Let $S$ be a relation on $P(\Bbb{R})$, such that $\displaystyle S=\{<A,B>\in\left(P(\Bbb{R})\right)^2.\left|A\triangle B\right|\le \aleph_0$
Is $S$ an equivalence relation?
My try:
$S$ is definitely reflexive as $\displaystyle \left|A\triangle A\right|=0 \le \aleph_0$, hence $<A,A>\in S$.
$S$ is symmetric. $A\triangle B=B\triangle A$, hence $\left|A\triangle B\right|=\left|B\triangle A\right| \le \aleph_0$, thus $<A,B>\in S \implies <B,A>\in S$.
$S$ is transitive. Let $<A,B> \in S \ \wedge <B,C> \in S$. Look on $L=(A\triangle B)\cup(B\triangle C)$.
Now $A\triangle C\subseteq L$. As $<A,B> \in S \ \wedge <B,C> \in S$ we know that $\left|A\triangle B\right|\le \aleph_0 \wedge \left|B\triangle C\right|\le \aleph_0$.
L is a finite or countable union of finite or countable sets, hence it is finite or countable, hence $|L|\le \aleph_0$. As $A\triangle C$ is a subset of $L$ we conclude that $\left|A\triangle C\right|\le \aleph_0$, hence $<A,C>\in S$, thus $(<A,B>\in S \wedge <B,C>\in S)\implies <A,C>\in S$.
We found that $S$ is reflexive, symmetric and transitive, hence $S$ is an equivalence relation.
Is it good? Thanks!
$\newcommand{\Z}{\mathbb{Z}}$$\newcommand{\R}{\mathbb{R}}$$\newcommand{\Supp}{\mathrm{Supp}}$An alternative way to look at transitivity is by noting that $(P(\R), \Delta)$ is isomorphic to $(\Z_{2}^{\R}, +)$, that is, to the set of group of functions from $\R$ to the field $\Z_{2} = \{0, 1\}$ with two elements, with pointwise sum.
On functions, the relations becomes $f S g$ iff $\lvert \Supp(f+g) \rvert \le \aleph_{0}$. (Here $\Supp(f) = \{ x \in \R : f(x) \ne 0 \}$.)
Clearly $\Supp(s + t) \subseteq \Supp(s) \cup \Supp(t)$, because if $0 \ne (s + t)(x) = s(x) + t(x)$ for some $x$, we cannot have $s(x) = 0 = t(x)$.
Now assume $\lvert \Supp(f+g) \rvert \le \aleph_{0}$ and $\lvert \Supp(g+h) \rvert \le \aleph_{0}$. Then $$ \Supp(f + h) = \Supp(f + g + g + h) \subseteq \Supp(f+g) \cup \Supp(g+h), $$ and we're done.
This is of course equivalent to yours $$ A \Delta C = (A \Delta B) \Delta (B \Delta C) \subseteq (A \Delta B) \cup (B \Delta C). $$