Prove that dimension is finite I want to prove that $\dim V/(X \cap Y)$ in finite, if $V$ be a vector space and $X$, $Y$ two sub spaces of $V$ such that $\dim V/Y$ and $\dim V/X$ are finite.
 A: Let $\{a_1 + Y,\ldots,a_r +Y\}$ be a basis of $V/Y$, where the $a_i$s are elements of $V$. Since $V/X$ is finite-dimensional, we have that $Y/(Y\cap X)$ is finite-dimensional (since we have maps $Y\to V\to V/X$ and the composition has kernel $Y\cap X$, so we have an injection $Y/(Y\cap X)\to V/X$, so $\dim Y/(Y\cap X)\le \dim V/X<\infty$), so let $\{b_1 + Y\cap X,\ldots,b_s + Y\cap X\}$ be a basis of $Y/(Y\cap X)$, where the $b_i$s are all elements of $Y$. Let $v+(X\cap Y)\in V/(X\cap Y)$. Now $v+Y\in V/Y$, so we can write $v + Y = \sum \lambda_i(a_i+Y)$, which means that $v = y+\sum \lambda_ia_i$ for some $y\in Y$. Then $y+Y\cap X\in Y/(Y\cap X)$, so we can write $y = \sum\nu_ib_i +z$ with $z\in Y\cap X$. Combining these together, we get that $v =  \sum \lambda_ia_i + \sum \nu_ib_i + z$, where $z\in X\cap Y$. Therefore $v + X\cap Y = \sum \lambda_i(a_i+X\cap Y) + \sum \nu_i(b_i + X\cap Y$.
Therefore, $\{a_1 + X\cap Y,\ldots,a_r+X\cap Y,b_1+X\cap Y,\ldots,b_s+X\cap Y\}$ span $V/(X\cap Y)$, so $V/(X\cap Y)$ is finite-dimensional. (Note that the spanning set is likely not a basis.)
A: A much easier argument - consider the map $V\to (V/X)\times (V/Y)$ given by the product of the two projection maps. The kernel of this map is $X\cap Y$. Therefore, we have an injection $V/(X\cap Y)\to (V/X)\times (V/Y)$. Therefore, $\dim V/(X\cap Y)\le \dim [(V/X)\times (V/Y)]=\dim(V/X)\dim(V/Y)<\infty$.
