1. The goal is to provide a formal mathematical definition for a subspace.
2. Provide a rigorous proof showing that if we take any collection of subspaces and look at the elements they all have in common (their intersection) , that resulting set still satisfies all the requirements of being a vector space.

Give a General Strategy for a Solution:
1. We list the three specific conditions a subset must satisfy to be considered a subspace (the "Subspace Test").
2. We will let `bbb"W"` be the intersection of an arbitrary collection of subspaces `{bbb"U"_i}_{i in I}` . Then we step through the three conditions to show that `bbb"W"` satisfies each one , relying on the fact that each individual `bbb"U"_i` is already known to be a subspace.

Write down the specific Theory used : A subspace `bbb"U" subseteq bbb"V"` is a subspace of `bbb"V"` if `bbb"U"` is itself a vector space under addition and scalar multiplication defined on `bbb"V"`.
The Subspace Criterion (The Subspace Test): A non-empty subset `bbb"U"` of a vector space `bbb"V"` is a subspace if and only if:
- Zero Vector: The zero vector of `bbb"V"` is in `bbb"U" \ (bb{0} in bbb"U")`.
- Closure under Addition: For every `bb{u} , bb{v} in bbb"U"` , the sum `bb{u} + bb{v} in bbb"U"`.
- Closure under Scalar Multiplication: For every `bb{u} in bbb"U"` and every `c in bbb"F"` , the product `c bb{u} in bbb"U"`.

Do the Exercise:
Definition: Let `bbb"V"` be a vector space over a field `bbb"F"`. A subset `bbb"U" subseteq bbb"V"` is called a subspace of `bbb"V"` if `bbb"U"` is non-empty and is closed under the operations of vector addition and scalar multiplication. More formally , it must satisfy:
- `bb{0 in bbb"U"}`
- `forall bb{u} , bb{v} in bbb"U" : bb{u} + bb{v} in bbb"U"`
- `forall c in bbb"F" , forall bb{u} in bbb"U" : c bb{u} in bbb"U"`

The Proof: Let `{bbb"U"_i}_{i in I}` be a collection of subspaces of `bbb"V"`. Let `bbb"W"` be their intersection: `bbb"W" = bigcap_{i in I} bbb"U"_i`. We must show that `bbb"W"` satisfies the three subspace criteria.

1. Zero Vector: Since each `bbb"U"_i` is a subspace , we know that `bb{0} in bbb"U"_i` for every `i in I`. By the definition of an intersection , if an element is in every set of the collection , it is in the intersection. Therefore , `bb{0} in bbb"W"`.

2. Closure under Addition: Let `bb{u} , bb{v} in bbb"W"`. By the definition of intersection it means `bb{u} in bbb"U"_i` and `bb{v} in bbb"U"_i` for every `i in I`. Since each `bbb"U"_i` is a subspace , it is closed under addition. Thus `(bb{u} + bb{v}) in bbb"U"_i` for every `i in I`. Because the sum is in every `bbb"U"_i` , it must be in their intersection: `(bb{u} + bb{v}) in bbb"W"`.

3. Closure under Scalar Multiplication: Let `bb{u} in bbb"W"` and `c in bbb"F"`. Since `bb{u} in bbb"W"` , then `bb{u} in bbb"U"_i` for every `i in I`. Since each `bbb"U"_i` is a subspace , it is closed under scalar multiplication. Thus , `c bb{u} in bbb"U"_i` for every `i in I`. Because the product is in every `bbb"U"_i` , it must be in their intersection: `c bb{u} in bbb"W"`.

Final Answer: Since `bbb"W"` contains the zero vector and is closed under addition and scalar multiplication , `bbb"W"` is a subspace of `bbb"V"`.
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