Jänich linear algebra, Question 2.3 solution clarification Let $ U_{1} $ and $ U_{2}$ be vector subspaces of a vector space $ V $. Prove if $ U_{1} \cup U_{2} = V $, then $ U_{1} = V $ or $ U_{2} = V $ or both. Attempt: $ U_{1} \cup U_{2} = V $ $\implies$ $ (U_{1} \cup U_{2} ) \setminus U_{2} = V \setminus U_{2} $. This is equivalent to $ U_{1} = V \setminus U_{2} $. But this is a contradiction as this means that $ 0 \notin U_{1} $. $ \square $

I don't think your proof works. As Huy's comment suggests, there's a mistake in your simplification: $$(U_1\cup U_2)\setminus U_2 = (U_1\setminus U_2)\cup (U_2\setminus U_2) = U_1\setminus U_2,$$ which is not necessarily equal to $U_1$.

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As a side note, I want to point out that if you're using a proof by contradiction, you have to use the negation of your conclusion ($U_1 \
eq V$ and $U_2 \
eq V$ in this case) somewhere.

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Here's a proof.

A union of two spaces is a subspace if and only if one subspace is contained in the other. (For a proof of this, see this question.)

$U_1\cup U_2=V$ is a subspace $\implies$ either $U_1 \subseteq U_2$ or $U_2 \subseteq U_1$.

Say, without loss of generality, that $U_1\subseteq U_2$.

But then $V=U_1\cup U_2=U_2$.