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We were given this problem for homework:

Let it be $A=\{a,b\}$.

a) Find the powerset $\mathcal P(A)$.

b) Construct an equivalence relation on $\mathcal P(A)$ so that it's quotient set has exactly two elements.

a) $\mathcal P(A)= \{\emptyset\ , \{a\},\{b\},\{a,b\}\}$

b) I constructed the following relation in which an empty set can be an element of ordered pair. (But I'm not sure if an empty set can be in relation. Can an empty set be an element of ordered pair?)

Let's say the relation is R. $$R = \{(\emptyset,\emptyset),(\{a\},\{a\}),(\{b\},\{b\}),(\{a,b\},\{a,b\}),(\{a\},\{b\}),(\{b\},\{a\}),(\emptyset,\{a,b\}),(\{a,b\},\emptyset)\}$$

I think this is correct because:

$[\{a\}] = \{\{a\},\{b\}\} = [\{b\}]$

$[\{\emptyset\}]=\{\emptyset,\{a,b\}\}=[\{a,b\}]$

which means that $\mathcal P(A)/_R=\{\{\{a\},\{b\}\}, \{\emptyset,\{a,b\}\}\}$.

Is this correct or should i remove the empty set from the relation?

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  • $\begingroup$ I do not see a problem. Its fine, I suppose. $\endgroup$ – hemu Nov 23 '17 at 18:55
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    $\begingroup$ It’s perfectly correct. Don’t get confused by having a set of sets. To become more at ease with that, abstract a bit: $X = \mathcal P (A)$ is just a set with four elements, the empty set being one of it, technically. If that gives you a headache, just rename the elements $x_1 = ∅$, $x_2 = \{a\}$, … and work with the renamed elements. $\endgroup$ – k.stm Nov 23 '17 at 18:57
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    $\begingroup$ By the way, the equivalence relation you have found can be described in words: Two sets $x, y ∈ \mathcal P (A)$ are equivalent (with respect to $R$) if and only if they are of the same parity. $\endgroup$ – k.stm Nov 23 '17 at 19:03
  • $\begingroup$ @k.stm Thanks! Also, I have another question. I have to prove that $A\neq \emptyset$ ( for $A=\{(a,b) \in \Bbb R^2 : x+y=c \}$, for every $c$). How do I prove this exactly? I understand why this is true but I don't know how to prove it. $\endgroup$ – i dont know much about algebra Nov 23 '17 at 19:05
  • $\begingroup$ You mean $A = \{(a,b) ∈ ℝ^2;~a + b = c\}$. Well fix a $c$ and find an element in $A$. You can always explicitly write one down. If you can’t manage that, try for $c = 0$ first and then see how you could generalise. $\endgroup$ – k.stm Nov 23 '17 at 19:07
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Yes, it is correct.

Empty set is an element of the power set and in fact you have to include it.

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