--- Commentary: $\textbf{Exercise 13.}$ Prove that the union of three subspaces of $V$ is a subspace of $V$ if and only if one of the subspaces contains the other two. $\textbf{Solution 13.}$ One direction of this exercise is trivial: if we have three subspaces of $V$, one of which contains the other two, then the union of these three subspaces equals the larger of them, which is a subspace of $V$. To prove the other direction, suppose $V_1, V_2, V_3$ are subspaces of $V$ such that $V_1 \cup V_2 \cup V_3$ is a subspace of $V$. We want to prove that one of these three subspaces contains the other two. First consider the case $V_1 \subseteq V_2 \cup V_3$. Then $V_2 \cup V_3$ equals $V_1 \cup V_2 \cup V_3$, which is a subspace of $V$. Exercise 12 now implies that $V_2 \subseteq V_3$ (and thus also $V_1 \subseteq V_3$) or $V_3 \subseteq V_2$ (and thus also $V_1 \subseteq V_2$). Either way, we have our desired conclusion that one of the subspaces $V_1, V_2, V_3$ contains the other two. Now consider the case $V_1 \nsubseteq V_2 \cup V_3$. Let $v \in V_1$ be such that $v \notin V_2 \cup V_3$. Suppose $V_2 \nsubseteq V_1$. Let $w \in V_2$ be such that $w \notin V_1$. For each $\lambda \in \mathbb{F}$, the vector $\lambda v + w$ is not in $V_1$ (because otherwise we would have $w \in V_1$). However, for each $\lambda \in \mathbb{F}$, the vector $\lambda v + w$ is in the subspace $V_1 \cup V_2 \cup V_3$ and thus is in $V_2 \cup V_3$. Thinking about three distinct values of $\lambda$, for each of which $\lambda v + w$ is in $V_2 \cup V_3$, we see that there are two distinct numbers $\lambda_1, \lambda_2 \in \mathbb{F}$ such that $\lambda_1v + w \in V_2 \text{ and } \lambda_2v + w \in V_2$ or $\lambda_1v + w \in V_3 \text{ and } \lambda_2v + w \in V_3.$ Subtracting the two vectors in the first case above, we have $(\lambda_1 - \lambda_2)v \in V_2$, which implies that $v \in V_2$, which is a contradiction. Subtracting the two vectors in the second case above, we have $(\lambda_1 - \lambda_2)v \in V_3$, which implies that $v \in V_3$, which is a contradiction. Either way, we have a contradiction to our assumption that $V_2 \nsubseteq V_1$. Thus $V_2 \subseteq V_1$. Similarly, $V_3 \subseteq V_1$. Thus we have shown that one of the subspaces contains the other two, as desired. $\textit{Commentary:}$ This exercise extends the result of Exercise 12 to the union of three subspaces. The proof is more involved than that of Exercise 12 and requires considering several cases. The key idea is to first handle the case where one subspace is contained in the union of the other two, reducing the problem to Exercise 12. In the other case, we choose a vector $v$ from the first subspace that is not in the other two, and a vector $w$ from the second subspace that is not in the first. Then, we consider linear combinations $\lambda v + w$ for different values of $\lambda$. Due to the subspace properties, certain pairs of these combinations must be in the same subspace, leading to a contradiction. This result further underscores the point that unions of subspaces are generally not subspaces. It's a strong property for a union of subspaces to be a subspace, and in the case of three subspaces, it's equivalent to one subspace containing the other two. $\textit{Example:}$ In $\mathbb{R}^3$, let $V_1$ be the $xy$-plane, $V_2$ be the $yz$-plane, and $V_3$ be the $z$-axis. Then $V_1 \cup V_2 \cup V_3$ is not a subspace of $\mathbb{R}^3$, because no subspace contains the other two. In the space of functions from $\mathbb{Z}_p$ to $\mathbb{Q}_p$, let $V_1$ be the subspace of constant functions, $V_2$ be the subspace of functions that are 0 at 0, and $V_3$ be the subspace of functions that are 0 at 1. Then $V_1 \cup V_2 \cup V_3$ is not a subspace, because no subspace contains the other two. In the space of polynomials over $\mathbb{F}_2$, let $V_1$ be the subspace of polynomials of degree at most 1, $V_2$ be the subspace of polynomials with constant term 0, and $V_3$ = $V_1$. Then $V_1 \cup V_2 \cup V_3$ is a subspace because $V_1 = V_3$ contains $V_2$. These examples demonstrate that for the union of three subspaces to be a subspace is a stringent condition that is rarely met in practice, unless one subspace trivially contains the others. The result also illustrates the interplay between the algebraic properties of the field and the geometric properties of the vector space.