I have two vector spaces $U$ and $V$. Can $U \cap V = \varnothing$? I was thinking about this.
I thought about a lot of vector spaces, but I always come to the conclusion, that $0$ always have to be in a vector space. So the intersection of two vector spaces can't be $\varnothing$ ...
Any ideas?
 A: Two vector spaces, as sets, can be disjoint: namely, the set of real numbers (with the obvious structure of $\Bbb R$-vector space) is disjoint from the set of functions $\{1,2\}\to\Bbb R$ (with the obvious structure of $\Bbb R$-vector space). Two vector subspaces of the same vector space $V$ will always both contain the $0$ element of $V$.
Added: The closest thing (I think) that we can write in the sense of ascertaining to what degree there is "a common element" to all vector spaces is that the category of $\Bbb F$-vector spaces has initial objects, id est, there are vector spaces $I$ such that for all vector spaces $V$ there is exactly one linear map $f^I_V:I\to V$. Such spaces are canonically isomorphic and they are exactly the zero-dimensional vector spaces. The maps $f^I_V$ also happen to be all injective. This is, however, a somewhat different notion than the one you are speaking of: it's really more along the lines of identifying a canonical element of each vector space, more than it is about identifying a common element.
A: By definition, a vector space is a nonempty set that follows the axioms, one of which is the 0 vector.    What this means is that if two vector spaces are BOTH subspaces of some common space, their intersection is nonempty and thus has the shared $0$ vector of the common bigger space.    However,  if the vector spaces are not both subspaces of some common space,   then their intersection will be the empty set (and thus not a vector space)
A: Considering the case of vector spaces which share a common superset, taking for example where both are subsets of $\mathbb R^3,$ I do think it's worth mentioning that there is a technicality under which they can possibly not share any common elements, so long as they are not subspaces of $\mathbb R^3$. This is possible if we allow the vector spaces to have an alternative definition of the $+$ and $\cdot$ operations.
Consider, for instance, the set $V_{\vec{a}} = \{\vec{a}\},$ for any vector $\vec{a} \in \mathbb R^3.$ If we allow the definitions $\vec{x} + \vec{y} = \vec{x}$ and $c \cdot \vec{x} = \vec{x}$ for all $\vec{x}, \vec{y} \in V_{\vec{a}}$ and $c \in \mathbb R,$ then $V_{\vec{a}}$ trivially satisfies all of the vector space axioms under these definitions. (noting that the commutativity of addition holds because for any pair of $\vec{x}$ and $\vec{y}$ in $V_{\vec{a}},$ it must be true that $\vec{x} = \vec{y}$)
In particular we can note that the zero vector under these operations can be any vector, so there is no issue here where the vector spaces must share a zero vector. Now, for any distinct $\vec{a}$ and $\vec{b}$ in $\mathbb R^3,$ we can construct two vector spaces $V_{\vec{a}}$ and $V_{\vec{b}}$ which share no common elements.
However, if we require that both of our vector spaces be subspaces of some parent vector space, then this construction fails to produce two disjoint vector spaces because the parent vector space must be a singleton in order for the commutativity property to hold, meaning that the single element of both of our vector spaces must be the same element, which is the zero vector of the parent vector space. There is another construction we can do where the vector spaces consider different operations, but this also fails for this case because a subspace of a vector space must consider the same operations as the parent vector space. Because of that restriction, we can prove, for the reason that you pointed out and the uniqueness of the zero vector in a vector space, that two subspaces of any vector space must at least share the parent vector space's zero vector.
