MHB How can you find the powerset of sets using simple identities?

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The discussion focuses on finding the powerset of the set containing the empty set and its singleton. The user calculates the powerset of the set $\{ \varnothing, \{ \varnothing \} \}$, resulting in four elements, and seeks to expand this to the powerset of that result, which should contain 16 elements. They reference identities related to subsets and the powerset but inquire about additional identities that could aid in finding the remaining elements. The conversation also touches on the combinatorial aspect of arrangements related to the binomial coefficient. Understanding these identities is crucial for accurately determining the complete powerset.
evinda
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Hello! (Wave)

I want to find the powerset $\mathcal{P} \mathcal{P} \{ \varnothing, \{ \varnothing \} \}$, which has $16$ elements.

I found that $\mathcal{P} \{ \varnothing, \{ \varnothing \} \}=\{ \varnothing, \{ \{ \varnothing \} \}, \{ \varnothing \}, \{ \varnothing, \{ \varnothing \} \} \}$.

Using the identities:

  • The empty set $\varnothing$ is a subset of each set.
    $$$$
  • If $A$ is a set, then $A \subset A \rightarrow A \in \mathcal{P} A$.
    $$$$
  • If $A$ is a set and $x \in A$, then:
    $$\{ x \} \subset A \rightarrow \{ x \} \in \mathcal{P} A$$

I found:

$$\mathcal{P} \{ \varnothing, \{ \{ \varnothing \} \}, \{ \varnothing \}, \{ \varnothing, \{ \varnothing \} \} \}=\{ \varnothing, \{ \varnothing \}, \{ \{ \{ \varnothing \} \} \}, \{ \{ \varnothing \} \}, \{ \{ \varnothing, \{ \varnothing \} \} \}, \{ \varnothing, \{ \{ \varnothing \} \}, \{ \varnothing \}, \{ \varnothing, \{ \varnothing \} \} \}$$

Which other identities could I use to find the other elements of the powerset? (Thinking)
 
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evinda said:
Which other identities could I use to find the other elements of the powerset?
Why do you need any identities?
\[
\mathcal{P}\{1,2,3,4\}=\{\varnothing,\{1\},\{2\},\{3\},\{4\},\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\},\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\}\}
\]
Replace in this identity 1, 2, 3, 4 with $\varnothing$, $\{ \varnothing \}$, $\{ \{ \varnothing \} \}$, $\{ \varnothing, \{ \varnothing \} \}$, respectively.
 
Evgeny.Makarov said:
Why do you need any identities?
\[
\mathcal{P}\{1,2,3,4\}=\{\varnothing,\{1\},\{2\},\{3\},\{4\},\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\},\{1,2,3\},\{1,2,4\},\{1,3,4\},\{2,3,4\},\{1,2,3,4\}\}
\]
Replace in this identity 1, 2, 3, 4 with $\varnothing$, $\{ \varnothing \}$, $\{ \{ \varnothing \} \}$, $\{ \varnothing, \{ \varnothing \} \}$, respectively.

I understand... (Nod) Is the number of different arrangements of $j, 1 \leq j \leq 4$ numbers equal to $\binom{4}{j}$ ? (Thinking)
 
First trick I learned this one a long time ago and have used it to entertain and amuse young kids. Ask your friend to write down a three-digit number without showing it to you. Then ask him or her to rearrange the digits to form a new three-digit number. After that, write whichever is the larger number above the other number, and then subtract the smaller from the larger, making sure that you don't see any of the numbers. Then ask the young "victim" to tell you any two of the digits of the...

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