To prove that $\text{Span}(S_1 \cup S_2) = \text{Span}(S_1) + \text{Span}(S_2)$, where $S_1$ and $S_2$ are arbitrary subsets of a vector space $V$, we proceed in two steps: showing both sides are contained within each other.

1. $\text{Span}(S_1 \cup S_2) \subseteq \text{Span}(S_1) + \text{Span}(S_2)$

Definition of Span

• The span of a set $S$, denoted as $\text{Span}(S)$, is the set of all linear combinations of elements in $S$.

Let $v \in \text{Span}(S_1 \cup S_2)$

• This means that $v$ can be expressed as a linear combination of vectors from $S_1 \cup S_2$, i.e., $v = a_1 v_1 + a_2 v_2 + \dots + a_n v_n$ where $v_i \in S_1 \cup S_2$ and $a_i \in \mathbb{F}$ (the field over which $V$ is defined).

• Since each $v_i \in S_1 \cup S_2$, for each $v_i$, either $v_i \in S_1$ or $v_i \in S_2$.

• We can break the linear combination into two parts: one for vectors in $S_1$ and one for vectors in $S_2$. Thus, we can write: $v = (a_1 v_1 + \dots + a_k v_k) + (a_{k+1} v_{k+1} + \dots + a_n v_n)$ where $v_1, \dots, v_k \in S_1$ and $v_{k+1}, \dots, v_n \in S_2$.

• The first part $a_1 v_1 + \dots + a_k v_k$ is a linear combination of vectors in $S_1$, so it is in $\text{Span}(S_1)$. Similarly, the second part is a linear combination of vectors in $S_2$, so it is in $\text{Span}(S_2)$.

• Hence, $v \in \text{Span}(S_1) + \text{Span}(S_2)$.

Thus, every vector in $\text{Span}(S_1 \cup S_2)$ is in $\text{Span}(S_1) + \text{Span}(S_2)$, which proves: $\text{Span}(S_1 \cup S_2) \subseteq \text{Span}(S_1) + \text{Span}(S_2).$

2. $\text{Span}(S_1) + \text{Span}(S_2) \subseteq \text{Span}(S_1 \cup S_2)$

Let $v \in \text{Span}(S_1) + \text{Span}(S_2)$

• By definition, $v$ can be written as $v = v_1 + v_2$, where $v_1 \in \text{Span}(S_1)$ and $v_2 \in \text{Span}(S_2)$.

• Since $v_1 \in \text{Span}(S_1)$, it is a linear combination of vectors in $S_1$. Similarly, $v_2 \in \text{Span}(S_2)$ is a linear combination of vectors in $S_2$.

• Therefore, $v_1 + v_2$ is a linear combination of vectors from $S_1 \cup S_2$, which means that $v \in \text{Span}(S_1 \cup S_2)$.

Thus, we have: $\text{Span}(S_1) + \text{Span}(S_2) \subseteq \text{Span}(S_1 \cup S_2).$

Conclusion:

Since we have shown both inclusions: $\text{Span}(S_1 \cup S_2) \subseteq \text{Span}(S_1) + \text{Span}(S_2)$ and $\text{Span}(S_1) + \text{Span}(S_2) \subseteq \text{Span}(S_1 \cup S_2),$ we conclude that: $\text{Span}(S_1 \cup S_2) = \text{Span}(S_1) + \text{Span}(S_2).$

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Would you like more details or have any questions?

Here are five related questions:

1. What is the definition of the span of a set in a vector space?
2. How does the span of the union of two sets relate to the sum of their individual spans?
3. Can you provide an example illustrating the sum of spans in a vector space?
4. What is the geometric interpretation of the span of vectors in $\mathbb{R}^3$?
5. What happens if $S_1$ and $S_2$ have an intersection in terms of their span?

Tip: When working with spans and linear combinations, always remember that the span of a set includes all possible linear combinations of its elements, not just the elements themselves.