What is the Relationship Between Union and Power Sets?

In summary, the conversation discusses the concepts of membership and subset in relation to different sets, with examples to illustrate the definitions. It is concluded that x can be a member of a set S, but not necessarily a subset of S, and that x can be a subset of P(x) if it is a transitive set. However, x is not a subset of P(x) if x is a set of multiple elements. It is also noted that x is not necessarily a set and may not always be a member of P({x}).
  • #1
Oxymoron
870
0
I want to make sure I understand the meaning of membership and subset.

For example, if I have a set x, then is x a member/subset of the set

S = {{y},x}

I came to the conclusion that x is a member of the set S because S contains x as an element, and x is also a subset of S because S contains the set x and x is a subset of x. This doesn't sound right, I've used the same argument for two different things?? Hmmm...

But if

S = {y,{x}}

then x would NOT be a member of S because the set S does not contain the set x as an element (but it does contain the set {x} as an element and {x} \neq x). And x is also NOT a subset of S for reasons I cannot think of.
 
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  • #2
Also, if I have the power set P(x), is x a member of P(x)?

If the set x is to be a member of the power set P(x) then P(x) must contain the set x. For example, let x = {a,b} then P(x) = {{},{a},{b},{a,b}}. And since if I remove the outer brackets I have {}, {a}, {b}, and {a,b} as members, and there is {a,b}! Since this will happen with whatever set x I start with, x will always be a member of P(x).

Is x a subset of P(x)?

I figured that since P(x) consists of all subsets of x and since x is a subset of itself, then x is a subset of P(x). Is it true that for every set x, x is always a subset of the power set P(x)?
 
  • #3
Oxymoron said:
I want to make sure I understand the meaning of membership and subset.

For example, if I have a set x, then is x a member/subset of the set

S = {{y},x}

I came to the conclusion that x is a member of the set S because S contains x as an element, and x is also a subset of S because S contains the set x and x is a subset of x. This doesn't sound right, I've used the same argument for two different things?? Hmmm...

Yes x is a member of S, but x is not a subset of S, however {x} is a subeset of S.

Oxymoron said:
But if

S = {y,{x}}

then x would NOT be a member of S because the set S does not contain the set x as an element (but it does contain the set {x} as an element and {x} \neq x). And x is also NOT a subset of S for reasons I cannot think of.

This is correct.
 
  • #4
Oxymoron said:
Also, if I have the power set P(x), is x a member of P(x)?

I take it that x is a set? Then yes x belongs to P(x).

Oxymoron said:
Is x a subset of P(x)?

I figured that since P(x) consists of all subsets of x and since x is a subset of itself, then x is a subset of P(x). Is it true that for every set x, x is always a subset of the power set P(x)?

P(x) is the SET of all subsets of x, so yes x is a subset of x and hence it will belong to P(x), but remember the conditions on a subset

A is a subset of B if for every element a in A a is also in B.

You should notice that none of the elements of x will belong to the power set of x and hence x is not a subset of P(X).
 
  • #5
Posted by d_leet

Yes x is a member of S, but x is not a subset of S, however {x} is a subeset of S.

Right. So {x} is a subset of S because every element of {x} is in S (namely the only element of {x} is x, and x is in S). However, not every element of x (say z) is in S because S contains only two distinct sets (namely x and {y}) and z is not one of these.
 
  • #6
Oxymoron said:
Right. So {x} is a subset of S because every element of {x} is in S (namely the only element of {x} is x, and x is in S). However, not every element of x (say z) is in S because S contains only two distinct sets (namely x and {y}) and z is not one of these.

Also that x is not necessarily a set.
 
  • #7
You should notice that none of the elements of x will belong to the power set of x and hence x is not a subset of P(x).
That's not strictly true; if x is a transitive set, then it will be a subset of its power set. (if and only if)
 
  • #8
Hurkyl said:
That's not strictly true; if x is a transitive set, then it will be a subset of its power set. (if and only if)

Thanks I didn't know that, is that the only case where a set will be a subset of its powerset?
 
  • #9
Posted by d_leet

I take it that x is a set?

Yes, I will always be assuming that x is a set.
 
  • #10
But what if my set was not P(x) but instead P({x})!?

P({x}) is the set of all subsets of {x}. But {x} is a set of one element, namely x, so P({x}) = {{},{x}} does it not? and x is not an member (but {x} would be!?).

Is x a subset of P({x})? No, would be my answer, because we cannot guarantee that a member of x is in P({x}) which contains only the empty set and the set containing the set x.
 
  • #11
Oxymoron said:
But what if my set was not P(x) but instead P({x})!?

P({x}) is the set of all subsets of {x}. But {x} is a set of one element, namely x, so P({x}) = {{},{x}} does it not? and x is not an member (but {x} would be!?).

