matrices are out of scope of this longish Answer. ]]

Which way I "Define" real numbers will then dictate which Sub-Sets $\sqrt{2}$ will have.

In this way we see that "Everything" is a SET !!!!

Prem
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    What are $A$ and $B$ and how does ${A, {A, B}}$ make division? – Peter Phipps Oct 20 '23 at 11:27
  • @PeterPhipps $A$ and $B$ are the sets representing your "numerator" and "denominator". The above description does not include algebraic structures. You have to define sum and multiplication of natural numbers in terms of sets. Then you define them for the new defined objects of the form ${A,{A,B}}$, where you identify ${A,{A,1}}$ with $A$, and you check that ${A,{A,B}}\times B=A$. In this sense ${A,{A,B}}$ implements division. – GReyes Oct 20 '23 at 11:47
  • @GReyes Thank you. – Peter Phipps Oct 20 '23 at 12:36