Prove that $Span(A\cup B)\subseteq Span(A)+Span(B)$

Question

Let $V$ be a vector space and $A$ and $B$ are two subsets of $V$. I have to prove that:

$Span(A\cup B)\subseteq Span(A)+Span(B)$

My attempt:

Let $z\in Span(A\cup B)$.

Therefore, $z=\sum_{i=1}^r\lambda_i u_i$, where $u_i\in A\cup B \forall i \implies u_i\in A$, or $u_i\in B$.

Let $u_i\in A$. Then $z=\sum_{i=1}^r\lambda_i u_i\in Span(A)$.

Since $Span(B)$ is a subspace, $0\in Span(B)$.

So we see $z\in Span(A), 0\in Span(B)$

Hence $z+0\in Span(A)+Span(B)$

$\implies z\in Span(A)+Span(B)$.

Similarly, for $u_i\in B$, we can show $z\in Span(A)+Span(B)$.

Since $z\in Span(A\cup B)\implies z\in Span(A)+Span(B)$

$Span(A\cup B)\subseteq Span(A)+Span(B)$

I know there are other easier methods to prove the above. I only want to know if there is any mistake in my attempt or whether the approach is okay. Thanks in advance.

Answer 1

What you wrote is false.

Let $u_i\in A$. Then $z=\sum_{i=1}^r\lambda_i u_i\in Span(A)$.

is not correct as each $u_i$ is in $A$ or $B$, but you can’t say that because one of them is in $A$ it is the case for all of them.

What you can write is

$$\sum_{i=1}^r \lambda_i u_i= \sum_{\substack{1\le i \le n\\u_i \in A}}\lambda_i u_i+\sum_{\substack{1\le i \le n\\ u_i \notin A}} \lambda_i u_i$$ and the RHS is indeed in $Span(A) + Span(B)$ as if $u_i \notin A$, $u_i \in B$.