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)
 
I was reading documentation about the soundness and completeness of logic formal systems. Consider the following $$\vdash_S \phi$$ where ##S## is the proof-system making part the formal system and ##\phi## is a wff (well formed formula) of the formal language. Note the blank on left of the turnstile symbol ##\vdash_S##, as far as I can tell it actually represents the empty set. So what does it mean ? I guess it actually means ##\phi## is a theorem of the formal system, i.e. there is a...
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