do not change anything from following solution just there are some syntax error fix those such as {a_1, a_2, \dots, a_n} etc only fix these error and nothing else keep this in the mind 
and give the new updated solution
We are given a universe ( U = {a_1, a_2, \dots, a_n} ) of ( n ) items and a family ( F ) of two-sized subsets of ( U ) (i.e., each subset in ( F ) contains exactly two elements from ( U )). For each set ( X \in F ), there is an associated weight ( w(X) \in \mathbb{N} ).

Our goal is to find a subset ( S \subseteq F ) satisfying the following conditions:

Connectivity Condition (a): For every partition ( S = S_1 \sqcup S_2 ) (i.e., ( S_1 \cap S_2 = \emptyset ) and ( S_1 \cup S_2 = S )), there exists sets ( Y \in S_1 ) and ( Z \in S_2 ) such that ( Y \cap Z \neq \emptyset ).
Coverage Condition (b): The union of the sets in ( S ) equals ( U ), i.e., ( \bigcup_{X \in S} X = U ).
Minimal Weight (c): Among all subsets satisfying conditions (a) and (b), ( S ) has the minimal possible total weight ( \sum_{X \in S} w(X) ).
Our task is to design an algorithm that finds such a subset ( S ) in polynomial time, prove its correctness, and analyze its running time in detail.

Understanding the Problem
Let's analyze the problem step by step, ensuring we understand each requirement and how they relate to familiar concepts in computer science and graph theory.

1. Interpreting Condition (a):
Condition (a) essentially states that in any partition of ( S ) into two non-overlapping subsets ( S_1 ) and ( S_2 ), there must be at least one element (from ( U )) that is shared between some set in ( S_1 ) and some set in ( S_2 ).
This condition prevents ( S ) from being "disconnected" in terms of the overlap of elements in ( U ) among the sets in ( S ).
This can be interpreted as a connectivity condition among the sets in ( S ).
2. Interpreting Condition (b):
Condition (b) requires that ( S ) covers the entire universe ( U ). Each element ( a_i \in U ) must be included in at least one set ( X \in S ).
3. Rewriting the Problem in Graph Terms:
Vertices: Each element in ( U ) can be considered a vertex in a graph.
Edges: Each pair ( X \in F ) (which contains two elements from ( U )) can be considered an edge connecting the two corresponding vertices.
Weights: The weight ( w(X) ) is associated with the edge between those two vertices.
Under this interpretation, ( F ) becomes the set of edges in a graph ( G = (V, E) ), where ( V = U ) and ( E = F ).
Designing the Algorithm
Given the graph interpretation, the problem reduces to finding a minimal-weight subset of edges ( S \subseteq E ) such that:

Spanning Condition: The edges in ( S ) cover all vertices in ( V ) (condition (b)), i.e., the graph ( G' = (V, S) ) spans all vertices.
Connectivity Condition: For any partition of ( S ) into ( S_1 ) and ( S_2 ), there is at least one vertex ( v \in V ) that is incident to edges in both ( S_1 ) and ( S_2 ) (condition (a)). In other words, the graph ( G' = (V, S) ) is connected.
Minimal Total Weight: The total weight ( \sum_{e \in S} w(e) ) is minimized (condition (c)).
Recognizing that we need a connected, spanning subgraph of ( G ) with minimal total edge weight, this problem is equivalent to the Minimum Spanning Tree (MST) problem.

