i am giving you the solution of following question , please explain the everything needed in the question detailed usng the given question 
3. Consider the following Intersecting Set problem. Given a universe U, a family F of disjoint
subsets of U and a family A of subsets of U. Each set X in A is associated with a budget
b(X). The goal is to find a subset S ⊆ U of minimum size such that the following holds: For
each set Y ∈ F, S ∩Y ̸= ∅ and for each set Z ∈ A, |Z ∩S| ≤ b(Z). Show that the Intersecting
Set problem is NP-complete. Prove the correctness of your hardness reduction.


the solution is 
To demonstrate that the Intersecting Set problem is NP-complete, we'll follow a structured approach:

Show that the problem is in NP.
Prove NP-hardness by reducing a known NP-complete problem, specifically the Vertex Cover problem, to it.
Ensure that the reduction preserves the correctness of solutions between the two problems.
Let's delve into each step in detail.

1. The Intersecting Set Problem is in NP
A problem is in NP (Nondeterministic Polynomial time) if, given a candidate solution, we can verify its correctness in polynomial time.

Verification Process
Given:

A universe ( U ).
A family ( F ) of disjoint subsets of ( U ).
A family ( A ) of subsets of ( U ), each with an associated budget ( b(X) ).
Candidate Solution: A subset ( S \subseteq U ).

We need to verify two conditions:

Intersection with ( F ): For every ( Y \in F ), ( S \cap Y \neq \emptyset ).
Budget Constraints: For every ( Z \in A ), ( |Z \cap S| \leq b(Z) ).
Verification Steps:
Check Intersections:
For each ( Y \in F ), ensure that at least one element of ( Y ) is in ( S ).
Since ( F ) consists of disjoint subsets, these checks are independent and can be performed in parallel or sequentially without overlap concerns.
Check Budget Constraints:
For each ( Z \in A ), count the number of elements from ( S ) that are in ( Z ).
Ensure this count does not exceed ( b(Z) ).
Both verification steps involve iterating through the families ( F ) and ( A ) and performing straightforward set operations, all achievable in polynomial time relative to the size of the input.

Conclusion: Since we can verify any candidate solution in polynomial time, the Intersecting Set problem is in NP.

2. The Intersecting Set Problem is NP-Hard
To establish NP-hardness, we'll reduce the Vertex Cover problem, a known NP-complete problem, to the Intersecting Set problem. This means we'll transform any instance of Vertex Cover into an instance of the Intersecting Set problem such that a solution to the latter corresponds to a solution of the former.

Vertex Cover Problem (VC)
Definition: Given a graph ( G = (V, E) ) and an integer ( k ), does there exist a subset ( C \subseteq V ) with ( |C| \leq k ) such that every edge in ( E ) is incident to at least one vertex in ( C )? In other words, ( C ) "covers" all edges.

Reduction Strategy
We'll map an instance of VC to an instance of the Intersecting Set problem, ensuring that:

Solutions correspond directly: A vertex cover ( C ) of size ( \leq k ) will map to a subset ( S ) satisfying the Intersecting Set conditions, and vice versa.
Preservation of constraints: The size and coverage constraints of VC should be mirrored in the Intersecting Set instance.
Constructing the Intersecting Set Instance
Given an instance of VC:

Graph: ( G = (V, E) )
Integer: ( k )
We construct an Intersecting Set instance as follows:

Universe ( U ): Let ( U = V ).
Each element in ( U ) corresponds directly to a vertex in ( G ).
Family ( F ): For each edge ( e = {u, v} ) in ( E ), define a subset ( Y_e = {u, v} ).
Thus, ( F = { Y_e \mid e \in E } ).
Note: To ensure that ( F ) consists of disjoint subsets, we need to adjust our construction since, in a typical graph, multiple edges can share vertices. To achieve disjointness, we'll use a technique called element duplication.
Handling Disjointness in ( F ):
Element Duplication: For each vertex ( u \in V ) and each edge ( e ) incident to ( u ), create a unique copy ( u_e ).
Revised Universe ( U ): ( U = { u_e \mid u \in V, e \in E, u \text{ is endpoint of } e } ).
Revised Family ( F ): For each edge ( e = {u, v} ), define ( Y_e = { u_e, v_e } ).
This ensures that each ( Y_e ) is a distinct pair of elements, making all subsets in ( F ) disjoint.
Family ( A ):
For each vertex ( u \in V ), define ( Z_u = { u_e \mid e \in E, u \text{ is endpoint of } e } ).
Set ( b(Z_u) = 1 ) for all ( Z_u \in A ).
This enforces that at most one copy of each vertex can be included in ( S ), effectively ensuring that selecting any ( u_e ) corresponds to selecting vertex ( u ) once in the vertex cover.
Budget Constraints:
Although the Intersecting Set problem as described doesn't directly impose a total size constraint on ( S ), we can simulate the Vertex Cover size constraint by adjusting the structure of ( A ). Specifically:
Introduce a universal set ( Z_{\text{all}} = U ) with ( b(Z_{\text{all}}) = k ).
This ensures that ( |S| \leq k ).
Final Intersecting Set Instance:

