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I'm working on these questions and I'm very confused about the difference between membership and inclusion.

If $a\in P(A∩B)\setminus C $ then:

  1. $(a\in D\setminus C)$ where $D = P(A \cap B)$;
  2. $a\subset (A \cap B)$ AND $(a \not\subset C)$;
  3. $a\subset(A \cap B)$ AND $(a \notin C)$;
  4. $a\subset A$ AND $a\subset B$.

Is the membership valid only between an element and a set or no? And in the inclusion valid only between sets? I searched a lot to do this exercise but i don't understand

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  • $\begingroup$ Every element is a set itself, so you can talk about membership and inclusion between any element and/or set $\endgroup$ – Exodd Sep 28 '15 at 20:17
  • $\begingroup$ @Exodd can you give me an example of answers 2 and 3 please? $\endgroup$ – dorino canciani Sep 28 '15 at 20:45
  • $\begingroup$ But you have to prove every of them, or it's a true/false? $\endgroup$ – Exodd Sep 28 '15 at 22:23
  • $\begingroup$ sorry it's true/false $\endgroup$ – dorino canciani Sep 29 '15 at 5:18
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In elemental set theory, every element is a set itself. The expression $$a\in P(A\cap B)$$ is the proof of it, since $a$ is a subset of $A$ and $B$. In simbols, $$a\subset (A\cap B)$$ For example, set $$A=\{ 0 , 1 , \{ 0, 1 \} \}$$ $$B=\{ 0 , 1 , \{ 0 \} \}$$ in this case $$A\cap B = \{ 0, 1 \}$$ $$P(A\cap B)= \{ \{ \}, \{ 0 \}, \{ 1 \} , \{ 0, 1 \}\}$$ If $$a= \{ 0, 1 \}$$ then the following statements are both true $$a\in A\qquad a\subset A$$

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  • $\begingroup$ Ok now it is better:) Last question: a ⊆ A ∩ B e a ̸⊆ C is true? And a ⊆ A ∩ B e a ̸∈ C is true? Because now for me both are true..am i right? $\endgroup$ – dorino canciani Sep 29 '15 at 22:08
  • $\begingroup$ Try with $C=B$ and answer by yourself $\endgroup$ – Exodd Sep 29 '15 at 22:45

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