$\{\{1\},\emptyset\}\setminus \{\emptyset\}=$?

If we find the set difference between the set $\{\{1\},\emptyset\}$ and the set $\{\emptyset\}$, what do we get? My best guess is $\{\{1\}\}$.

• It is correct . – Masacroso Mar 16 '16 at 20:21
• Thanks for your quick response! – ctkw Mar 16 '16 at 20:22
• Thinking $\{1\}$ and $\emptyset$ are just $2$ elements rather than $2$ sets, it might help. – SiXUlm Mar 16 '16 at 20:37

Yes, you're correct: $A\setminus B=\{x: x\in A\text{ and } x\notin B\}$.
Let $A=\{\{1\},\emptyset\},B=\{\emptyset\}$ as $\{1\}$ is in $A$ but not in $B$, it is in $A\setminus B$, but as $\emptyset\in B$, it's not in $A\setminus B$, thus $A\setminus B= \{\{1\}\}$.