Choice correspondence notation
"1." is wrong and "2." is correct, because (see e.g. Wolfram):

$$A \setminus B = \{x: x \in A \text{ and } x \notin B\}$$

Say for example we have $S = \{a, b, c\}$ and $T = \{b,c\}$.

Then we'd write $S \setminus \{a\} = T$. We do NOT write $S \setminus a = T$.

And so in general, for any set $S$, $S\setminus \emptyset =S$.

We write $2^{X}\backslash \{\emptyset\}$ because $\emptyset \in 2^X$. In contrast, we have $2^{X}\backslash \emptyset = 2^{X}$.

Example. The set $S$ contains four elements: the empty set, an element called $a$, the set containing the empty set, and the set containing $a$.

$$
S=\Bigg\{\emptyset,a,\{\emptyset \},\{a \}\Bigg\}.
$$

Then we have: 

$$ \begin{align} S \setminus \emptyset &amp; = S 
\\ S\setminus\{\emptyset\}&amp; = \Bigg\{a,\{\emptyset \},\{a \}\Bigg\}, 
\\ S\setminus\{\{\emptyset\}\}&amp; = \Bigg\{\emptyset,a,\{a \}\Bigg\},
\\ S\setminus\{a\}&amp; = \Bigg\{\emptyset,\{\emptyset \},\{a \}\Bigg\},
\\ S\setminus\{a,\{\emptyset\}\}&amp; = \Bigg\{\emptyset,\{a \}\Bigg\},
\\ S\setminus\{\{a,\emptyset\}\}&amp; = S.
\end{align} $$

The last equation is true because $S$ does not contain the set $\{a,\emptyset\}$.

See Enderton, in particular p.3, top paragraph -- it's well worth spending a little time if only on the first chapter.