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Let $S$ be the equivalence relation defined on $\wp(\{1, 2, 3, 4\})$ defined by: $$XSY\text{ if and only if } |X|\equiv|Y|\;\mod 2$$ Write down the equivalence classes of S.

I understand that equivalence classes have relations where it is reflexive, symmetric and transitive but how are you supposed to write equivalence classes?

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  • $\begingroup$ It is not the absolute value. It is the cardinality, i.e. (for finite sets) the number of its elements. $\endgroup$ Apr 30, 2019 at 7:57
  • $\begingroup$ Presumably $|X|$ means the number of elements of $X$. If $X$ is in the power set of $A$ then its elements are some (or all or none) of the elements of $A$ $\endgroup$
    – Henry
    Apr 30, 2019 at 7:58

2 Answers 2

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What you're calling the "absolute value of a set" is actually referred to as its cardinality. For finite sets, which is the relevant case here, cardinality is how many elements are in the set. It gets a bit murkier with infinite sets but I won't bog you down with the details here.

So in short: you look at the power set of $\{1,2,3,4\}$, i.e. the set of subsets of that. You then define the relation $S$ on these sets, where two sets, $X,Y$ are related if their cardinalities satisfy $|X| \equiv |Y| \pmod 2$.

Equivalently, let $X$ have $n$ elements and $Y$ have $m$ elements, where $X,Y \in P(\{1,...,4\})$. Then $XSY$ if and only if $n \equiv m \pmod 2$.

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In particular, there are only two equivalence classes of $S$. One is, denoted by $\overline{0}$, $$\{ \emptyset, \{1, 2\}, \{ 1, 3\}, \{1, 4\}, \{2,3\},\{2,4\},\{3,4\},\{1,2,3,4\} \}.$$ And the other one is, denoted by $\overline{1}$ $$ \{ \{1\},\{2\},\{3\},\{4\}, \{2,3,4\},\{1,3,4\},\{1,2,4\},\{1,2,3\}\}.$$

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    $\begingroup$ I do not quite understand how you got those values as the answer, what does the line above the number mean? Sorry I'm a beginner in discrete maths. $\endgroup$
    – Andrea
    Apr 30, 2019 at 8:54
  • $\begingroup$ You can see @Eevee Trainer's answer, $\overline{0}$ contains all the subsets of cardinalities 0,2 and 4, which are $0 \pmod{2}$. and $\overline{1}$ contains all the subsets of cardinalities 1 and 3, which are $1 \pmod{2}$. That's why I use the two notations, $\overline{0}$ and $\overline{1}$. $\endgroup$ Apr 30, 2019 at 8:58

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