Equivalence relation involving finiteness of symmetric difference of sets

Question

I'm posting this message so as to know if it would possible to get a hint so as to solve the following problem: A subset $A$ of $\mathbb N$ is related to a subset $B$ of $\mathbb N$ (A%B) if the symmetric difference of the two sets is a finite set.

I managed to prove the reflexive and symetric property of this relation but I'm stuck for proving the transitive proporty. As a matter of fact, I don't know how to deal with it when the sets are infinite. For example, I notice that if the set A = {2.k , $k \in \mathbb N$} - {2} is related to the set B = { 2.k | $k . \in \mathbb N$} $\cup$ {3}. This would mean that the set $C$ has to be of the form C {2.k | $k \in \mathbb N$}. Nevertheless this is just an example, I don't know how to generalize it if it's transitive (it may seem that yes it's).

Thank you in advance for your feedback.

Answer 1

The symmetric difference of A and B is finite,
iff A and B differ in a finite number of points.
Is this sufficient insight to prove transitivity?

Answer 2

I'll denote the symmetric difference of $A$ and $B$ by $A \Delta B$, as is customary.

Then $$A \Delta C \subseteq (A \Delta B) \cup (B \Delta C)$$

Proof: let $x \in A \Delta C$. Case 1: $x \in A$ and $x \notin C$, then if $x \in B$ then $x \in B \setminus C$ so $x \in B \Delta C$ and if $x \notin B$, then $x \in A \setminus B$ so $x \in A \Delta B$. Case 2, where $x \in C$ and $x \notin A$ is similar, again two subcases depending on $x \in B$ or not.

If then $A \sim B$ and $B \sim C$, then the above inclusion shows that $A \Delta C $ is a subset of a union of two finite sets, hence finite and so $A \sim C$ too.

In your example, $A \Delta B = \{2,3\}$ and many sets $C$ are related to $B$, e.g. all sets of the form $\{2n: n \in \mathbb{N}\} \cup F$, where $F$ is finite. $C$ need not be of the exact form you mention.

Answer 3

Here is a proof almost without words, using a Venn diagram, where symbol "F" means "finite". We use the fact that a subset of a finite set is itself a finite set.

Answer 4

I find the next solution more simple than the mentioned earlier, but it's only matter of taste:
Note that $ A \triangle C = (A \cup C)\setminus (A \cap C)$ . Now, I'll try to convince that $A\setminus C $ is finite, and the explenation is the same for $C\setminus A$. Together, we will get what we want, by the equivalence of $ A \triangle C$ just mentioned.
Note that $A \setminus C$ is the set of elements in A that aren't in C. These elements either in B or not. The elements from B are in the set B \setminus C , which is finite since $ B \triangle C$ finite. The elements which aren't in B are in $ A \triangle B$ , which is finite as well.

All the above based on $ A \triangle C =(A \setminus C) \cup (C \setminus A)$