Algorithm Steps:
Construct Graph ( G = (V, E) ):
Vertices: ( V = U ) (elements ( a_1, a_2, \dots, a_n )).
Edges: ( E = F ) (pairs of elements from ( U )).
Weights: Each edge ( e \in E ) has weight ( w(e) = w(X) ) for ( X \in F ).
Compute the Minimum Spanning Tree (MST) of ( G ):
Use a standard MST algorithm, such as Kruskal's Algorithm or Prim's Algorithm.
Extract the Set ( S ):
Let ( S ) be the set of edges in the computed MST.
Verify Conditions:
Condition (a): The MST is connected by definition, satisfying the connectivity requirement.
Condition (b): The MST spans all vertices ( V ), ensuring that all elements in ( U ) are covered.
Condition (c): The MST has minimal total weight among all connected spanning subgraphs, satisfying the minimal weight requirement.
Detailed Explanation of the Algorithm
1. Graph Construction:
We begin by representing the problem using a graph:

Vertices (Elements of ( U )):
Each element ( a_i \in U ) becomes a vertex ( v_i ) in the graph ( G ).
Edges (Pairs in ( F )):
Each pair ( X = {a_i, a_j} \in F ) becomes an edge ( e_{ij} ) connecting vertices ( v_i ) and ( v_j ).
Edge Weights:
The weight assigned to edge ( e_{ij} ) is ( w(e_{ij}) = w(X) ).
Example:

Suppose ( U = {a_1, a_2, a_3} ) and ( F = {{a_1, a_2}, {a_2, a_3}, {a_1, a_3}} ) with weights ( w({a_1, a_2}) = 2 ), ( w({a_2, a_3}) = 3 ), ( w({a_1, a_3}) = 1 ).

The graph ( G ) would have vertices ( v_1, v_2, v_3 ) and edges ( e_{12}, e_{23}, e_{13} ) with corresponding weights.

2. Computing the Minimum Spanning Tree:
We apply a standard MST algorithm:

Kruskal's Algorithm:

Step 1: Sort all edges in ( E ) in non-decreasing order of their weights.
Step 2: Initialize a forest where each vertex is its own tree.
Step 3: For each edge ( e ) in the sorted list:
If ( e ) connects two different trees, add ( e ) to the MST ( T ) and merge the two trees.
If adding ( e ) would create a cycle (i.e., both endpoints are in the same tree), skip ( e ).
Step 4: Continue until all vertices are connected in a single tree.
Prim's Algorithm:

Step 1: Start with an arbitrary vertex ( v ) and consider it the initial MST.
Step 2: While there are vertices not included in the MST:
Select the edge ( e ) with the minimal weight that connects a vertex in the MST to a vertex outside the MST.
Add ( e ) and the new vertex to the MST.
Our Choice: Either algorithm is acceptable; Kruskal's Algorithm is often simpler to implement when the edges can be easily sorted.

Example Using Kruskal's Algorithm:

Step 1: Sort edges: ( e_{13} ) (weight 1), ( e_{12} ) (weight 2), ( e_{23} ) (weight 3).
Step 2: Initialize trees: ( {v_1}, {v_2}, {v_3} ).
Step 3: Add ( e_{13} ) to MST; merge ( {v_1} ) and ( {v_3} ).
Step 4: Add ( e_{12} ) to MST; merge ( {v_1, v_3} ) and ( {v_2} ).
Step 5: All vertices are connected; MST consists of edges ( e_{13}, e_{12} ).
3. Extracting ( S ):
Let ( S ) be the set of edges in the computed MST.

In the example, ( S = {e_{13}, e_{12}} ).

4. Verifying the Conditions:
Condition (b) - Coverage:

The MST spans all vertices ( V ), meaning every vertex ( v_i ) (element ( a_i )) is connected by some edge in ( S ).
Since every element in ( U ) corresponds to a vertex in ( V ), and the MST covers ( V ), the union ( \bigcup_{X \in S} X ) equals ( U ).
Condition (a) - Connectivity:

The MST is connected; there is a path between any two vertices in ( V ) via edges in ( S ).
For any partition ( S = S_1 \sqcup S_2 ):
Removing the edges in ( S_2 ) will disconnect the MST into subtrees.
However, there will always be edges in ( S_1 ) and ( S_2 ) that share a common vertex at the point of disconnection.
This satisfies the requirement that there exists ( Y \in S_1 ) and ( Z \in S_2 ) such that ( Y \cap Z \neq \emptyset ).
Condition (c) - Minimal Weight:

