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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.

May someone please validate this proof:

Let $V_1,V_2$ be two subspaces of $V$. Suppose $V_1 \cup V_2$ is a subspace of $V$. Then if $x_1 \in V_1$ and $x_2 \in V_2$ then we must have $x_1 \in V_1 \cup V_2$ and $x_2 \in V_1 \cup V_2$ so that we must have $x_1+x_2 \in V_1 \cup V_2$. But this by definition means $x_1+x_2 \in V_1$ or $x_1+x_2 \in V_2$. Either way, by existence of additive inverses and closure properties for subspaces we have:

$$(x_1+x_2)+(-x_1) \in V_1$$

Or

$$(x_1+x_2)+(-x_2) \in V_2$$

By associative/commutative properties we have:

$x_2 \in V_1, \text{or} ,x_2 \in V_1$

Edit:

Thus we have shown if $x_1 \in V_1$ and $x_2 \in V_2$ then $x_1 \in V_2$ or $x_2 \in V_1$. If $x_1 \in V_2$ for all $x_1 \in V_1$ then we have $V_1 \subseteq V_2$. If $x_2 \in V_1$ for all $x_2 \in V_2$ then $V_2 \subseteq V_1$. In either case we see one subspace is a subset of the other.

Ahmed S. Attaalla
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    So, you've proved $(\forall x_1 \in V_1) (\forall x_2 \in V_2) (x_2 \in V_1 \vee x_1 \in V_2)$. Why does that imply $V_1 \subseteq V_2 \vee V_2 \subseteq V_1$? – Daniel Schepler Jun 16 '17 at 20:55
  • May you please take a look at my edit @DanielSchepler – Ahmed S. Attaalla Jun 16 '17 at 21:14
  • Your edit still doesn't quite nail it. You don't seem to be appreciating that if, for each element z in a set, either P(z) holds or Q(z) holds then this does not mean that either P(z) holds for all z or Q(z) holds for all z. – Pete L. Clark Jun 16 '17 at 21:15
  • So how can I fix, I am struggling to understand @PeteL.Clark – Ahmed S. Attaalla Jun 16 '17 at 21:17
  • @https://math.stackexchange.com/questions/334405/if-a-group-is-the-union-of-two-subgroups-is-one-subgroup-the-group-itself: Um, I answered the question. But I think your real question is a logical one. – Pete L. Clark Jun 16 '17 at 21:18

1 Answers1

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[Edited!] As pointed out in @Daniel Schepler's comment, there is a quantifier error here. You need to show that either

$\bullet$ $\forall x_1 \in V_1$, $x_1 \in V_2$

or

$\bullet$ $\forall x_2 \in V_2$, $x_2 \in V_1$.

The basic idea you give is sound; you just need to ensure this. Suggestion: we're done unless there is $x_1 \in V_1 \setminus V_2$ and $x_2 \in V_2 \setminus V_1$. Now use what you've done.

By the way: essentially the same argument shows that the union of two subgroups of a group is a subgroup iff one contains the other.

Pete L. Clark
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  • To be clear, this uses proof by contradiction, along with the fact that the negation of $(\forall x_1 \in V_1) (x_1 \in V_2) \vee (\forall x_2 \in V_2) (x_2 \in V_1)$ is $(\exists x_1 \in V_1) (x_1 \notin V_2) \wedge (\exists x_2 \in V_2) (x_2 \notin V_1)$. – Daniel Schepler Jun 16 '17 at 21:01
  • @Daniel: Indeed. I hope that comment will be helpful to the OP. – Pete L. Clark Jun 16 '17 at 21:03
  • And as an addendum to the last sentence in the answer: invertibility is essential to the proof. For example, in the monoid $\mathbb{N}$, ${ 0, 2, 4, 5, \ldots }$ and ${ 0, 3, 4, 5, \ldots }$ are both submonoids, and their union ${ 0, 2, 3, 4, 5, \ldots }$ is also a submonoid; but neither is contained in the other. – Daniel Schepler Jun 16 '17 at 21:08
  • @Daniel: Nice example! I'm beginning to wish that I hadn't randomly popped in to (too?) quickly answer this question -- your answer would have been better, I think. – Pete L. Clark Jun 16 '17 at 21:09