Proof of $X\cup Y\neq V$ Suppose $X,Y$ are subspaces of dimension $n-k$ of the vector space $V$ of dimension $n$. Why is it always true that $X\cup Y\neq V$? 
I can show this by arguing that if $X=Y$ then clearly by the difference in dimension, that $X\cup Y\neq V$. If $X\neq Y$ then there is a member, $x$, in the basis of $X$ that in not in $Y$ and vice versa. Then $x+y\not\in X\cup Y$ but is in $V$.
Is this argument valid? Is there a more elegant/simpler argument?
Thanks!
 A: The argument is correct and seems pretty elegant already, up to some minor issues of exposition.
Maybe it is also elegant to discuss generalizations?

*

*The exact number of proper subspaces needed to cover any vector space (of any dimension greater than or equal to $2$, including any infinite dimension) over any field $F$ is computed in this note.  (It was also computed independently by A. Khare at around the same time.) Namely, the number of subspaces required is $\# F + 1$ unless $F$ and $\operatorname{dim}_F V$ are both infinite, in which case it is $\aleph_0$.  In particular, since any field has at least two elements, this linear covering number is always at least $3$.  Moreover it is equal to three iff $F = \mathbb{F}_2$.


*No group $G$ is a union of two proper subgroups $H_1$, $H_2$.  The OP's proof works to prove this as well!  We are clearly in trouble if either $H_1$ is contained in $H_2$ or $H_2$ is contained in $H_1$, so assume not and let $h_1 \in H_1 \setminus H_2, \ h_2 \in H_2 \setminus H_1$.  But then $h_1 h_2$ does not lie in either $H_1$ or $H_2$, so $H_1 \cup H_2 \subsetneq G$.  (Actually we proved something stronger: the union of two subgroups of a group is not itself a subgroup except in the trivial case in which one subgroup contains the other.)
There are some interesting relations between 1) and 2).  For instance, 1) shows that there are infinitely many groups which can be written as a union of $3$ subgroups: namely, for any index set $I$ with at least two elements, we may take $V_I = \bigoplus_{i \in I} \mathbb{Z}/2\mathbb{Z}$, a special case of 1) above.  The proof of 1) in this case is just the observation that $V_2 = (\mathbb{Z}/2\mathbb{Z})^2$ can be covered by three proper subgroups, and any group $V_I$ as above has $V_2$ as a quotient, so can also be covered by three proper subgroups...namely the preimages of the covering subgroups under the quotient map.
In other words, we have shown that a group which has the Klein four-group $V_2$ as a quotient can be covered by $3$ proper subgroups.  As 1) above may suggest, the converse is also true: these are the only groups which can be covered by $3$ proper subgroups.  As I recall, the proof of this is still rather short and elementary.
Much more is true along the above lines.  For any positive integer $n$, there is a unique minimal finite set $S(n)$ of isomorphism classes of finite groups such that any group $G$ can be covered by $n$ proper subgroups (and not by fewer) iff $G$ has a quotient isomorphic to some $K \in S(n)$.  Trivially $S(1) = \varnothing$.  Above we showed that $S(2) = \varnothing$ and claimed that $S(3) = \{ V_2\}$.  The above claim is proved in this recent paper of M.(ira) Bhargava.  She also gives $S(n)$ explicitly up to $n = 7$.
