Let's analyze the statement step by step: **Statement:** For all sub-vector spaces \( U_1, U_2 \) of a vector space \( V \), if \( V = U_1 \cup U_2 \) then \( V = U_1 \) or \( V = U_2 \). --- ### **Step 1: Understanding the Union of Subspaces** The union \( U_1 \cup U_2 \) of two subspaces is generally **not** a subspace unless one is contained in the other. The union of two subspaces is a subspace only if one is contained in the other. **Explanation:** If \( U_1 \nsubseteq U_2 \) and \( U_2 \nsubseteq U_1 \), then there exist vectors \( u_1 \in U_1 \setminus U_2 \) and \( u_2 \in U_2 \setminus U_1 \). Their sum \( u_1 + u_2 \) may not belong to \( U_1 \cup U_2 \), violating vector space closure. --- ### **Step 2: Applying the Statement to the Whole Space** If \( V = U_1 \cup U_2 \), then for every \( v \in V \), \( v \in U_1 \) or \( v \in U_2 \). But if neither \( U_1 \subseteq U_2 \) nor \( U_2 \subseteq U_1 \), consider \( v_1 \in U_1 \setminus U_2 \) and \( v_2 \in U_2 \setminus U_1 \). Their sum \( v_1 + v_2 \) must still be in \( V \), so it must be in \( U_1 \) or \( U_2 \). However, this leads to a contradiction unless one subspace contains the other. **Explanation:** If \( v_1 + v_2 \in U_1 \), then \( (v_1 + v_2) - v_1 = v_2 \in U_1 \), contradicting the assumption \( v_2 \notin U_1 \). A similar argument applies for \( U_2 \). Thus, one must be contained in the other. --- ### **Step 3: Conclude with the Logical Implication** Therefore, if \( V = U_1 \cup U_2 \), then either \( U_1 = V \) or \( U_2 = V \). **Explanation:** This is because if neither subspace equals \( V \), then their union cannot be the whole space unless one is contained in the other, in which case the larger one must be \( V \). --- ### **Final Answer:** **True.** If \( V = U_1 \cup U_2 \), then \( V = U_1 \) or \( V = U_2 \).