Is x a subset of P({x})? No, would be my answer, because we cannot guarantee that a member of x is in P({x}) which contains only the empty set and the set containing the set x.

Yes that would be correct again.
 
  • #12
I don't know if it will help, but (I think) if you assume the axiom of foundation, you can actually represent sets as trees. For example, the set {x, {y}} can be represented as:

Code:
   *
  / \
 /   \
x     *
      |
      |
      y

In this representation, an asterisk denotes the top of set, and all of the children of an asterisk are the elements of that set. (If x and y were themselves sets, you could substitute their trees into the above diagram)


I say "I think", because I'm pretty sure that the axiom of foundation guarantees every set can be drawn like this, but not 100% sure.
 
  • #13
Is x a subset of P({x})? No, would be my answer, because we cannot guarantee that a member of x is in P({x}) which contains only the empty set and the set containing the set x.
Technicalities again. The answer is yes if x is the empty set, or the set containing the empty set. (Or, if we reject the axiom of foundation, and allow x = {{x}} or x = {{}, {x}})
 
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  • #14
Perhaps this will help: I will always assume that x is non-empty, x is not an member of itself, and y is not x.
 
  • #15
The Axiom of Foundation has not been covered in lectures yet, so I am pretty sure I am not meant to use it. However, your idea may come in handy never-the-less...I like it.
 
  • #16
Well, allowing recursion, and sets with transfinite depth is a sort of trappy thing, and not really needed for doing ordinary mathematics. So for most purposes, you're going to want to assume foundation. I mainly just want you to be aware of its necessity. :smile:
 
  • #17
What about U P(x), that is the union of the power set?

I figured that since the union is "like" bracket removal, x will not be a member. For example, take x = {1,2}. Then

U P(x) = U {{},{1},{2},{1,2}} = {} U {1} U {2} U {1,2} = ?? not sure what goes here.
 
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  • #18
I was thinking, (assuming that x is not a member of itself and is non-empty), x is not a member of x BUT x is a member of {x}. So, would x be both a member AND an element of the set {{x},x}?
 
  • #19
Note that, in your particular case, U P(x) = {1, 2} = x. Can you prove the general case?


So, would x be both a member AND an element of the set {{x},x}?
Yes... but member and element are synonyms.

I think you meant to ask:

Would {x} be an element and a subset of {{x},x}?

And the answer would be yes.
 
  • #20
...you read my mind! ;)
 
  • #21
Posted by Hurkyl:

Note that, in your particular case, U P(x) = {1, 2} = x. Can you prove the general case?

No, but I am sure it is easy. Could you simply use the definition of "big"-union? The membership for the union of the power set P(X) is that x is in A_i for at least one i in the index set I. Note that A_i are the subsets of X. (so, say A_1 = {}, A_2 = {1}, A_3 = {2}, and A_4 = {1,2}). Then you could just argue that there does not exist an A_i which contains an element not in X. Therefore the union of the collection of all subsets of X (i.e. the power set of X) is exactly X.
 
  • #22
That's the idea. Now just clean it up! It's often easier to organize proofs like this as either proving:

a is in U P(x) iff a is in x

which is usually done by proving the if part and the only if part separately. I often find this more convenient than showing

x <= U P(x) and U P(x) <= x

(where <= means subset)
 
  • #23
(<=)
Suppose a is in x. Then a is in at least one of the subsets in the power set of x, since P(x) consists of all subsets of x. Then by the definition of union, a is in U P(x) because a is in at least one of the subsets. Therefore a is in U P(x).

(=>)
Suppose a is in U P(x). Then a is in at least one of the subsets of P(x). But the power set P(x) of a set x is the set of all subsets of x, therefore if a is in at least one of the subsets of P(x) is must be in x itself. Therefore a is in x.

Therefore a is in U P(x) <=> a is in x. Therefore U P(x) = x.

Does this look ok?
 

What is set theory?

Set theory is a branch of mathematics that deals with the study of collections of objects, called sets, and the relationships between them.

What is the basic concept of sets?

The basic concept of sets is that they are a collection of distinct objects, called elements, that are grouped together based on specific characteristics or properties.

What is the difference between a set and an element?

A set is a collection of elements, while an element is a single object within that set. For example, a set of fruits would include apples, oranges, and bananas as elements.

What are the operations in set theory?

The basic operations in set theory are union, intersection, and complement. Union combines two sets to form a new set that contains all elements from both sets. Intersection creates a new set with only the elements that are in both sets. Complement creates a new set with all the elements that are not in a given set.

How is set theory used in other fields of science?

Set theory is used in various fields of science, including computer science, linguistics, and physics. It provides a foundation for understanding and analyzing complex systems and relationships between objects. For example, set theory is used in computer science to understand and organize data, and in linguistics to study language structure and syntax.

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