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So I need to prove that for the union of $n$ subspaces to be a subspace, each subspace must be a subset of another one of the subspaces. My thought process so far is that I need to prove that it is possible for a union of $n$ subspaces to be a subspace space (which I proved). and that I need to prove that if an element is in some subspace of $w_j$ and not in the union, then the union is not a subspace. (I can't seem to figure out how to prove this part). Any hints or tips, I have been working on it for a while now, I feel like i am missing an obvious solution.

Yiorgos S. Smyrlis
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user123429
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  • The underlying field is infinite right? I am unable to follow your thought process. For example, what does "in some subspace of wj and not in the union" mean? – anon Jan 29 '14 at 19:19
  • I apologize, yes it is finite. Let me try to rephrase. If I have a vector space V with subspaces w1, w2, w3,....wn, I need to prove that the only way that Union of those subspaces is itself a subspace is if each subspace is contained by another subspace (with the exception of the "biggest" subspace). Is that more clear? – user123429 Jan 29 '14 at 19:24
  • @atomic: The question is to show that if the union is a subspace show... – copper.hat Jan 29 '14 at 19:25
  • @user127.0.0.1 You've shown if there is a chain of inclusions then their union is a subspace. The problem is to prove the other direction: if a union of subspaces is a subspace, then there is a chain of inclusions. – anon Jan 29 '14 at 19:26
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    @anon: Not quite. The union (in $3$-space) of the $x$-axis, $y$-axis, and $xy$-plane is the $xy$-plane, but the inclusions do not comprise a chain. – Cameron Buie Jan 29 '14 at 19:30
  • Yeah, was going to say "bunch" of inclusions, duplicated the word "chain" instead. – anon Jan 29 '14 at 19:31
  • Here is a related result for the finite dimensional case: http://math.stackexchange.com/a/10803/27978. – copper.hat Jan 31 '14 at 08:41

2 Answers2

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Let $\mathcal F$ be a finite family of subspaces of $V$, a vector field over an infinite field $F$.

We want to prove $\bigcup\limits_{W\in \mathcal F}W\neq \bigoplus\limits_{W\in \mathcal F}W$.

We shall prove if $\mathcal J$ is a finite family of proper subspaces of $\bigoplus\limits_{W\in F}W$ then $\bigcup\limits_{W\in \mathcal J}W\neq \bigoplus\limits_{W\in \mathcal F}W$. By induction over $|\mathcal J|$, base case is trivial. Note we can take $\mathcal J$ so no subspace contains any other.

Inductive step:

Take $W\in \mathcal J$, and take $w\in W$ so that it is not in any of the other subspaces (possible by inductive step). Take a nonzero vector $v\not\in W$, then the set $A=\{fw+v|f\in F\}$ is infinite since $F$ is infinite.

Moreover any $U\in \mathcal J$ contains at most one element of $A$. Clearly $W$ contains none of them. If another subspace contained $fw+v$ and $gw+v$ it would contain $(f-g)w$ and so it would contain $w$, a contradiction.

Hence every subspace contains at most one element of $A$, since $A$ is infinite then this means some elements of $A$ are not in $\bigcup\limits_{W\in \mathcal J}W$. Therefore $\bigcup\limits_{W\in \mathcal J}W\neq \bigoplus\limits_{W\in \mathcal F}W$ as desired.

Asinomás
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Let's assume the negative of the statement, each subspace must be a subset of another subspace (except for the "biggest" one).

Then there exists a subspace $W_i$, where $W_i \not \subset W_j$ for all $j \neq i$ and $W_j \not \subset W_i$ for at least one j, denoted by $W_k$. If this last condition didn't hold then $W_i$ would just be the "biggest" set as noted in your comment.

Now the first condition says that for each pair $W_i, W_j$, there is an element, call it $w_{ij}$ such that $w_{ij} \in W_i$ and $w_{ij} \not \in W_j$. The second says that there is a subspace $W_k$ which has an element, call it $w_k$, which is not in $W_i$.

Now note that $\sum_{j \neq i} w_{ij} \in W_i$ since $W_i$ is a subspace. However, $w_{ij} \not \in W_j$ for all $j \neq i$, and so $\sum_{j \neq i} w_{ij} \not \in W_j$ for all $j \neq i$. (If the sum was in $w_j$, then $w_{ij}$ would be in $W_j$ because $W_j$ is a subspace, which is a contradiction of our definition of $w_{ij}$). Therefore $\sum_{j \neq i} w_{ij} \not \in \cup \,W_j$.

Now take $v = w_k + \sum_{j \neq i} w_{ij}$. This is not in $W_i$, because this would imply that $w_k$ is in $W_i$, which contradicts our earlier assumption. It is also not in $\cup \, W_j$, because this would imply $\sum_{j \neq i} w_{ij}$ would be in $\cup \, W_j$ (this assumes that $\cup \, W_j$ is a subspace, which must be true otherwise we are done anyway). So we have shown that v is not in $\cup \, W_j$ for all j, which means that the union of the subspaces is not closed under addition, which means it is not a subspace.

clem
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