The MST has the minimal total weight among all connected spanning subgraphs of ( G ).
Since any set ( S' ) satisfying conditions (a) and (b) must be connected and cover all vertices, ( S ) minimizes the total weight ( \sum_{X \in S} w(X) ).
Proving Correctness
Lemma 1: The set ( S ) obtained from the MST satisfies conditions (a) and (b).
Proof:

Coverage (Condition (b)):
By definition, an MST spans all vertices in ( V ) (elements of ( U )). Thus, ( \bigcup_{X \in S} X = U ).
Connectivity (Condition (a)):
In an MST, the edges form a tree, which is a connected acyclic graph.
For any partition ( S = S_1 \sqcup S_2 ):
Because the tree is connected, removing the edges in ( S_2 ) will partition the tree into subgraphs.
The edge(s) in ( S_2 ) that connect different parts of the tree must connect to vertices present in the subgraphs formed by ( S_1 ).
Therefore, there exist ( Y \in S_1 ) and ( Z \in S_2 ) such that ( Y ) and ( Z ) share a common vertex (element of ( U )).
Thus, condition (a) is satisfied.
Lemma 2: The set ( S ) obtained from the MST minimizes the total weight among all subsets satisfying conditions (a) and (b).
Proof:

Suppose there exists another subset ( S' \subseteq F ) satisfying conditions (a) and (b) with a total weight less than that of ( S ).
Then ( S' ) must also be connected and span all vertices in ( V ), forming a connected spanning subgraph of ( G ) with a total weight less than that of the MST.
This contradicts the minimality of the MST computed by Kruskal's or Prim's algorithm.
Therefore, no such ( S' ) exists, and ( S ) minimizes the total weight.
Detailed Runtime Analysis
The algorithm consists of two main steps:

Graph Construction:
Time Complexity:
The graph ( G ) has ( n ) vertices (since ( |V| = |U| = n )).
The number of edges ( |E| ) is ( |F| ). Let's denote ( m = |F| ).
Constructing ( G ) involves creating ( n ) vertices and ( m ) edges.
Runtime:
( O(n + m) ). Vertex creation is ( O(n) ), and edge creation is ( O(m) ).
Computing the MST:
Using Kruskal's Algorithm:
Edge Sorting:
Sorting ( m ) edges based on their weights.
Time Complexity: ( O(m \log m) ).
Union-Find Data Structure:
Initializing the Union-Find structure takes ( O(n) ).
Performing up to ( m ) find and union operations, each in near-constant amortized time using path compression and union by rank.
Time Complexity: ( O(m \alpha(n)) ), where ( \alpha(n) ) is the inverse Ackermann function, which grows extremely slowly and can be considered constant for all practical purposes.
Total Time Complexity for Kruskal's Algorithm:
( O(m \log m) ).
Using Prim's Algorithm:
Priority Queue Operations:
Initialize the priority queue with ( n ) vertices.
Each edge insertion and extraction takes ( O(\log n) ) time.
Time Complexity:
Total of ( O(m \log n) ) operations.
Total Time Complexity:
Overall, the dominant term is from the MST computation.
Using Kruskal's Algorithm: ( O(m \log m) ).
Using Prim's Algorithm: ( O(m \log n) ).
Since ( m ) could be up to ( O(n^2) ) (the maximum number of pairs in ( F )), ( \log m ) and ( \log n ) are ( O(\log n) ) and ( O(\log n) ), respectively.
Conclusion:
The algorithm runs in polynomial time.
If ( m ) is ( O(n^2) ), the time complexity becomes ( O(n^2 \log n) ).
Final Algorithm Pseudocode

Algorithm SubsetPairProblem(U, F, w):

Input:
- U: Set of elements {a_1, a_2, ..., a_n}
- F: Family of subsets of U, where each subset has size 2
- w: Weight function w: F -> N

