If $U\dotplus V = U\dotplus Y = V \dotplus Z = X$ does it necessary follow that $Y\dotplus Z = X$? Suppose that $X$ is a vector space and $U,V,Y,Z$ are subspaces such that $U\dotplus V = U\dotplus Y = V \dotplus Z = X$ where $\dotplus$ is the direct sum. Does it follow that $Y\dotplus Z = X$? If $X$ is finite dimensional then the answer is yes, but what about $\dim X = \infty$?
 A: Note that the following statement is true when $X$ is finite dimensional.  "Let $X$ be a vector space with subspaces $U$, $V$, $Y$, and $Z$ such that $U\cap V=\{0\}$, $U\cap Y=\{0\}$, $V\cap Z=\{0\}$, and $Y\cap Z=\{0\}$.  If $X=U\oplus V=U\oplus Y=V\oplus Z$, then $X=Y\oplus Z$."  
This can be proven by dimension counting.  If $n=\dim X$ and $k=\dim U$, then $\dim V=n-k$, $\dim Y=n-k$, and $\dim Z=k$.  Since $Y\cap Z=\{0\}$, we get that the dimension of $Y\oplus Z$ is equal to $k+(n-k)=n$, so $Y\oplus Z=X$.
The claim is no longer true if $X$ is infinite dimensional.  Here is a counterexample.  Let $X$ be the set of all sequences $(x_1,x_2,x_3,\ldots)$ with finitely many nonzero terms and let $e_k$ be the sequence $$(0,0,\ldots,0,1,0,0,\ldots),$$ where $1$ appears only at the $k$-th term.  Then, note that $X$ is spanned by $e_1$, $e_2$, $e_3$, $\ldots$.  Take $U$ to be the span of $e_k$ where $k$ is odd, $V$ the span of $e_k$ where $k$ is even, $Y$ the span of $e_k+e_{k+1}$ where $k$ is even, and $Z$ the span of $e_{k}+e_{k+1}$ where $k$ is odd.  Then, $$U\cap V=U\cap Y=V\cap Z=Y\cap Z=\{0\}$$ and $$X=U\oplus V=U\oplus Y=V\oplus Z,$$ but $Y\oplus Z$ is properly contained in $X$ since $e_1\notin Y\oplus Z$.
