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Given a nonempty set S, we select a fixed subset $S_0\subseteq S$ and use it to define two functions on the set of all subsets of $S$:

  1. $f(A)=A\cup S_0$

  2. $g(A)=A\setminus S_0$

For what choices of $S_0$ do these functions commute?

  1. Only for $S_0=\varnothing$? or
  2. Only for $S_0=S$? or
  3. For all $S_0$?
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    $\begingroup$ You have three options. Test each option with an example. $\endgroup$ – Newb Dec 23 '13 at 21:58
  • $\begingroup$ first option: f(A)=A , g(A) = A $\endgroup$ – tim Dec 23 '13 at 22:03
  • $\begingroup$ second option: f(A) = A ∪ S, g(A) = A / S, i am confused this with the third option actually, $\endgroup$ – tim Dec 23 '13 at 22:04
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HINT:

Recall that the functions commute if for every $A$, $f(g(A))=g(f(A))$, that is $$(A\cup S_0)\setminus S_0=(A\setminus S_0)\cup S_0.$$

The choices for which the functions don't commute can be found by testing the case where $S=\{a,b,c\}$. The choice where the functions do commute is easily provable.

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  • $\begingroup$ thank you for the detailed answer, i followed your explanation, the answer would be the first one, is that correct? thank you $\endgroup$ – tim Dec 23 '13 at 22:06
  • $\begingroup$ Yes, the first one is true, the others false. $\endgroup$ – Asaf Karagila Dec 23 '13 at 22:22
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$gf(A)=g(A\cup S_0)=(A\cup S_0)\setminus S_0=A\setminus S_0$

$fg(A)=(A\setminus S_0)\cup S_0=A$

1 -true

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    $\begingroup$ Your calculation of $fg(A)$ is false. $A=\varnothing$ and $S_0\neq\varnothing$ gives $fg(A)=S_0\neq A$. $\endgroup$ – Asaf Karagila Dec 23 '13 at 22:11

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