How to write $\pi$ as a set in ZF? I know that from ZF we can construct some sets in a beautiful form obtaining the desired properties that we expect to have these sets. In ZF all is a set (including numbers, elements, functions, relations, etc...). 
For example we can define a copy of $\mathbb{N}$, using the empty set axiom and others, called sometimes $\omega$ and defining,
\begin{align}
0_\mathbb{N}&=\{\}\\
1_\mathbb{N}&=\{\{\}\}\\
2_\mathbb{N}&=\{\{\},\{\{\}\}\}\\
3_\mathbb{N}&=\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\\
4_\mathbb{N}&=\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}, \{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\\
&\ \vdots
\end{align}
Well, ZF allow us to build this type of sets that are kind of arrangements of brackets and commas. We can follow with $\mathbb{Z}$ since we can define $(a,b)=\{\{a\},\{a,b\}\}$:
$\mathbb{Z}$ is defined as the set of equivalence classes $\mathbb{Z}=(\mathbb{N}\times\mathbb{N})\big/\sim$ where
$$\sim\, =\{\big((m,n),(h,k)\big)\in(\mathbb{N}\times\mathbb{N})\times (\mathbb{N}\times\mathbb{N}):(m+_\mathbb{N} k)= (h+_\mathbb{N} n)\}$$
here, the integers are sets more complicated than natural numbers. For example,
\begin{align}
-2_\mathbb{Z}&=\{(1_\mathbb{N},3_\mathbb{N}),(2_\mathbb{N},4_\mathbb{N}),(3_\mathbb{N},5_\mathbb{N}),\ldots,(n_\mathbb{N},(n+2)_\mathbb{N}),\ldots\}\\
&=\{\{\{1_\mathbb{N}\},\{1_\mathbb{N},3_\mathbb{N}\}\},\{\{2_\mathbb{N}\},\{2_\mathbb{N},4_\mathbb{N}\}\},\ldots\}\\
-2_\mathbb{Z}&=\{\{\{\{\{\}\}\},\{\{\{\}\},\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\},\\
&\quad\{\{\{\{\},\{\{\}\}\}\},\{\{\{\},\{\{\}\}\},\{\{\},\{\{\}\},\{\{\},\{\{\}\}\}, \{\{\},\{\{\}\},\{\{\},\{\{\}\}\}\}\}\}\},\ldots\}\\
\end{align}
Here we note the importance of notations. We continue with  $\mathbb{Q}=(\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\})\big/\sim$ where
$$\sim\, =\{\big((m,n),(h,k)\big)\in(\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\})\times (\mathbb{Z}\times(\mathbb{Z\setminus\{0_\mathbb{Z}\})}\}):m\odot_\mathbb{Z}k=h\odot_\mathbb{Z}n\}$$
And for example:
$$(0.2)_\mathbb{Q}=\{(1_\mathbb{Z},5_\mathbb{Z}),(2_\mathbb{Z},10_\mathbb{Z}),(3_\mathbb{Z},15_\mathbb{Z}),\ldots,(n_\mathbb{Z},(5n)_\mathbb{Z}),\ldots\}$$
Imagine if we write the integer numbers as before and write the ordered pairs in unabridged form ($(0.2)_\mathbb{Q}$ is a nice abbreviation for this monster). However we can. 
Finally we define $\mathbb{R}$ as the set of all Dedekind cuts, for example:
$$(0.2)_\mathbb{R}=\{x\in\mathbb{Q}:x<_\mathbb{Q} (0.2)_\mathbb{Q}\}$$
Note that $(0.2)_\mathbb{R}$ is even more monstrous than $(0.2)_\mathbb{Q}$. Also I can write $(\sqrt{2})_\mathbb{R}$ showing its elements in a simple form,
$$(\sqrt{2})_\mathbb{R}=\{x\in\mathbb{Q}:(x^2<_\mathbb{Q} 2_\mathbb{Q}) \lor (x<_\mathbb{Q} 0_\mathbb{Q})\}$$
But I don't know how to do it with $\pi_\mathbb{R}$, since
$$\pi=\lim_{k\to\infty}\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}$$
I only know that
$$(\pi)_\mathbb{R}=\bigcup_{k=1}^{\infty} \left(\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}\right)_\mathbb{R}$$
Since monotonically converge to $\pi$ we have
$$(\pi)_\mathbb{R}=\bigcup_{k=1}^{\infty} \left\{x\in\mathbb{Q}:x<_\mathbb{Q} \left(\sum_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}\right)_\mathbb{Q}\right\}$$
Is there any way to avoid infinite union (and
the choice of a particular convergent sequence
) as the case of $(\sqrt{2})_\mathbb{R}$? If not, why?
