assume K, L are proper subspaces. Prove $K\cap L$ is a subspace of V, but $K\cup L$ is never a subspace. Solution:

if $v_1,v_2\in K$, then $c_1v_1+c_2v_2 \ in K$ [because K is a subspace]

if $v_1,v_2\in L$, then $c_1v_1+c_2v_2 \ in L$ [because L is a subspace]

Hence, $c_1v_1+c_2v_2 \forall c_1,c_2\in\mathbb{R} $ are common elements and thus belong to $k \cap L $. showing $k \cap L$ is closed under multiplication is direct. Thus $K\cap L$ is a subspace.

assume $v_1 \in K, v_1 \notin L$ and $v_2 \in L , v_2 \notin K$.

Suppose for the sake of contradiction $(v_1+v_2) \in K\cup L$. then its either in K or L. Assume its in K. since K is a subspace then its closed under addition. hence, since $v_1+v_2\in K$ and $v_1 \in K$, then $v_1+v_2-v_1=v_2\in K$ a contradiction!

Is my proof correct?

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    $\begingroup$ You seem to have a $U$ that should have been an $L$. Also, you need to assume that neither subspace is contained in the other for this argument to work. $\endgroup$ – Tobias Kildetoft Apr 29 '15 at 7:50
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    $\begingroup$ The claim is slightly wrong: $\\;W\cup U\;$ is a subspace iff $\;W\subset U\;$ , or $\;U\subset W\;$ . This is why you needed to assume what you did in the middle of your proof. $\endgroup$ – Timbuc Apr 29 '15 at 7:50
  • $\begingroup$ Corrected the U, L typo. $\endgroup$ – user3382078 Apr 29 '15 at 7:53
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    $\begingroup$ yes, if I assumed one is contained in the other than the claim is false. $\endgroup$ – user3382078 Apr 29 '15 at 7:54
  • $\begingroup$ But more strongly than this, the converse also holds: the line $v_1 \in K, v_1 \not\in L; v_2 \in L, v_2 \not \in K$ in the middle of your proof is possible only if $K \not\subset L$ and $L \not \subset K$. $\endgroup$ – Circonflexe Apr 29 '15 at 8:09

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