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.