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this is from Velleman's How To Prove It, exercise 4.6.14.b. The exercise reads as in the title and $A\vartriangle Y$ is the symmetric difference between $X$ and $Y$, so $X\vartriangle Y = (X \setminus Y) \cup (Y \setminus X)$. I have written a proof and would appreciate it if someone could check it. To prove uniqueness I use the equivalence $\exists !xP(x) \equiv \exists x(P(x)\wedge \forall y (P(y) \to y=x))$. Are there any other ways to prove it besides using proof by contradiction twice? Maybe using identities on the symmetric difference operation? Maybe using the equivalence $\exists!xP\left(x\right)\equiv\exists xP\left(x\right)\wedge\forall y\forall z\left(\left(P\left(y\right)\wedge P\left(z\right)\right)\to y=z\right)$? By the way, the relation $R$ is proven to be an equivalence relation in exercise 4.6.14.a.

Suppose $B \subseteq A$, and define a relation $R$ on $\mathscr{P}(A)$ as $$R=\left\{ \left(X,Y\right)\in\mathscr{P}\left(A\right)\times\mathscr{P}\left(A\right)\mid\left(X\vartriangle Y\right)\subseteq B\right\}.$$ Prove that for every $X\in\mathscr{P}\left(A\right)$, there is exactly one $Y\in\left[X\right]_{R}$ such that $Y\cap B=\varnothing$.

Let $Y=X\setminus B$. Then $X\vartriangle\left(X\setminus B\right)=\left(X\setminus\left(X\setminus B\right)\right)\cup\left(\left(X\setminus B\right)\setminus X\right)=\left(X\cap B\right)\subseteq B$, so $Y\in\left[X\right]_{R}$. Since $\left(X\setminus B\right)\cap B=\varnothing$, $Y$ has the desired properties. To see that $Y$ is unique, suppose $Z$ is an arbitrary set such that $Z\in\left[X\right]_{R}$ and $Z\cap B=\varnothing$. We will prove that $Z=X\setminus B=Y$.

Let $x\in Z$ and, seeking a contradiction, suppose $x\notin X$. Then $x\in X \vartriangle Z$. Since $Z\in\left[X\right]_{R}$, $Z\vartriangle X\subseteq B$, so $x\in B$. But $Z\cap B=\varnothing$, so $x\notin B$ and thus we can conclude that $x\in X$. Hence $x\in X\setminus B$, and since $x$ was an arbitrary element of $Z$, it follows that $Z\subseteq X\setminus B$.

Now let $x\in X\setminus B$ and, seeking a contradiction once more, suppose $x\notin Z$. Then $x\in X\setminus Z$, so $x\in X\vartriangle Z\subseteq B$. But we know $x\notin B$, so it must be the case that $x\in Z$. Since $x$ was arbitrary, $X\setminus B\subseteq Z$, and therefore $Z = X\setminus B = Y$.

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  • $\begingroup$ It may be easier if you observe that $(X,Y)\in R \iff X$ \ $B=Y$ \ $B.$ So if $(X,Y)\in R$ and $Y \cap B=\phi$ then $Y=Y$ \ $B=X$ \ $B.$ It often pays to look for equivalent but perhaps simpler versions of a given def'n or hypothesis. $\endgroup$ – DanielWainfleet Feb 1 '17 at 22:22
  • $\begingroup$ This is exactly what I was looking for. That equivalence makes a lot of sense intuitively, but I wonder if proving it could also get a little verbose as my proof above. Do you know if the proof for it is simple, or was the equivalence just obvious to you? $\endgroup$ – MuchToLearn Feb 1 '17 at 23:49
  • $\begingroup$ I tried to picture what $X$ and $Y$ look like if $(X,Y)\in R.$ In particular what $X$ \ $B$ and $Y$ \ $B$ would be....I noticed that if $p\in (X$ \ $B)$ or $p\in (Y$ \ $B)$ then $p$ must belong to both $X$ \ $B$ and $Y$ \ $B,$ otherwise $p$ would belong to $X\Delta Y$ but not to $B$.... Without referring to members, we can say : $(X \backslash B)\Delta (Y\backslash B) = (X \Delta Y) \backslash B$ for any $X,Y,B.$... So if $(X,Y)\in R$ we have $\phi=(X\Delta Y)$ \ $B=(X$ \ $B)\Delta (Y$ \ $B),$ and this last expression is empty iff $X$ \ $B= Y$ \ $B.$ (Because $\phi =P\Delta Q\iff P=Q.$) $\endgroup$ – DanielWainfleet Feb 2 '17 at 4:47
  • $\begingroup$ Excellent. Would have accepted it as the answer. Thank you! $\endgroup$ – MuchToLearn Feb 2 '17 at 15:56
  • $\begingroup$ You are welcome. $\endgroup$ – DanielWainfleet Feb 3 '17 at 3:15
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I think your proof is fine. Here is another way to express it, although it is essentially the same as yours. I don't know if it is any clearer, since justifying each step will end up re-doing what you have already done...


The first part of your first paragraph establishes $Y=X \setminus B$ satisfies $Y \in [X]_R$ and $Y \cap B = \varnothing$.


It remains to prove the converse: if some arbitrary $Y$ satisfies $Y \in [X]_R$ and $Y \cap B = \varnothing$, then $Y=X \setminus B$.

Note that $X \triangle Y = (X \setminus Y) \cup (Y \setminus X) \subseteq B$ and $Y \cap B = \varnothing$ together imply $Y \setminus X = \varnothing$. Thus $Y \subseteq X$. Combining this with $Y \cap B = \varnothing$ yields $Y \subseteq X \setminus B$.

Thus, $$B = X \setminus (X \setminus B) \subseteq X \setminus Y = X \triangle Y \subseteq B,$$ so all these subsets are actually equalities, yielding $Y=X \setminus B$.

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