# How to show that one of the $W_i$ is under another [duplicate]

This question already has an answer here:

perhaps this question was asked before, but if not give me a hand for it. My question is how to prove that if $W_1$ and $W_2$ be sub-spaces of vector space $V$ with the union of $W_1$ and $W_2$ is also a subspace, then one of the subspace $W_i$ is contained in other ?

Please give me a hint for start. Where should I look here?

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## marked as duplicate by Gerry Myerson, Babak S., Zev ChonolesFeb 21 '13 at 12:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Duplicate of math.stackexchange.com/q/71872/264 – Zev Chonoles Feb 21 '13 at 11:46
@BasilR More specifically, see the answer of Gerry Myerson to the above-linked question. – user1551 Feb 21 '13 at 12:13
Thaak you. ok i will... – Basil R Feb 21 '13 at 14:31

## 2 Answers

Suppose that $W_1$ has a basis $v_1, \ldots , v_n$, and that $W_2$ has a basis $u_1, \ldots, u_m$.

If neither space is a subspace of the other then there is a $u_i$ not in $W_1$, and a $v_j$ not in $W_2$. But since the union is a subspace that contains $u_i$ and $v_j$ then it must contain $v_j + u_i$. But this implies the sum is in either $W_1$ or $W_2$ and hence that either $u_i \in W_1$ or $v_j \in W_2$, a contradiction.

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Hint: What would be happen if for example $W_1$ is not contained in $W_2$ and $W_2$ is not contained in $W_1$? Think of the point that with this assumption the set $W_1\cup W_2$ would not be a subspace of whole space $V$. This is a nice contradicton!

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Sharp observation! – amWhy Feb 21 '13 at 13:32