# Finding example of sets that satisfy conditions

give examples of sets such that:

i)$A\in B$ and $A\subseteq B$

My answer : $B=\mathcal{P(A)}=\{\emptyset,\{1\},\{2\},\{1,2\}\}$ and $A=\{1,2\}$ then $A\in B$ and $A\subseteq B$

ii) $|(C\cup D)\setminus(C\cap D)|=1$

My answer is: $C=\{1,2,3\}$, $D=\{2,3\}$ then $C\cup D=\{1,2,3\}$ and $C\cap D=\{2,3\}$ so $(C\cup D)\setminus(C\cap D)=\{1\}$ and $|(C\cup D)\setminus(C\cap D)|=1$

Can we find sets A and B such that $A\in B$ and $B\subseteq A$? My answer is no.

• Yes. FYI the last answer follows from the ZF axiom of regularity. – Angela Richardson May 23 '13 at 17:49

(i) This doesn’t quite work, unfortunately, because $A\nsubseteq B$, i.e., $\{1,2\}\nsubseteq\{\varnothing,\{1\},\{2\},\{1,2\}\}$. In order for $A$ to be a subset of $B$, each element of $A$ must be an element of $B$. The elements of $A$ are $1$ and $2$, and neither of them is an element of $B$. It’s true that $\{1\}\in B$ and $\{2\}\in B$, but that’s very different from having $1\in B$ and $2\in B$. Try the same idea with $A=\varnothing$.

(ii) This is fine.

(iii) Your answer is correct: if $A\in B\subseteq A$, then $A\in A$, which is ruled out by the axiom of regularity (also called foundation).

• Ah that was a stupid mistake, thanks for pointing it out. If I say, take $A=\emptyset$ and if $B=\{\emptyset,\{1\}\}$ then $A\subseteq B$ right? – H.E May 23 '13 at 18:14
• @Heidi: Yes, because $\varnothing$ is a subset of every set. You could also take $B=\wp(\varnothing)=\{\varnothing,\{\varnothing\}\}$. – Brian M. Scott May 23 '13 at 18:16
• @@Brian: So when we write for example $A=\{1,2\}$ then we somehow abuse the notation as we should write $A=\{\emptyset, 1,2\}$ ? – H.E May 23 '13 at 18:19
• @Heidi: No, you confuse $\subseteq$ and $\in$. The empty set is a subset of every set, it's not an element of every set. – Asaf Karagila May 23 '13 at 18:23
• You are right, thank you both! – H.E May 23 '13 at 18:25

Your answer is incorrect. Because $1,2\in A$, but $1,2\notin B$. Your second answer is correct.

To the last question, the answer is again correct (assuming $\sf ZF$), because $A\in B$ and $B\subseteq A$ would imply that we have $A\in A$, which is impossible due to the axiom of regularity.

To correct the first answer, consider the empty set.