Let \(W_1\) and \(W_2\) be subspaces of a vector space \(V\) such that the set-theoretic union of \(W_1\) and \(W_2\) is also a subspace. Prove that one of the spaces \(W_i\) is contained in the other.
Solution
1. Suppose for contradiction that neither \(W_1 \subseteq W_2\) nor \(W_2 \subseteq W_1\).
assume the conclusion fails.
2. Then there exist \(w_1 \in W_1 \setminus W_2\) and \(w_2 \in W_2 \setminus W_1\).
1.
3. Since \(W_1 \cup W_2\) is a subspace, \(w_1 + w_2 \in W_1 \cup W_2\).
closure under addition.
4. If \(w_1 + w_2 \in W_1\), then \(w_2 = (w_1 + w_2) - w_1 \in W_1\), contradicting \(w_2 \notin W_1\).
3 and closure of \(W_1\) under subtraction.
5. If \(w_1 + w_2 \in W_2\), then \(w_1 = (w_1 + w_2) - w_2 \in W_2\), contradicting \(w_1 \notin W_2\).
3 and closure of \(W_2\) under subtraction.
6. Both cases contradict, so one of \(W_1, W_2\) is contained in the other.
4 and 5.
7. Q.E.D.
6.