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

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

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