Universe: ( U = { u_e \mid u \in V, e \in E, u \text{ is endpoint of } e } )
Family ( F ): ( F = { Y_e = { u_e, v_e } \mid e = {u, v} \in E } )
Family ( A ): ( A = { Z_u = { u_e \mid e \in E, u \text{ is endpoint of } e } \mid u \in V } ) and ( Z_{\text{all}} = U ), with budgets ( b(Z_u) = 1 ) for all ( u \in V ) and ( b(Z_{\text{all}}) = k ).
Visualization of the Reduction
Consider a simple graph:


    u1
   /  \
 e1    e2
   \  /
    u2
Vertices: ( u1, u2 )
Edge: ( e1 = {u1, u2}, e2 = {u1, u2} )
Universe ( U ):

( u1_{e1}, u2_{e1}, u1_{e2}, u2_{e2} )
Family ( F ):

( Y_{e1} = { u1_{e1}, u2_{e1} } )
( Y_{e2} = { u1_{e2}, u2_{e2} } )
Family ( A ):

( Z_{u1} = { u1_{e1}, u1_{e2} } ), ( b(Z_{u1}) = 1 )
( Z_{u2} = { u2_{e1}, u2_{e2} } ), ( b(Z_{u2}) = 1 )
( Z_{\text{all}} = { u1_{e1}, u2_{e1}, u1_{e2}, u2_{e2} } ), ( b(Z_{\text{all}}) = k )
A valid ( S ) might include ( u1_{e1} ) and ( u1_{e2} ), but due to ( b(Z_{u1}) = 1 ), only one can be chosen, effectively selecting ( u1 ) once to cover both edges.

Ensuring Correctness of the Reduction
We need to verify two directions:

If G has a Vertex Cover ( C ) of size ( \leq k ), then the Intersecting Set instance has a corresponding subset ( S ) satisfying its conditions.
Construction of ( S ):
For each vertex ( u \in C ), select any one of its copies ( u_e ) (for some edge ( e ) incident to ( u )).
Due to the budget constraints ( b(Z_u) = 1 ), only one copy per vertex can be selected.
Verification:
Since ( C ) covers all edges, for each ( Y_e ), at least one of ( u_e ) or ( v_e ) will be in ( S ).
The total number of elements in ( S ) will be ( \leq k ), adhering to ( b(Z_{\text{all}}) = k ).
If the Intersecting Set instance has a subset ( S ) satisfying its conditions, then G has a corresponding Vertex Cover ( C ) of size ( \leq k ).
Construction of ( C ):
For each ( u_e \in S ), include vertex ( u ) in ( C ).
Due to ( b(Z_u) = 1 ), each vertex is included at most once.
Verification:
Since ( S ) intersects every ( Y_e ), for each edge ( e ), at least one of its endpoints is in ( C ).
The total size of ( C ) is ( \leq k ), as enforced by ( b(Z_{\text{all}}) = k ).
Conclusion: The reduction correctly maps Vertex Cover instances to Intersecting Set instances, preserving the existence of solutions. Therefore, the Intersecting Set problem is NP-hard.

3. NP-Completeness of the Intersecting Set Problem
Since we've established that:

The Intersecting Set problem is in NP.
The Intersecting Set problem is NP-hard via a reduction from the NP-complete Vertex Cover problem.
Therefore, the Intersecting Set problem is NP-complete.

Summary
Interjecting Set in NP: Any candidate solution ( S ) can be verified efficiently.
NP-Hardness via Vertex Cover: A polynomial-time reduction ensures that solving Intersecting Set would solve Vertex Cover.
Correctness: The reduction accurately preserves the solution's validity between the two problems.
This comprehensive reduction and verification confirm that the Intersecting Set problem is indeed NP-complete.