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This is a follow-up of this question

Prove that $[A^c]_\sim = \{ Y \space| \space Y^c \in [A]_\sim \}$

I tried to prove this by using the definition of equivalence class. So take an arbitrary set $B \in [A^c]_\sim$, then $A^c \sim B$. By the equivalence relation we see that $A^c\Delta B$ is finite, but $B^c\Delta A$ is also finite as it is symmetric, so $B^c \sim A$ and $B^c \in [A]_\sim$.

Is this proof sufficient?

NimaJan
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  • There seems to be something wrong here. The equivalence relation is $\sim$, not $\Delta$. So I suppose you mean $A^c\Delta B$ is finite iff $B^c\Delta A$ is finite (according to the definition of $\sim$ in the linked question). – amrsa Jan 10 '20 at 20:20
  • @amrsa Yes thank you, I meant finite, I will edit it. – NimaJan Jan 10 '20 at 20:23
  • The proof is not sufficient. In particular I don't understand the «but $B^c\Delta A$ is also finite as it is symmetric». Even if it was correct it would prove only that the first set is included in the second. – Gribouillis Jan 10 '20 at 20:39
  • @Gribouillis what proof is sufficient? – NimaJan Jan 10 '20 at 20:41

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Hint \begin{equation} A^c\setminus B^c = \{x\in X, x\not\in A \text{ and } x\in B\} = B\setminus A \end{equation}

Gribouillis
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