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

  • Thread starter Thread starter evinda
  • Start date Start date
  • Tags Tags
    Elements
evinda
Gold Member
MHB
Messages
3,741
Reaction score
0
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)
 
Physics news on Phys.org
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)
 
Namaste & G'day Postulate: A strongly-knit team wins on average over a less knit one Fundamentals: - Two teams face off with 4 players each - A polo team consists of players that each have assigned to them a measure of their ability (called a "Handicap" - 10 is highest, -2 lowest) I attempted to measure close-knitness of a team in terms of standard deviation (SD) of handicaps of the players. Failure: It turns out that, more often than, a team with a higher SD wins. In my language, that...
Hi all, I've been a roulette player for more than 10 years (although I took time off here and there) and it's only now that I'm trying to understand the physics of the game. Basically my strategy in roulette is to divide the wheel roughly into two halves (let's call them A and B). My theory is that in roulette there will invariably be variance. In other words, if A comes up 5 times in a row, B will be due to come up soon. However I have been proven wrong many times, and I have seen some...
Back
Top