Subtracting empty set from another I am considering two sets $A = \{\emptyset, \{\emptyset\}\}\setminus\{\emptyset\}$, $B = \{\emptyset, \{\emptyset\}\}\setminus\{\{\emptyset\}\}$. From the definition of the set difference, I have to consider only those elements of the first set that don't belong to the second one. So my guess is $B = \{ \emptyset \}$. $A$ is more problematic - the empty set doesn't have any elements. My guess is there are no elements to remove. Then, is $A = \{\emptyset, \{\emptyset\}\}$ the correct result?
 A: It is true that subtracting the empty set gives you back the
set you started with:
$$
A - \emptyset = A.
$$
But $\{ \emptyset \}$ is not the empty set; it has one element, namely,
$\emptyset$.
A: Let $e:=\varnothing, f:=\{\varnothing\}=\{e\}$.
Then
$$A = \{e, f\}\setminus\{e\}=\{f\}=\{\{\varnothing\}\}\\
B = \{e, f\}\setminus\{f\}=\{e\}=\{\varnothing\}$$
A: $A = \{\varnothing, \{\varnothing\}\}\setminus\{\varnothing\}$
$B = \{\varnothing, \{\varnothing\}\}\setminus\{\{\varnothing\}\}$
I think that seeing the $\varnothing$ somehow confuses your perception of what is going on. Towards that end, I wonder if this will help.
Let  $x = \varnothing$ and 
$y = \{\varnothing\}$.
Then $A  = \{\varnothing, \{\varnothing\}\}\setminus\{\varnothing\}
     =\{x,y\} \setminus y
     = \{x,y\} \setminus \{x\} = \{y\} = \{\{\varnothing\}\}$
and 
$B  = \{\varnothing, \{\varnothing\}\}\setminus\{\{\varnothing\}\} 
    = \{x,y\}\setminus\{y\} = \{x\} = \{\varnothing\}$

You could also try this.
$A = \{\varnothing, \{\varnothing\}\}\setminus\{\varnothing\}
   = (\{\varnothing\} \cup \{\{\varnothing\}\})\setminus\{\varnothing\}
   = \{\{\varnothing\}\}$.
$B = \{\varnothing, \{\varnothing\}\}\setminus\{\{\varnothing\}\}
   = (\{\varnothing\} \cup \{\{\varnothing\}\}) \setminus\{\{\varnothing\}\}
   = \{\varnothing\}$
A: You are not removing the empty set. You are removing a set whose only element is the empty set, as
$$\emptyset\ne\{\emptyset\}$$
So,
$$A\setminus\{\emptyset\}=\{\{\emptyset\}\}$$
Your claim would be true if the set being removed were indeed empty, i.e.,
$$A\setminus\emptyset=A$$
A: Put $\{\varnothing\}=a$ and $\varnothing=b$. Hence, 
$$
A=\{b,a\}\setminus\{b\}=\{a\}=\{\{\varnothing\}\}
$$
$$
B=\{b,a\}\setminus\{a\}=\{b\}=\{\varnothing\}
$$
A: Subtracting empty set means you got the null,But {null } is not the empty set; it has one element.
