Set Theory: Subsets & Power Set of A - Joe

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The discussion centers on the misunderstanding of the relationship between subsets and elements of a power set. The statement "If B is a subset of A, then {B} is an element of the Power Set A" is identified as false; rather, it is B (not {B}) that is an element of the power set of A. The distinction between a set and a set containing a single member is crucial, as {B} represents a different entity than B itself. Examples illustrate that while B can be a subset of A, {B} does not necessarily belong to the power set unless B is also an element of A. This highlights the importance of understanding set notation in set theory discussions.
Agent M27
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I am currently covering Set Theory from the book, A Transition to Abstract Mathematics (Douglas Smith) and have a question about subsets and an implication. The statement reads as follows:

If B is a subset of A, then {B} is an element of the Power Set A.

I choose this to be true. By definition the Power Set A is comprised of all subsets of A. Given the first condition that B is a subset of A I can't really see how this is false, which the book gives as the correct answer. Does it have something to do with the braces around B? The way I interpreted the statement is: If B is a subset of A, then the set B is an element of the Power Set A. I know, for example, that when x is an element of A this does not automatically imply it is also an element of the Power Set A, this is one of the cases from which the confusion arises. Thanks in advance.

Joe
 
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P(A) is the set of all subsets of A. B is a subset of A, so if {B} is an element of the power set of A, then {B} must be a subset of A. This is easily seen to be false in general. Pick e.g. A = empty set, then B = empty set is a subset of A. But the power set of A is {empty set}, which does not contain {B} = {the empty set} as an element.

But more interestingly, can it be true for any set A? Assume such an A exists.
A is a subset of A, so {A} is an element of P(A). But then {A} must be a subset A, so A is an element of A. This violates the axiom of regularity, so it cannot be true in ZFC.
 
Thanks a lot Jarle, the example of A= Empty Set is just what I needed to see the justification. Take care.

Joe
 
Agent M27 said:
I am currently covering Set Theory from the book, A Transition to Abstract Mathematics (Douglas Smith) and have a question about subsets and an implication. The statement reads as follows:

If B is a subset of A, then {B} is an element of the Power Set A.
That statement,as written, is false. What is true is that if B is a subset of A, then B (not {B} which means the set containing a single member, B) is a member of the power set of P.

I choose this to be true. By definition the Power Set A is comprised of all subsets of A. Given the first condition that B is a subset of A I can't really see how this is false, which the book gives as the correct answer. Does it have something to do with the braces around B?
Yes. "B" is not the same as "{B}".

The way I interpreted the statement is: If B is a subset of A, then the set B is an element of the Power Set A.
As I said before, {B} is the set that has the set B as its only member.

I know, for example, that when x is an element of A this does not automatically imply it is also an element of the Power Set A, this is one of the cases from which the confusion arises.
That's the same confusion in reverse! Suppose A= {1, 2, 3, 4}. Then "2" is a member of A but certainly not a subset of A (it is not a set at all- it is a number) and so not a member of the power set of A. B= {2} is a subset of A (it is the set whose only member is the number "2") and so a member of the power set of A. {B}, in that case, would be the set {{2}}, the set whose only member is the set whose only member is "2". Since B= {2} is not itself a member of A, it is {B}= {{2}} is NOT a member of the power set of A.

Thanks in advance.

Joe
 
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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