Let $U$ and $V$ be subspaces of the vector space $\R^n$.
If neither $U$ nor $V$ is a subset of the other, then prove that the union $U \cup V$ is not a subspace of $\R^n$.

Since $U$ is not contained in $V$, there exists a vector $\mathbf{u}\in U$ but $\mathbf{u} \not \in V$.
Similarly, since $V$ is not contained in $U$, there exists a vector $\mathbf{v} \in V$ but $\mathbf{v} \not \in U$.

Seeking a contradiction, let us assume that the union is $U \cup V$ is a subspace of $\R^n$.
The vectors $\mathbf{u}, \mathbf{v}$ lie in the vector space $U \cup V$.
Thus their sum $\mathbf{u}+\mathbf{v}$ is also in $U\cup V$.
This implies that we have either
\[\mathbf{u}+\mathbf{v} \in U \text{ or } \mathbf{u}+\mathbf{v}\in V.\]

If $\mathbf{u}+\mathbf{v} \in U$, then there exists $\mathbf{u}’\in U$ such that
\[\mathbf{u}+\mathbf{v}=\mathbf{u}’.\]
Since the vectors $\mathbf{u}$ and $\mathbf{u}’$ are both in the subspace $U$, their difference $\mathbf{u}’-\mathbf{u}$ is also in $U$. Hence we have
\[\mathbf{v}=\mathbf{u}’-\mathbf{u} \in U.\]

However, this contradicts the choice of the vector $\mathbf{v} \not \in U$.

Thus, we must have $\mathbf{u}+\mathbf{v}\in V$.
In this case, there exists $\mathbf{v}’ \in V$ such that
\[\mathbf{u}+\mathbf{v}=\mathbf{v}’.\]

Since both $\mathbf{v}, \mathbf{v}’$ are vectors of $V$, it follows that
\[\mathbf{u}=\mathbf{v}’-\mathbf{v}\in V,\]
which contradicts the choice of $\mathbf{u} \not\in V$.

Therefore, we have reached a contradiction. Thus, the union $U \cup V$ cannot be a subspace of $\R^n$.

Related Question.

In fact, the converse of this problem is true.

Problem. Let $W_1, W_2$ be subspaces of a vector space $V$. Then prove that $W_1 \cup W_2$ is a subspace of $V$ if and only if $W_1 \subset W_2$ or $W_2 \subset W_1$.

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