MHB Proper Subsets and Relations of Sets

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Proper subsets of the set S = {1, 2, 3, 4} include the empty set, single-element subsets, two-element subsets, and three-element subsets, totaling 15 proper subsets. For the relation R defined on S = {1, 2, 5, 6}, pairs (a, b) are included if the product a*b is even, which can be demonstrated with various combinations. The relation R on the set S = {x, y, z} is defined with specific pairs, and for R to be reflexive, it must include (x,x), (y,y), and (z,z). The discussion emphasizes understanding proper subsets and the properties of relations, particularly reflexivity. Overall, the thread provides insights into set theory and relational properties.
saaddii
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Q1: Write all proper subsets of S = {1, 2, 3, 4 }.

Q2: Let S = {1,2,5,6 }
Define a relation R on S of at least four order pairs, as (a,b)  R iff a*b is even (i.e. a multiply by b is even)

Q3: Let S = {x, y, z } and R is a relation defined on S such that
R={(y,y),(x,z),(z,x),(x,x),(z,z),(x,y),(y,x)}
Show that R is reflneed proper solution
 
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saaddii said:
Q1: Write all proper subsets of S = {1, 2, 3, 4 }.

Q2: Let S = {1,2,5,6 }
Define a relation R on S of at least four order pairs, as (a,b)  R iff a*b is even (i.e. a multiply by b is even)

Q3: Let S = {x, y, z } and R is a relation defined on S such that
R={(y,y),(x,z),(z,x),(x,x),(z,z),(x,y),(y,x)}
Show that R is reflneed proper solution

Q1:
A proper subset of $S$ is any subset of $S$ that is not equal to $S$.
So we have...
$\emptyset$
$\left\{1\right\}, \left\{2\right\}, \left\{3\right\}, \left\{4\right\}$
$\left\{1,2\right\}, \left\{1,3\right\}, \left\{1,4\right\}, \left\{2,3\right\}, \left\{2,4\right\}, \left\{3,4\right\}$
$\left\{1,2,3\right\}, \left\{2,3,4\right\}, \left\{1,3,4\right\}, \left\{1,2,4\right\}$

Q3:
R is reflective if for every element $s$ of $S$, $(s,s)$ is in $R$.
So what do you need to check?
For R to be reflexive, it must have elements $(x,x)$, $(y,y)$, and $(z,z)$.
 
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...

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