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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.

Stefan Hamcke
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Johnny Apple
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  • The product of two discrete spaces is discrete, so you’re fine. If $\langle x,y\rangle$ is a point of the product, ${\langle x,y\rangle}={x}\times{y}$, which is open in the product topology. – Brian M. Scott Aug 28 '15 at 19:59
  • That's what I thought. Do you know how this is an "outside the box" problem? – Johnny Apple Aug 28 '15 at 20:00
  • Not really, no, unless it’s simply the fact that one doesn’t usually endow a vector space with the discrete topology. – Brian M. Scott Aug 28 '15 at 20:01
  • Scalar multiplication is a map $K\times X\to X$ (where $K$ is the scalar field), not $X\times X\to X$. – Eric Wofsey Aug 28 '15 at 20:12
  • See also http://math.stackexchange.com/questions/492483/topological-vector-space-with-discrete-topology-is-the-zero-space – Stefan Hamcke Aug 28 '15 at 20:23

2 Answers2

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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.

Eric Wofsey
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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.

Disintegrating By Parts
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