Counterexample in topological vector space.

Question

Let's $X$ be a topological vector space. So it has two operations $+$ and $\times$.

As we know these operations could be continous. But can it be that one of this operation is continous, but the second isn't?

UPD. Continuity means : if $(x,y) \to x + y$ and $(\alpha,x) \to \alpha x$ are continous.

Answer 1

Note: Usually the definition of a topological vector space says the operations are continuous. Here is an example of a vector space with a topology that makes addition but not multiplication continuous. It is not a topological vector space under the standard definition.

Let $V$ be the space $C(0,1)$ of continuous functions on the open interval under uniform convergence. The addition is continuous but scalar multiplication isn't.

For example consider the sequence of constants $c_n = 1/n$. Then $c_n \to 0$ and continuity would imply $c_n f \to 0$ (uniformly) for any fixed $f \in V$. It's an exercise to show this fails for $f(x) = 1/x$.

Answer 2

In topological vector space addition and scalar multiplication are continuous by definition, so you probably are asking for vector space with topology where one of them is continuous and other isn't.

@Daron's answer gives an example when addition is continuous but multiplication isn't. For example of space with continuous multiplication but not addition take $\mathbb R^2$ with only open sets been $\varnothing$, $\mathbb R^2$ and line $x = 0$.

For Hausdorff example, take topology generated by open subsets of lines passing through origin (so set is open if it's intersection with any line crossing origin is open set plus may be origin).