Question about vector spaces with the discrete topology Is it true that every vector space with the discrete topology is a topological vector space? (That is, a topological space with continuous addition and scalar multiplication whose singletons are closed?)
Obviously singletons are closed in such a space by discreteness. I would also guess that addition/scalar multiplication are continuous since I thought every map from $X$ $\times$ $X$ to $X$ would be continuous, again by discreteness.
The only issue is that I'm not sure I'm allowed to say that every set in $X$ $\times$ $X$ is open. I'm questioning even very basic things I thought I knew since my professor asked us to do this problem as an "outside the box" problem. 
By the way, it's not homework, but he wants to discuss it next class.
 A: Scalar multiplication will not be continuous unless the topology on the scalar field $K$ is also discrete (or unless $X=0$).  Indeed, if $x\in X$ is any nonzero vector, $a\mapsto a\cdot x$ is a continuous injection $K\to X$, as the restriction of the scalar multiplication map $K\times X\to X$ to the subspace $K\times\{x\}\cong K$.  If $X$ has the discrete topology, it follows that $K$ must also have the discrete topology.  In particular, if $K$ is $\mathbb{R}$ or $\mathbb{C}$ with the usual topology, a topological vector space $X$ over $K$ cannot be discrete unless $X=0$.
On the other hand, if $K$ does have the discrete topology, the discrete topology can be given to any vector space to get a topological vector space, for as noted in the comments, a product of discrete spaces is discrete.
A: Every Hausdorff topology on a real or complex finite-dimensional topological vector space is equivalent to the usual topology. So, without doing any checking: No, your topology cannot result in a TVS.
