Prove that $\sim$ is an equivalence relation on $\mathcal{P}(X)$

Question

Given $X$ a non-empty set and $\mathcal{P}(X)$ with relation $\sim$ on $\mathcal{P}(X)$: $$\forall A, B \subset X: A \sim B \Leftrightarrow A \Delta B \text{ is finite}$$

Prove that $\sim$ is an equivalence relation on $\mathcal{P}(X)$.

I know that I have to prove that the relation is reflexive, symmetric and transitive. But I have no idea on how to start.

Good. What does "$\sim$ is reflexive" mean in the context of this relation? — Matthew Leingang, Jan 09 '20 at 17:17
@MatthewLeingang That means that $A \Delta A$ is finite. I am struggling to show that it is finite, which definition should I use to prove it? — NimaJan, Jan 09 '20 at 17:31
@LucasHenrique https://en.wikipedia.org/wiki/Symmetric_difference — NimaJan, Jan 09 '20 at 17:31
By the definition of $\Delta$, $A\Delta A = \emptyset$, which is finite. — Lucas Henrique, Jan 09 '20 at 17:34

score 2 · Accepted Answer · answered Jan 09 '20 at 17:18

2

Just apply the definition. For reflexive, is $A \Delta A$ finite? For symmetric, does $A \Delta B \text{ finite} \implies B\Delta A \text { finite}?$ The purpose of questions like this is to make sure you understand the definition.

answered Jan 09 '20 at 17:18

Ross Millikan

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Prove that $\sim$ is an equivalence relation on $\mathcal{P}(X)$

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