For any set $x$ there is a set $y$ whose elements are precisely the elements of the elements of $x$.
Sounds more complicated than it actually is. A simple example. Union says that, if there is a set $\{\{1,2\},\{3\}\}$, then there is also another set, namely $\{1,2,3\}$. It is seen that the former set is a pair and the latter set contains more than just two elements. We can proceed in this way, forming the pair and the union $\{\{1,2,3\},\{4\}\} \Longrightarrow \{1,2,3,4\}$, and so on and so forth. It is noted once more that we will not dwell upon the well known set theoretic properties of union and intersection. For the obvious reason that it can all be found in many books about common set theory (e.g. Halmos' book).
More systematically now. Starting with the empty set.
$$
\{\} = 0
$$
With the empty set alone, we can make - employing Pairing and Extension -
the singleton that contains it. It is associated with the next natural number:
$$
\{\;\{\}\;\} = \{0\} = 1
$$
With the sets numbered as $0$ and $1$, we can proceed and
create the following sets - employing Pairing and Extension (and substitution):
$$
2 = \{1\} = \{\{0\}\} = \{\{\{\}\}\}
$$ $$
3 = \{0,1\} = \{\{\},\{\{\}\}\}
$$
Next we do:
\begin{eqnarray*}
4 = \{2\} = \{ \{\{\{\}\}\} \}\\
5 = \{0,2\} = \{ \{\} , \{\{\{\}\}\} \}\\
6 = \{1,2\} = \{ \{\{\}\} , \{\{\{\}\}\} \}
\end{eqnarray*}
But here we are stuck, because the axiom of Pairing forbids us to do something
like this:
$$
7 = \{0,1,2\}
$$
So we need the axiom of Union for the first time:
$$
\{3,\{2\}\} = \{\{0,1\},\{2\}\} \quad \Longrightarrow \quad
\{0,1,2\} = 7 = \{ \{\} \{\{\}\} \{\{\{\}\}\} \}
$$
Where it is noticed that it would become more and more difficult to decide if a
set has "equal members" eventually.
Especially if we leave out the commas, which may be essential for readability,
but not for definiability.
It is thus argued, once more, that the axiom of Extension might be not so
trivial as it looks like.
Things we have found so far can also be written as follows, while skipping the
first four results:
\begin{eqnarray*}
4 = \{2\}\\ 5 = 1 \cup 4\\ 6 = 2 \cup 4\\ 7 = 3 \cup 4
\end{eqnarray*}
Where it is noted that actually elements of a Pair (e.g.
$\{1,4\}$ ) are taken for making a Union with $\cup$ .
Continuing this story, we find:
\begin{eqnarray*}
8 = \{3\} \\ 9 = 1 \cup 8 \\ 10 = 2 \cup 8 \\ 11 = 3 \cup 8 \\
12 = 4 \cup 8 \\ 13 = 5 \cup 8 \\ 14 = 6 \cup 8 \\ 15 = 7 \cup 8
\end{eqnarray*}
General pattern:
$$
2^n = \{n\}
$$ $$
2^n + k = k \cup 2^n
\qquad \mbox{where} \quad 1 \le k \le 2^n-1
$$
This is a recursive pattern. There are precisely enough numbers $k$ already
defined with every jump from $2^n$ to the next power of $2$.
So, for the moment being, we end up with only four out of the nine axioms of
Zermelo Fraenkel (and axiom of Choice, eventually: ZFC) set theory: