Prove that any subspace of $V$ that contains both $W_{1}$ and $W_{2}$ must also contain $W_{1}+W_{2}$.

Question

Let $W_{1}$ and $W_{2}$ be subspaces of a vector space $V$.

(a) Prove that $W_{1}+W_{2}$ is a subspace of $V$ that contains both $W_{1}$ and $W_{2}$.

(b) Prove that any subspace of $V$ that contains both $W_{1}$ and $W_{2}$ must also contain $W_{1}+W_{2}$.

Here it is what I've tried. (EDIT)

(a) To start with, $0\in W_{1}+W_{2}$, because $0 = 0 + 0$ and $0\in W_{1}$ as well as $0\in W_{2}$.

If $w\in W_{1} + W_{2}$, then $w = w_{1} + w_{2}$ where $w_{1}\in W_{1}$ and $w_{2}\in W_{2}$.

Consequently, $aw = aw_{1} + aw_{2}$ where $aw_{1}\in w_{1}$ and $aw_{2}\in W_{2}$. Hence $aw\in W_{1} + W_{2}$.

Finally, if $w\in W_{1} + W_{2}$ and $z\in W_{1} + W_{2}$, then $w = w_{1} + w_{2}$ and $z = z_{1} + z_{2}$, where $w_{1},z_{1}\in W_{1}$ and $w_{2},z_{2}\in W_{2}$.

Hence $w + z = (w_{1} + z_{1}) + (w_{2} + z_{2}) \in W_{1} + W_{2}$, and we are done.

Now it remains to prove that $W_{1}+W_{2}\supseteq W_{1}$ and $W_{1}+W_{2}\supseteq W_{2}$.

Indeed, if $w_{1}\in W_{1}$, then $w_{1} + 0 \in W_{1}+W_{2}$. Similarly, if $w_{2}\in W_{2}$, then $0 + w_{2}\in W_{1} + W_{2}$.

(b) Let us suppose that $W\supseteq W_{1}$ and $W\supseteq W_{2}$.

If $w = w_{1} + w_{2}\in W_{1} + W_{2}$, where $w_{1}\in W_{1}\subseteq W$ and $w_{2}\in W_{2}\subseteq W$, then $w\in W$ because $W$ is a linear subspace.

Any comments on my solution are appreciated.

Answer 1

Your solution is almost correct, except it seems you've only shown that $W_1+W_2$ is indeed a subspace in part a), when it also asks you to show it contains $W_1$ and $W_2$. This is straightforward to show and I'll leave that for you to try.

Answer 2

For part (b), it might be better to start with:

Let us suppose that $W \subseteq V$, and $W$ is a subspace of $V$, where $W\supseteq W_{1}$ and $W\supseteq W_{2}$.

(explicitly stating that $W$ is a subspace of $V$), and also it may be better to split the definition of what $w$ is, from the deduction, trivial though it may be, that $w_1$ and $w_2$ are in $W$, like so:

If $w = w_{1} + w_{2}\in W_{1} + W_{2}$, where $w_{1}\in W_{1}$ and $w_{2}\in W_{2}$, then $w_{1}\in W$ and $w_{2}\in W$ by definition of $W$, so because $W$ is a linear subspace of $V$, we have $w_{1} + w_{2}\in W$, and so $w\in W$. Hence $W$ contains $W_{1} + W_{2}$.