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

  • Thread starter Thread starter evinda
  • Start date Start date
  • Tags Tags
    Elements
Click For Summary
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
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)
 
There is a nice little variation of the problem. The host says, after you have chosen the door, that you can change your guess, but to sweeten the deal, he says you can choose the two other doors, if you wish. This proposition is a no brainer, however before you are quick enough to accept it, the host opens one of the two doors and it is empty. In this version you really want to change your pick, but at the same time ask yourself is the host impartial and does that change anything. The host...

Similar threads

  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
18
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K