Can we write the set representing $\pi$ listing its elements as we do with integers or rationals (as $\mathbb{Q}$ is countable I guess that should be able to do, but I don't know how to do)?
If in ZF all is a set, is so surprising the fact that so many things can be defined, then my last question is
How many more things can be built using ZF?, ZF could define us what is a derivative, an integral, a limit or a measure?
Thanks in advance.
 A: Note that when we define the natural numbers we have a good sense of addition and multiplication (ordinal arithmetics), and from those we can define the operations on $\Bbb Z$ and $\Bbb Q$ and then by using Dedekind cuts construction we can extend these to $\Bbb R$ as well.
So we have that $\Bbb R$ has the operations $+,\cdot$ and they all satisfy all the things we know they do from the times we did mathematics without writing all the sets explicitly.
Now we can use these things to start and define anything else that we desire using the $+$ and $\cdot$ and whatnot as our stones. For example you can define $\pi$ to be the length of the semi-circle of radius $1$.
How do we do that? We define what is an integral, and a path integral, and so on. All from the sets which are addition and multiplication and so on, and then we can define $\pi$ in a painfully tedious way.
The whole point of using set theory, and in this case $\sf ZF$, as our foundation is that we can do things, once we can define the real numbers with their basic properties we have formulas which define things from that structure, and we don't have to write everything in set-form explicitly.
Once we have the real numbers (with the order) it is easy to define the collection of open intervals, and then it is easy to define the standard topology (the smallest collection containing the intervals and having certain properties), from there we can define the Borel sets, the Lebesgue sets, and the Lebesgue measure (being the unique function from the Lebesgue sets into the real numbers which satisfies certain properties), then we can define integration and with respect to the measure, and we can define derivation.
All these things end up being immensely long and complicated formulas, but the point is that we can write them up. And all this with just $\in$ and the axioms of $\sf ZF$. (Although we may want to add $\sf DC$ or even $\sf AC$ if we discuss measure theory.)
But if you do want to insist on $\pi$ being written in set form:
$$\pi = \left\{x\in\Bbb Q\mathrel{}\middle|\mathrel{} x<_\Bbb Q0\lor \left(x\geq_\Bbb Q0\land\exists k\in\Bbb N:\frac{x^2}6<_\Bbb Q\sum_{n=1}^k\frac1{n^2}\right)\right\}$$
A: Let me try this $$(\pi)_{\mathbb{R}}=\left\{x\in \mathbb{Q}:\;\exists k\in\mathbb{N}.\;x< \sum\limits_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}\right\}$$ This would be the Dedekind construction of the reals, and note that there is no union here.
Or, I would complete the rationals first by taking the quotient of ring of fundamental sequences with maximal ideal of null sequences $\mathcal{F}_Q/\mathcal{N}_Q$ to obtain a complete field $\mathbb{R}$, then I can safely say that $$\pi=\lim\limits_{k\to\infty}\sum\limits_{n=0}^{k}\cfrac{2^{n+1} n!^2}{(2n + 1)!}$$ (or define exp first, then define sin, then define $\pi$)  
A: It is hard to tell what "Avoid infinite union" means. Of course all dedekind cuts are infinite sets of elements of $\mathbb Q$, which are themselves infinite sets and so on.
If you want to write an expression for $\pi_{\mathbb R}$ that avoids blobs of ink looking like $\bigcup_{k=1}^\infty$, you can try to define $\pi$ as smallest positive root of any nonzero function that has a root at zero and is the negative of its second derivative. It may turn out as fun to formulate all this using only rationals.
A: I wrote the original equation in Word and added it here

