# Equivalence relation by the symmetric difference of sets

Let $$A, B$$ subsets of $$X$$ and $$\mathbb P(X)$$ the power set, we define the following equivalence relation on $$\mathbb P(X)$$:

Let $$S\subseteq X$$ a fixed subset of $$X$$ and $$A$$~$$B$$ $$\iff A△B \subseteq S$$

Prove that is is an equivalence relation and find the class of $$X$$ and $$S$$

My work:

I have already shown that the relationship satisfies reflexivity and symmetry, all this is justified respectively by the fact that the empty set is a subset of any set and the symmetric difference is commutative.

My problem is with transitivity, I do not know how to do it, that is when I try to use it for the definition of symmetric difference I fall in many cases. There is some way to test transitivity using only operations between sets. And with respect to the equivalence class of $$S$$, I showed that they are all subsets of $$X$$ contained in $$S$$. But with respect to the equivalence class of $X4 I do not see what it is. Any help would be useful. Thank you! • @EthanBolker You're right, my mistake! – Hendrik Matamoros May 23 '19 at 21:54 ## 5 Answers Two hints about ways to go. • Draw a Venn diagram for $$A, B, C, X$$ showing $$A \Delta B$$ and so on. • Know or show that $$\Delta$$ is associative. That and the fact that $$B \Delta B$$ is empty leads to an algebraic proof. Hint: By definition, $$A\sim B$$ means $$A-B$$ and $$B-A\subset S$$. So you have to show that, if $$A-B, B-A, B-C, C-B\subset S$$, then both $$A-C$$ and $$C-A$$ are subsets of $$S$$. Consider first an element $$x\in A-C$$. Either it is in $$B$$, or it is not in $$B$$. What can you deduce from the hypotheses in each case? • Thanks! My problem is now, what is the equivalence class of$X$? – Hendrik Matamoros May 23 '19 at 22:01 • Well, it seems to be made up of the subsets of$X$which contain$X-S$. – Bernard May 23 '19 at 22:16 We need to show theat $$A\sim B$$ and $$B \sim C$$ give $$A \sim C$$. This can easily be seen by visualising $$A, B, C$$ in a Venn diagram (try it yourself!) To put the Venn diagram proof formally, consider any element $$x$$ in $$A$$ but not in $$C$$. If $$x$$ is in $$B$$, it lies in the symmetric difference of $$B$$ and $$C$$ and so is in $$S$$. If $$x$$ is not in $$B$$, it lies in the symmetric difference of $$A$$ and $$B$$ and so is in $$S$$. By symmetry any element in $$C$$ but not $$A$$ is in $$S$$, completing the proof. If $$A \Delta B \subseteq S$$ and $$B \Delta C \subseteq S$$, then $$A \Delta C= (A \Delta B) \Delta (B \Delta C) \subseteq S$$ as well. • Must be, subset of$S$. Any help about the equivalence class of$X$? – Hendrik Matamoros May 23 '19 at 21:59 • @HendrikMatamoros: if$A$is a subset of$X$, then$A\Delta X=A^c$, the complement of$A$in$X\$. This should help – Taladris May 24 '19 at 12:13

Below, a proof using the fact that (1) symmetrixc difference can be defined using exclusive OR ( w) (2) that exclusive OR is the negation of the biconditional operator, and (3) that the biconditional operator is transitive