MHB Transitive Sets: Prove, Show With $n$ Elements

  • Thread starter Thread starter Also sprach Zarathustra
  • Start date Start date
  • Tags Tags
    Sets
Click For Summary
A transitive set is defined as one where all elements are subsets of the set itself. It is proven that if \( A \) is a transitive set, then \( A \cup \{A\} \) is also transitive because the only new member added, \( A \), is a subset of the original set. To demonstrate the existence of transitive sets with \( n \) elements, induction is used, starting with \( A_1 = \{\emptyset\} \) and defining \( A_{n+1} = A_n \cup \{A_n\} \). This construction ensures that for every natural number \( n \), there is a corresponding transitive set. The discussion highlights the foundational properties of transitive sets and their construction through induction.
Also sprach Zarathustra
Messages
43
Reaction score
0
Hello, I need a help with the following:

1. Let $A$ be a transitive set, prove that $A\cup \{A \}$ is also transitive.
2. Show that for every natural $n$ there is a transitive set with $n$ elements.
 
Physics news on Phys.org
Also sprach Zarathustra said:
Hello, I need a help with the following:

1. Let $A$ be a transitive set, prove that $A\cup \{A \}$ is also transitive.
2. Show that for every natural $n$ there is a transitive set with $n$ elements.
For 2., use induction. Let $A_1 = \{\emptyset\}$. For $n\geqslant1$, let $A_{n+1} = A_n\cup \{A_n\}$ and use 1.
 
A transitive set is one in which all elements are subsets, now for 1. you have that the only new member that you have introduced is $A$ and it is a subset so the set is transtitve.

Imagine the tansitive set to be $A=\{1,2,3,4,5\}$ where these are defined in the usual way (in terms of the empty set).

Then the new set would be $B=\{1,2,3,4,5,A\}$ now then we can see that $A\in B$ but also that $\{1,2,3,4,5\}\subset B$ and so $A$ is a subset of B and so the set is transitive
 
The standard _A " operator" maps a Null Hypothesis Ho into a decision set { Do not reject:=1 and reject :=0}. In this sense ( HA)_A , makes no sense. Since H0, HA aren't exhaustive, can we find an alternative operator, _A' , so that ( H_A)_A' makes sense? Isn't Pearson Neyman related to this? Hope I'm making sense. Edit: I was motivated by a superficial similarity of the idea with double transposition of matrices M, with ## (M^{T})^{T}=M##, and just wanted to see if it made sense to talk...

Similar threads

  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 35 ·
2
Replies
35
Views
5K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K