91Ó°ÊÓ

A subset \(S\) of a vector space \(V\) is a subspace if it satisfies the following three conditions: 1. The zero vector of \(V\) is in \(S\). 2. If \(u\) and \(v\) are elements of \(S\), then the sum \(u+v\) is an element of \(S\). 3. If \(u\) is an element of \(S\) and \(c\) is a scalar, then the product \(cu\) is an element of \(S\). Since we are given that \(U \cup W\) is a subspace, we will assume it satisfies these three conditions.

To show that either \(U \subseteq W\) or \(W \subseteq U\), we will use contradiction. We will assume that neither \(U \subseteq W\) nor \(W \subseteq U\), and then we will show that under this assumption, \(U \cup W\) cannot be a subspace, contradicting the given condition. Assume neither \(U \subseteq W\) nor \(W \subseteq U\). Then there exist vectors \(u \in U\) and \(w \in W\) such that \(u \notin W\) and \(w \notin U\). Consider the sum of these two vectors: \(u + w\). Since \(U\) is a subspace and \(u \in U\), the vector \(u\) forms a subspace by itself, and \(U = \{u\}\) must be a subspace. Now, since we are assuming that \(U \cup W\) is a subspace, it must be the case that \(u + w \in U \cup W\), satisfying Condition 2 for subspaces above. There are two possibilities for where \(u + w\) is in the union \(U \cup W\): 1. \(u + w \in U\). 2. \(u + w \in W\). We will show that both of these possibilities lead to contradictions:

If \(u + w \in U\), then by Condition 3 for subspaces (\(cu \in U\) for any scalar \(c\)), we have that \((u + w) + (-u) \in U\). This simplifies to \(w \in U\), which is a contradiction, since we assumed \(w \notin U\).

If \(u + w \in W\), then by Condition 3 for subspaces (\(cw \in W\) for any scalar \(c\)), we have that \((u + w) + (-w) \in W\). This simplifies to \(u \in W\), which is a contradiction, since we assumed \(u \notin W\). Since both possibilities lead to contradictions, our original assumption that neither \(U \subseteq W\) nor \(W \subseteq U\) must be false. Therefore, either \(U \subseteq W\) or \(W \subseteq U\).