Prove or disprove union of two subspaces of a vector space V is a subspace of V .

To determine whether the union of two subspaces of a vector space \( V \) is itself a subspace of \( V \), we need to consider the properties that define a subspace.

### Subspace Properties Recap
A subset \( W \) of a vector space \( V \) is a subspace if:
1. **The zero vector of \( V \) is in \( W \).**
2. **\( W \) is closed under vector addition.**
3. **\( W \) is closed under scalar multiplication.**

### Given: \( W_1 \) and \( W_2 \) are subspaces of \( V \).

We need to check if \( W_1 \cup W_2 \) (the union of \( W_1 \) and \( W_2 \)) is a subspace of \( V \).

### 1. Zero Vector
Since \( W_1 \) and \( W_2 \) are subspaces, both contain the zero vector \( \mathbf{0} \). Thus, \( \mathbf{0} \in W_1 \cup W_2 \), so the union contains the zero vector.

### 2. Closure Under Addition
To be a subspace, \( W_1 \cup W_2 \) must be closed under addition. Consider two vectors \( \mathbf{u} \) and \( \mathbf{v} \) that belong to \( W_1 \cup W_2 \).

There are three possible cases:
1. **Case 1:** \( \mathbf{u}, \mathbf{v} \in W_1 \).
   - Then \( \mathbf{u} + \mathbf{v} \in W_1 \), so \( \mathbf{u} + \mathbf{v} \in W_1 \cup W_2 \).
2. **Case 2:** \( \mathbf{u}, \mathbf{v} \in W_2 \).
   - Then \( \mathbf{u} + \mathbf{v} \in W_2 \), so \( \mathbf{u} + \mathbf{v} \in W_1 \cup W_2 \).
3. **Case 3:** \( \mathbf{u} \in W_1 \) and \( \mathbf{v} \in W_2 \) (or vice versa).

   Here, \( \mathbf{u} + \mathbf{v} \) might not necessarily belong to \( W_1 \) or \( W_2 \). If \( W_1 \) and \( W_2 \) are not the same subspace, \( \mathbf{u} + \mathbf{v} \) may not belong to either \( W_1 \) or \( W_2 \), and therefore might not belong to \( W_1 \cup W_2 \).

Thus, \( W_1 \cup W_2 \) is **not necessarily closed under addition** unless \( W_1 \subseteq W_2 \) or \( W_2 \subseteq W_1 \).

### 3. Closure Under Scalar Multiplication
Suppose \( \mathbf{u} \in W_1 \cup W_2 \). There are two cases:
1. **Case 1:** \( \mathbf{u} \in W_1 \).
   - For any scalar \( c \in \mathbb{R} \), \( c\mathbf{u} \in W_1 \), so \( c\mathbf{u} \in W_1 \cup W_2 \).
2. **Case 2:** \( \mathbf{u} \in W_2 \).
   - For any scalar \( c \in \mathbb{R} \), \( c\mathbf{u} \in W_2 \), so \( c\mathbf{u} \in W_1 \cup W_2 \).

Thus, \( W_1 \cup W_2 \) is closed under scalar multiplication.

### Conclusion
While \( W_1 \cup W_2 \) contains the zero vector and is closed under scalar multiplication, it is **not necessarily closed under vector addition** unless \( W_1 \) and \( W_2 \) are nested within each other (i.e., one is a subset of the other).

Therefore, **the union of two subspaces of \( V \) is not necessarily a subspace of \( V \)**.

Would you like more details or have any questions? Here are some related questions for further exploration:

1. Can you give an example of two subspaces whose union is not a subspace?
2. What is the intersection of two subspaces, and is it always a subspace?
3. How do the concepts of subspaces relate to linear independence?
4. Can the union of two subspaces ever be a subspace? Under what conditions?
5. What is the sum of two subspaces, and how does it differ from the union?
6. How can the dimension of the sum of two subspaces be calculated?
7. What are direct sums of subspaces, and how do they relate to vector spaces?
8. How does the union of subspaces differ in infinite-dimensional vector spaces?

**Tip:** Remember, the union of two sets is not necessarily closed under addition, which is why the union of two subspaces is not generally a subspace. Instead, consider the sum of subspaces, which is always a subspace.

Explore whether the union of two subspaces of a vector space V forms a subspace. Understand the conditions under which the union satisfies the subspace criteria, including closure under addition and scalar multiplication.

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