## 证明V的两个子空间的并是V的子空间当且仅当其中一个子空间包含另一个子空间

Question:

Prove that the union of two subspaces of $$V$$ is a subspace of $$V$$ if and only if one of the subspaces of $$V$$ is contained in the other.

Solution:

Assume two set A, B are subspaces of $$V$$.

Part 1:

Assume $$A \cup B = A$$ or $$B$$.

Clearly $$A \cup B = A$$ or $$B \in V$$.

Therefor if one subspace of $$V$$ is contained in the other, the union of two subspaces is a subspace of $$V$$.

Part 2:

Assume $$A \cup B \neq A$$ or $$B$$.

Now we take $$a \in A$$ but not in $$B$$, and $$b \in B$$ but not in $$A$$.

Since $$a \in A$$, we know $$-a \in A$$.

Assume $$a + b \in A$$.

As we know, $$A$$ is a subspace of $$V$$. So $$A$$ is closed under addition.

Therefor, $$(a + b) + (-a)$$ should be in A.

But in fact, $$(a + b) + (-a) = b$$. We have assumed that $$b \notin A$$. We have reached a contradiction on the assumption that $$a + b \in A$$. Thus $$a + b \notin A$$.

If we assume $$a + b \in B$$, we will also reach a contradiction like this.

Therefor, $$a + b$$ is not in A and B.

i.e. $$a + b \notin A \cup B$$.

Then the set $$A \cup B$$ isn’t closed under addition.

Therefor if $$A \cup B \neq A$$ or $$B$$, $$A \cup B$$ is not a subspace of V.

Now we have proved that the union of two subspaces of $$V$$ is a subspace of $$V$$ if and only if one of the subspaces of $$V$$ is contained in the other.