- 11. Prove that the intersection of every collection of subspaces of 𝑉 is a subspace of 𝑉. --- SOLUTION Suppose ${\huge \{𝑉_\alpha\} _{\alpha\in\Gamma}}$ is a collection of subspaces of 𝑉; here $\Gamma$ is an arbitrary index set. We need to prove that ${\huge\cap_{\alpha \;\in \;\Gamma}} \; \in V_{\alpha}$ ,which equals the set of vectors that are in $𝑉_{\alpha}$ for every 𝛼 ∈ Γ , is a subspace of 𝑉. The additive identity 0 is in $𝑉_{\alpha}$ for every 𝛼 ∈ Γ (because each $𝑉_{\alpha}$ is a subspace of 𝑉). Thus 0 ∈ ${\huge\cap_{\alpha \;\in \;\Gamma}} \; \in V_{\alpha}$ In particular, $\cap_{\alpha \;\in \;\Gamma} \; \in V_{\alpha}$ is a nonempty subset of 𝑉. Suppose 𝑢, 𝑣 ∈ $\cap_{\alpha \;\in \;\Gamma} \; \in V_{\alpha}$ Then 𝑢, 𝑣 $\in V_{\alpha}$ for every 𝛼 ∈ Γ . Thus 𝑢 + 𝑣 ∈ 𝑉܄ for every 𝛼 ∈ Γ (because each 𝑉܄ is a subspace of 𝑉). Thus 𝑢 + 𝑣 ∈ $\cap_{\alpha \;\in \;\Gamma} \; \in V_{\alpha}$ Thus ${\huge\cap_{\alpha \;\in \;\Gamma}} \; \in V_{\alpha}$ is closed under addition. Suppose 𝑢 ∈ ${\huge\cap_{\alpha \;\in \;\Gamma}} \; \in V_{\alpha}$ and 𝑎 ∈ 𝐅. Then 𝑢 ∈ $𝑉_{\alpha}$ for every 𝛼 ∈ Γ . Thus 𝑎𝑢 ∈ $𝑉_{\alpha}$ for every 𝛼 ∈ Γ (because each 𝑉܄ is a subspace of 𝑉). Thus 𝑎𝑢 ∈ $\cap_{\alpha \;\in \;\Gamma} \; \in V_{\alpha}$ Thus ${\huge\cap_{\alpha \;\in \;\Gamma}} \; \in V_{\alpha}$ is closed under scalar multiplication. Because $\cap_{\alpha \;\in \;\Gamma} \; \in V_{\alpha}$ is a nonempty subset of 𝑉 that is closed under addition and scalar multiplication, ${\huge\cap_{\alpha \;\in \;\Gamma}} \; \in V_{\alpha}$ is a subspace of 𝑉. $\color{red}comment$ For many students, the hardest part of this exercise is understanding the meaning of an arbitrary intersection of sets. Instructors who do not want to deal with this issue should change the exercise to “Prove that the intersection of every finite collection of subspaces of 𝑉 is a subspace of 𝑉.” Many students will then prove that the intersection of two subspaces of 𝑉 is a subspace of 𝑉 and use induction to get the result for finite collections of subspaces. --- Commentary: $\textbf{Exercise 11.}$ Prove that the intersection of every collection of subspaces of $V$ is a subspace of $V$. $\textbf{Solution 11.}$ Suppose ${V_\alpha}_{\alpha \in A}$ is a collection of subspaces of $V$; here $A$ is an arbitrary index set. We need to prove that $\bigcap_{\alpha \in A} V_\alpha$, which equals the set of vectors that are in $V_\alpha$ for every $\alpha \in A$, is a subspace of $V$. The additive identity $0$ is in $V_\alpha$ for every $\alpha \in A$ (because each $V_\alpha$ is a subspace of $V$). Thus $0 \in \bigcap_{\alpha \in A} V_\alpha$. In particular, $\bigcap_{\alpha \in A} V_\alpha$ is a nonempty subset of $V$. Suppose $u, v \in \bigcap_{\alpha \in A} V_\alpha$. Then $u, v \in V_\alpha$ for every $\alpha \in A$. Thus $u + v \in V_\alpha$ for every $\alpha \in A$ (because each $V_\alpha$ is a subspace of $V$). Thus $u + v \in \bigcap_{\alpha \in A} V_\alpha$. Hence $\bigcap_{\alpha \in A} V_\alpha$ is closed under addition. Suppose $u \in \bigcap_{\alpha \in A} V_\alpha$ and $a \in \mathbb{F}$. Then $u \in V_\alpha$ for every $\alpha \in A$. Thus $au \in V_\alpha$ for every $\alpha \in A$ (because each $V_\alpha$ is a subspace of $V$). Thus $au \in \bigcap_{\alpha \in A} V_\alpha$. Hence $\bigcap_{\alpha \in A} V_\alpha$ is closed under scalar multiplication. Because $\bigcap_{\alpha \in A} V_\alpha$ is a nonempty subset of $V$ that is closed under addition and scalar multiplication, $\bigcap_{\alpha \in A} V_\alpha$ is a subspace of $V$. $\textit{Commentary:}$ This exercise extends the result of Exercise 10 from the intersection of two subspaces to the intersection of any collection of subspaces. The proof follows the same structure as in Exercise 10, but uses the concept of an arbitrary index set to handle the potentially infinite collection of subspaces. The key idea is that a vector is in the intersection of all the subspaces if and only if it's in each individual subspace. This allows us to reduce the problem of proving that the intersection is a subspace to proving that each individual subspace satisfies the necessary conditions. This result is a powerful tool in linear algebra, as it allows us to construct new subspaces from existing ones. For instance, it's used in the study of invariant subspaces (subspaces that are preserved under a linear transformation), where the invariant subspace is the intersection of all subspaces that are preserved by the transformation. $\textit{Example:}$ In the space of sequences $\mathbb{Q}^{\infty}$, for each positive integer $n$, let $V_n$ be the subspace of sequences whose $n$-th term is 0. Then $\bigcap_{n=1}^{\infty} V_n$ is the subspace of sequences that are eventually 0. In the space of functions from $\mathbb{R}$ to $\mathbb{R}$, for each $a \in \mathbb{R}$, let $V_a$ be the subspace of functions that are differentiable at $a$. Then $\bigcap_{a \in \mathbb{R}} V_a$ is the subspace of everywhere differentiable functions. In any vector space $V$ over a field $\mathbb{F}$, the intersection of all subspaces of $V$ is the trivial subspace ${0}$, and the intersection of no subspaces of $V$ is $V$ itself (by convention). These examples demonstrate how the intersection of an arbitrary collection of subspaces can be used to define subspaces with specific properties. The key is to define a family of subspaces, each of which encodes a particular property, and then take their intersection to obtain the subspace of vectors satisfying all of those properties simultaneously.