Proving Existence/Uniqueness of a Set.

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In summary: But then U \setminus A = U \cap B, so A = B, and hence B is unique.In summary, for every A ∈ P(U), there is a unique B ∈ P(U) such that for every C ∈ P(U), C \ A = C ∩ B.
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Homework Statement


Let U be any set.
(a) Prove that for every A ∈ P (U) there is a unique B ∈ P (U) such
that for every C ∈ P (U), C \ A = C ∩ B.
Remark: P(U) = the Power set of U, i.e. A ∈ P (U) then A⊆U

Homework Equations


The Attempt at a Solution


The question's form is as follows:
∀A(A∈P(U)⇒∃!B(Q(B)))
Where Q(B) = B∈P(U)∧∀C(C∈P(U)⇒C\A=C∩B)
I first let A be an arbitrary element, then
A∈P(U)⇒∃!B(Q(B))

Suppose A⊆U, then I want to prove ∃!B(Q(B))

Existence: I want to prove ∃B(Q(B))
Uniqueness: I want to prove that, ∀r∀s(Q(r)∧(Q(s)⇒r=s)
I start with existence,

To find a B such that Q(B) is true, I consider all subsets of U and their relation to A in order to find a suitable B using the hint C\A=C∩B
Let u be an arbitrary subset of U

Since both A and the subsets of U are both subsets of U, then there are two possibilities: either A∩u=∅∨A∩u≠∅
In the first case u\A is just u, therefore B can be any set such that u⊆B since u∩B=u =u\A
In the second case u\A = some subset of u, say u’ since u∩A≠∅. We want to find the choices for B such that u∩B = u’. Therefore B can be any set that does not contain u∩A and contains u’; For example: U\A, or U\(u∩A)

Even considering these choices for B, I cannot narrow it down to a choice where B satisfies Q(B) since the possibilities for B in both cases are case-specific and do not apply to both cases. This is where I’m stuck, any hints or even the solution would be much appreciated. I feel like I’m overlooking something simple and that I might be over-complicating things.
 
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pandaBee said:

Homework Statement


Let U be any set.
(a) Prove that for every A ∈ P (U) there is a unique B ∈ P (U) such
that for every C ∈ P (U), C \ A = C ∩ B.
Remark: P(U) = the Power set of U, i.e. A ∈ P (U) then A⊆U

Homework Equations





The Attempt at a Solution


The question's form is as follows:
∀A(A∈P(U)⇒∃!B(Q(B)))
Where Q(B) = B∈P(U)∧∀C(C∈P(U)⇒C\A=C∩B)
I first let A be an arbitrary element, then
A∈P(U)⇒∃!B(Q(B))

Suppose A⊆U, then I want to prove ∃!B(Q(B))

Existence: I want to prove ∃B(Q(B))
Uniqueness: I want to prove that, ∀r∀s(Q(r)∧(Q(s)⇒r=s)
I start with existence,

To find a B such that Q(B) is true, I consider all subsets of U and their relation to A in order to find a suitable B using the hint C\A=C∩B
Let u be an arbitrary subset of U

Since both A and the subsets of U are both subsets of U, then there are two possibilities: either A∩u=∅∨A∩u≠∅

Your statement here is simply "either A is empty or it isn't".

In the first case u\A is just u, therefore B can be any set such that u⊆B since u∩B=u =u\A
In the second case u\A = some subset of u, say u’ since u∩A≠∅. We want to find the choices for B such that u∩B = u’. Therefore B can be any set that does not contain u∩A and contains u’; For example: U\A, or U\(u∩A)

Even considering these choices for B, I cannot narrow it down to a choice where B satisfies Q(B) since the possibilities for B in both cases are case-specific and do not apply to both cases. This is where I’m stuck, any hints or even the solution would be much appreciated. I feel like I’m overlooking something simple and that I might be over-complicating things.

The best way to prove existence of B is to find a B which works. So consider B = U \ A. Then for any subset C, we can partition C into disjoint subsets [itex]C \cap A[/itex] and [itex]C \cap B[/itex]. Then [itex]C \setminus A = C \cap B[/itex] as required.

It remains to show that B = U \ A is the only possibility. That's straightforward: [itex]C \setminus A = C \cap B[/itex] must hold for all [itex]C \subset U[/itex], so in particular it holds for [itex]C = U[/itex].
 

Related to Proving Existence/Uniqueness of a Set.

What is the importance of proving existence/ uniqueness of a set in science?

Proving the existence or uniqueness of a set is crucial in science because it provides evidence and justification for the validity of a theory or hypothesis. It allows scientists to confidently assert the presence or uniqueness of a certain phenomenon or entity, which can then be further studied and analyzed. Additionally, proving existence or uniqueness helps to establish the boundaries and limitations of a set, which can aid in future research and experimentation.

What is the difference between proving existence and proving uniqueness of a set?

Proving existence of a set means demonstrating that there is at least one element within the set that satisfies certain conditions. On the other hand, proving uniqueness means showing that there is only one element in the set that satisfies those conditions. In other words, proving existence is about demonstrating that something exists, while proving uniqueness is about demonstrating that there is only one of that thing.

How do scientists go about proving existence/ uniqueness of a set?

There are various methods that scientists use to prove the existence or uniqueness of a set, depending on the nature of the set and the hypothesis being tested. These methods can include mathematical proofs, experimental data analysis, and observation of natural phenomena. Scientists also rely on established principles and theories in their respective fields to support their claims.

Can a set have both existence and uniqueness?

Yes, a set can have both existence and uniqueness. This means that there is at least one element that satisfies certain conditions, and that element is the only one that satisfies those conditions. In other words, the set contains only one unique element that exists. This is commonly seen in mathematical proofs and in scientific laws, where a specific entity or phenomenon is shown to exist and be unique.

What happens if a set cannot be proven to have existence or uniqueness?

If a set cannot be proven to have existence or uniqueness, it may indicate that the theory or hypothesis being tested is not valid. In this case, further research and experimentation may be necessary to refine the conditions and criteria for proving existence or uniqueness. Alternatively, it could also mean that the set does not actually exist or is not unique, and further investigation is needed to determine the reason behind this.

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