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

This is my proof by contradiction in the forward direction; I'm not sure it's correct.

(=>) Let $U_1, U_2,$ and $U_3$ be subspaces and $U_1\cup U_2\cup U_3$ is a subspace.

Suppose by contradiction that one of the subspaces does not contain the other two, i.e., negate the following ($\forall x \in U_1 \cup U_2, x \in U_3$)

$x\in U_1\backslash U_2 \backslash U_3$, $y\in U_2 \backslash U_1 \backslash U_3$, meaning these elements live in $U_1 \cup U_2$ and $U_1 \cup U_2 \cup U_3$.

So, $x+y\in U_1\cup U_2\cup U_3$ by closure under addition since each element lives in the union of all three subspaces.

$x+y\in U_1$ or $x+y\in U_2$ or $x+y\in U_3$

$y$ can't be in $U_1$, x can't be in $U_2$, and $x+y$ can't be in $U_3$ as the elements chosen are not in $U_3$.

I'll end it there as I don't feel confident this was the correct approach.

Indecisive
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h weber
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  • $x \notin U$ and $y \notin U$ does not imply $x+y \notin U$. Consider the real plane, $U = {(a,0) : a \in \mathbb{R}}$, $x=(1,1), y=(1,-1)$. – aschepler Aug 10 '24 at 04:24
  • Better use latex to type your question out, here’s the tutorial: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – user1176409 Aug 10 '24 at 04:28
  • @aschepler I see the error in my logic. Thank you. Is there any way to salvage this? – h weber Aug 10 '24 at 04:38
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    The result is false in the general: it fails for the two dimensional vector space over the field of two elements. So the argument must use the cardinality of the field in some way. – Arturo Magidin Aug 10 '24 at 05:01

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