Let's see what the most simple hereditary sets look like.
In the subsection about the empty set, we have shown that the empty set is
equivalent with an integer number, namely zero (without the overlines here):
$$
\{\} = 0
$$
Applying the recursion rule once, starting with the empty set, we obtain:
$$
\{\} = 0 \quad \Longrightarrow \quad \{\{\}\} = \{0\}
$$
But the set containing only a zero can be bitmapped upon a sequence of bits
where only the bit at position zero is up:
.. 0000000001
And the numerical (integer) equivalent of this is the number $1$ :
$$
\{\} = 0 \quad \Longrightarrow \quad \{\{\}\} = \{0\} = 1
$$
We can proceed in this way:
$$
\{ \, \{\{\}\} \, \} = \{ 1 \} = \mbox{?}
$$
Bitmap:
.. 0000000010
Hence:
$$
\{ \, \{\{\}\} \, \} = \{ 1 \} = 2
$$
Next example:
$$
\{ \, \{\{\}\} , \{\} \, \} = \{ 1,0 \} = 3
$$
Bitmap:
.. 0000000011
The general pattern is supposed to be clear now. Proceeding in this way, we subsequently have:
$$
\begin{array}{l}
0 = 000 = [\; ] = \{\} \\
1 = 001 = [0] = \{\{\}\} \\
2 = 010 = [1] = \{\{\{\}\}\} \\
3 = 011 = [0\; 1] = \{\{\}\{\{\}\}\} \\
4 = 100 = [2] = \{\{\{\{\}\}\}\} \\
5 = 101 = [0\; 2] = \{\{\}\{\{\{\}\}\}\} \\
6 = 110 = [1\; 2] = \{\{\{\}\}\{\{\{\}\}\}\} \\
7 = 111 = [0\; 1\; 2] = \{\{\}\{\{\}\}\{\{\{\}\}\}\} \\
\cdots
\end{array}
$$
Where the curly braces in the middle have been replaced by square brackets and where commas and blanks on the right
are left out for the purpose of unreadability (:-) Or is it because they might rather disturb automatic processing?
Anyway, what we have obtained now, in a very specific sense, is a hitherto missing string of curly braces
representation of sets. Herewith, our sequence of set representations (number - bitmap - array - string) is complete.