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Consider these three sets $(0\notin\mathbb{N}$ for the sake of this example):

$M_1:= \{n\in\mathbb{N}:n $ is even$\}$

$M_2:= \{n\in\mathbb{N}:n $ is odd$\}$

$M_3:= \{1,2,3,4,5\}$


I need to find $(M_1\cap M_2)\cup(M_3\cap M_2)$ which I did, but I have some problems with the notation regarding the empty set.

Version 1:

$(M_1\cap M_2)\cup(M_3\cap M_2)=\emptyset\ \cup \{1,3,5\}=\{\emptyset,1,3,5\}$

Version 2:

$(M_1\cap M_2)\cup(M_3\cap M_2)=\emptyset\ \cup \{1,3,5\}=\{1,3,5\}$

Version 3:

$(M_1\cap M_2)\cup(M_3\cap M_2)=\emptyset\ \cup \{\{1\},\{3\},\{5\}\}=\{\emptyset,\{1\},\{3\},\{5\}\}$

Which is the correct notation?

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  • $\begingroup$ Version 2 is the correct one. $\endgroup$ – Arnaud D. Oct 16 '16 at 11:38
  • $\begingroup$ If $S$ is always false then $(S$ or $T)\iff (T).$ Since $x\in \emptyset$ is always false we have $x\in (\emptyset \cup A)\iff (x\in \emptyset$ or $x\in A)\iff (x\in A)....$ So $\; \emptyset \cup A=A.$ $\endgroup$ – DanielWainfleet Oct 16 '16 at 21:42
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Only version 2 is correct. For any set $A$ we have $A\cup \emptyset=A$.

Also note: $M_3\cap M_2$ is a set of numbers (because $M_2$ and $M_3$ are). Hence the elements of $M_3\cap M_2$ are numbers, not (singleton) sets of numbers. This makes version 3 wrong.

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