Suppose $U_1$ and $U_2$ are subspaces of $V$. Prove that $U_1 \cap U_2$ is a subspace of $V$.

By the definition of subspaces we have:

1. Additive Identity: $\vec{0} \in U.$

2. Closed Under Addition: $u,w \in U \implies u + w \in U.$

3. Closed Under Scalar Multiplication: $a \in \mathbb{F}, u \in U \implies au \in U.$

From the hypothesis we know $\vec{0}$ is an element of $U_1$ and $U_2$ via condition 1, and that $\vec{0}$ is an element of $U_1 \cap U_2$ from the definition of intersection.

Labeling $U_1 \cap U_2$ as $U_3$, we can say $\vec{0} \in U_3$, satisfying condition 1.

In the event that $U_3 = \{\vec{0}\}$, we can consider it a subspace of $V$, since $\{\vec{0}\}$ easily satisfies conditions 1, 2, and 3. If my mathematical parlance is correct, I believe this is the trivial case of $U_1 \cap U_2$ being a subspace of $V$.

If $U_3 \neq \{\vec{0}\}$, however, then we have additional work to do, to show it satisfies conditions 1, 2, and 3, with regard to the other elements it may contain. We've already demonstrated, however, that $U_3 = U_1 \cap U_2$ contains at least $\vec{0}$, so we don't need to prove that it satisfies condition 1, in the event that there are additional elements.

Moving forward, assume two elements $u, w$ that are common to $U_1$, and $U_2$: from the hypothesis, we know $u + w \in U_1$, and $u + w \in U_2$, via condition 2, meaning that $u + w \in U_3 = U_1 \cap U_2$, for any $u, w$, in $U_1$, and $U_2$, so that $U_3$ satisfies condition 2.

Next taking some element $u$, common to $U_1$ and $U_2$, we know that for some $a \in \mathbb{F}$, we have $au \in U_1$, and $au \in U_2$, via condition 3, so that $au \in U_3$, meaning $U_3 = U_1 \cap U_2$ satisfies condition 3.

Also, since our subspaces, and their union satisfy condition 3, if we choose our $a \in F$ to be $-1$, and select some $u$, common to $U_1$ and $U_2$, then $(-1)u = -u \in U_3$, so that $U_3$ has an additive inverse. Since $U_3 = U_1 \cap U_2$ satisfies conditions 1, 2 and 3, of being a subspace, $U_1 \cap U_2$ is a subspace of $V$. Additionally, I decided to investigate the case where there are both elements common to $U_1$, and $U_2$, and elements that aren't, because I wanted to convince myself whether or not the stated intersection would still be a valid subspace.

Assuming $u, w \in U_1$ and $u, v \in U_2$: The fact that $u + w, u + v \notin U_3 =U_1 \cap U_2$, doesn't violate the conditions for being a subspace, since the definition of closure under addition requires $v, w \in U_3$, which would contradict the definition of intersection, and $U_3$ is an intersection. The fact that $av, aw \notin U_3$, for some $a \in \mathbb{F}$ doesn't violate the conditions because the definition of closure under scalar multiplication again requires $v, w \in U_3$, which again contradicts the definition of intersection. This also means that the lack of an additive inverse for $v, w$ in $U_3$ doesn't violate the conditions for being a subspace, because the additive inverse of $v$, $-v = (-1)v$, which is just scalar multiplication, and even if we wanted to use addition, say $v + (-2v)$, we'd still need $v \in U_3$, which we already ruled out.