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So, I'm trying to prove the following result.

Let $V$ be a vector space over $F$. Let $U_1$ and $U_2$ be subspaces of $V$. Then, the following is true:

$U_1 \cup U_2 = V \implies [(U_1 = V) \lor (U_2 = V)]$


Proof Attempt:

Since $U_1 \cup U_2$ is a vector space, we have previously shown that $U_1 \subset U_2 \lor U_2 \subset U_1$. Hence:

$[U_1 \cup U_2 = U_1] \lor[ U_1 \cup U_2 = U_2] \implies (U_1 = V) \lor (U_2 = V)$.

Where we have made use of the fact that $A \subset B \implies A \cup B = B$. This proves the result.

Can someone check my proof above and see if it's correct? Also, how would I prove this without having proved the statement right at the beginning? Like, how might I prove this from first principles?

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  • $\begingroup$ You meant $(U_1=V) \lor (U_2=V)$ instead of $(U_1=V) = (U_2=V)$ , right? Otherwise, the proof seems fine to me. $\endgroup$
    – amator2357
    Feb 13, 2020 at 9:28

2 Answers 2

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\begin{equation} U_1 \cup U_2 = V \iff ( \textbf{v} \in U_1 \cup U_2 \iff \textbf{v} \in V) \iff ((\textbf{v} \in U_1 \lor \textbf{v} \in U_2) \iff \textbf{v} \in V) \iff (\textbf{v} \in U_1 \iff \textbf{v} \in V) \lor (\textbf{v} \in U_2 \iff \textbf{v} \in V) \iff (U_1 = V) \lor (U_2=V) \end{equation}

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  • $\begingroup$ Thank you! That's interesting, I didn't think the equivalence could be distributed like that. $\endgroup$ Feb 13, 2020 at 10:23
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Prove by contradiction. Suppose $U_1 \neq V$ and $U_2 \neq V$. Then there exist vectors $x \in V \setminus U_1$ and $y \in V \setminus U_2$. Consider $x+y$. Either $x+y \in U_1$ or $x+y \in U_2$. In the first case $y \notin U_1$ [because $y \in U_1$ would imply that $x=(x+y)-y\in U_1$]. But then $y $ is neither in $U_1$ nor in $U_2$ which is a contradiction.

In the second case $x \notin U_2$ [because $x \in U_2$ would imply that $y=(x+y)-x\in U_2$]. But then $y $ is neither in $U_1$ nor in $U_2$ which is a contradiction.

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  • $\begingroup$ Oh nice, thanks. That's lovely. $\endgroup$ Feb 13, 2020 at 10:23

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