Output:
- S: Subset of F satisfying conditions (a), (b), and (c)

Steps:
1. Construct graph G = (V, E):
   a. V = U
   b. E = F with weights assigned by w

2. Compute the Minimum Spanning Tree (MST) of G:
   a. Use Kruskal's Algorithm:
      i. Sort all edges in E by non-decreasing weight
      ii. Initialize Union-Find data structure for V
      iii. Initialize empty set S
      iv. For each edge e in sorted E:
          - If adding e does not form a cycle:
            - Add e to S
            - Union the sets of its endpoints
          - Else:
            - Discard e
      v. Continue until |S| = n - 1 (since MST has n - 1 edges)

3. Return S
Example Walkthrough
Suppose we have:

( U = {a_1, a_2, a_3, a_4} )
( F = {{a_1, a_2}, {a_2, a_3}, {a_3, a_4}, {a_1, a_4}, {a_1, a_3}} )
Weights:
( w({a_1, a_2}) = 1 )
( w({a_2, a_3}) = 2 )
( w({a_3, a_4}) = 3 )
( w({a_1, a_4}) = 4 )
( w({a_1, a_3}) = 5 )
Algorithm Execution:

Construct G:
Vertices: ( V = {v_1, v_2, v_3, v_4} )
Edges with weights:
( e_{1,2} ) with weight 1
( e_{2,3} ) with weight 2
( e_{3,4} ) with weight 3
( e_{1,4} ) with weight 4
( e_{1,3} ) with weight 5
Compute MST using Kruskal's Algorithm:
Sorted Edges:
( e_{1,2} ) (weight 1)
( e_{2,3} ) (weight 2)
( e_{3,4} ) (weight 3)
( e_{1,4} ) (weight 4)
( e_{1,3} ) (weight 5)
Initialize Union-Find:
Disjoint sets: ( {{v_1}, {v_2}, {v_3}, {v_4}} )
Edge Processing:
Edge ( e_{1,2} ):
Connects ( v_1 ) and ( v_2 ) in different sets.
Add to MST (( S )).
Union sets ( {v_1} ) and ( {v_2} ).
Edge ( e_{2,3} ):
( v_2 ) and ( v_3 ) are in different sets.
Add to MST.
Union sets containing ( v_1, v_2 ) and ( v_3 ).
Edge ( e_{3,4} ):
( v_3 ) and ( v_4 ) are in different sets.
Add to MST.
Union all vertices into a single set.
Edge ( e_{1,4} ):
Both vertices already connected; adding would create a cycle.
Discard.
Edge ( e_{1,3} ):
Both vertices already connected; adding would create a cycle.
Discard.
MST Complete:
Edges in MST: ( S = {e_{1,2}, e_{2,3}, e_{3,4}} )
Verify and Return ( S ):
Total Weight: ( 1 + 2 + 3 = 6 )
Conditions:
Coverage: All vertices ( v_1, v_2, v_3, v_4 ) are connected.
Connectivity: The MST is connected.
Minimum Weight: Since MST is minimal, ( S ) has the minimal total weight among connected spanning subgraphs.
Conclusion
The algorithm effectively reduces the given Subset Pair problem to the well-known Minimum Spanning Tree problem. By constructing a graph where elements of ( U ) are vertices and pairs in ( F ) are edges with weights, we leverage efficient polynomial-time algorithms to find the optimal subset ( S ).

Key Points:

Graph Interpretation: Transforming the problem into graph terms simplifies understanding and solution design.
MST Algorithms: Kruskal's and Prim's algorithms are standard, efficient solutions for finding minimal spanning trees.
Conditions Alignment: The properties of spanning trees inherently satisfy the problem's conditions.
Polynomial Time: The algorithm runs in ( O(m \log n) ) time, which is polynomial with respect to the input size.
Therefore, the designed algorithm correctly and efficiently solves the Subset Pair problem in polynomial time.