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.

证明V的两个子空间的并是B的子空间当且仅当其中一个子空间包含领一个子空间。

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.