I'm facing the following problem. Let $\sim\subseteq\mathcal{P}(\mathbb{N})^2$ be a equivalence relation with the following definition:
$(P,Q)\in\sim$ if $|P\Delta Q|\in\mathbb{N}$ with $P,Q\subseteq\mathbb{N}$
That is, two sets are related by $\sim$ if their symmetric difference is finite.
It is apparent the relation $=_1$, i.e the relation where $(P,Q)\in =_1$ if and only if $|P\Delta Q|=1$ is contained in $\sim$. How can I prove that $\sim$ is the finest such relation containing $=_1$.
My idea was to provide a proof by contradiction by assuming an even finer equivalence relation $\sim'$ where it would then hold that $\forall P\subseteq\mathbb{N}:|[P]_\sim| < |[P]_{\sim'}|$ where $[P]_R$ is the equivalence class of $P$ over $R$.
Let $\approx$ denote an equivalence relation on $\wp(\mathbb N)$ with:$$|P\Delta Q|=1\implies P\approx Q$$
It is our aim to prove on base of that: $$P\Delta Q\text{ is finite}\implies P\approx Q\tag1$$
Because $\approx$ is reflexive we also have:$$|P\Delta Q|=0\implies P\approx Q$$ and with induction we aim to prove that for every $n\in\{0,1,2,\dots\}$ we have:$$|P\Delta Q|=n\implies P\approx Q\tag2$$
If that's settled then we are ready.
Assume that $(2)$ is true for $n\in\{0,1,\dots,m\}$ for positive integer $m$ and that $|P\Delta Q|=m+1$.
It is enough now to prove that $P\approx Q$.
WLOG we may assume that the set $Q\setminus P$ contains some element $q$.
Then we have $|P\Delta(Q-\{q\})|=m$ so that $P\approx Q-\{q\}$.
It is evident that $|(Q-\{q\})\Delta Q|=1$ hence $Q-\{q\}\approx Q$.
Then $P\approx Q$ because $\approx$ is transitive.