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Find ℙ(ℙ(ℙ(∅))).

I know that ℙ(∅) = {∅}. Then, ℙ(ℙ(∅)) = {∅, {∅}, {∅,{∅}}? so, ℙ(ℙ(ℙ(∅))) = {∅,{∅, {∅}, {∅,{∅}}}? Is it? Will it be ok if someone explain to me this concept?

eLg
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2 Answers2

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I think it's easier if you consider $P(\varnothing)$ to be a one-element set, say $\{1\}$. Then it's pretty clear that $P(P(\varnothing))$ has to be precisely $\{ \varnothing, \{1\}\}$. So $P(P(P(\varnothing))) = \{\varnothing, \{\varnothing\}, \{\{1\}\}, \{\varnothing, \{1\}\}\}$.

Machine Caliber
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  • How do you know the value of the next element? – eLg Mar 30 '15 at 03:45
  • What do you mean by "next"? – Machine Caliber Mar 30 '15 at 03:45
  • For ℙ(ℙ(∅)) = {∅, {∅}}, how did you know the value will be {∅}, and for ℙ(ℙ(ℙ(∅))) ={∅,{∅},{{1}},{∅,{1}}}? – eLg Mar 30 '15 at 03:47
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    If you have a one-element set, the only possible subsets are those that contain the element or not, which is how you get $P(P(\varnothing))$. For the next step, you now have just a two-element set, so you can have:
    1. Neither element; 2. Element A; 3. Element B; 4. Both elements. That's what you see in the answer.
     – Machine Caliber Mar 30 '15 at 03:55
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    A few notes: We have $P(\emptyset)={\emptyset}$. The set-theoretic definition would be $0=\emptyset$ and $1={0}={\emptyset}$ (not ${1}$). Finally, the substitution of ${\emptyset}$ has been made only two out of three times, so, writing $1$ instead of ${1}$, the answer would be ${\emptyset,1,{1},{\emptyset,1}}$. – Matija Oct 17 '22 at 16:30
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P(ø) = {ø}

P(P(ø)) = P({ø}) = {{ø}, ø}

P(P(P(ø))) = P(P({ø})) = P({{ø}, ø}) = {{{ø}, ø}, {{ø}}, {ø}, ø}

hahardee
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    What does this add to the accepted answer? – José Carlos Santos Oct 17 '22 at 15:22
  • I think the answer is fine, as opposed to the accepted answer this one stresses how one can compute the power sets using nothing but (string) substitution. In my experience, this is exactly in the spirit of set theory. – Matija Oct 17 '22 at 16:07