Topology on vector space such that vector operations are not continuous everyone
I am reading about distribution theory and the first concept is topological vector space. The definition says it is a topological vector space if its topology keeps the two vector operations, i.e., the vector addition and scalar multiplication continuous. I am trying to find a topology on a vector space that does not keeps this continuity property. Unluckily I don't find a convincing one. 
 A: Consider the topology on $\Bbb R$ in which the set is open iff it's empty or contains $1$.
To prove scalar multiplication is discontinuous, let $\alpha$ be any scalar different from $0,1$. Then letting $f:x\mapsto \alpha x$, the preimage of open set $\{1\}$ is $\{\frac{1}{\alpha}\}$, which isn't open.
To prove addition is discontinuous, consider $f:(x,y)\mapsto x+y$. Then the preimage of $\{1\}$ under $f$ is the set $\{(x,y)\in\Bbb R^2:x+y=1\}$. We need to show the last set is not open. But it's easy to see that in the product topology, if a set $A\subseteq\Bbb R^2$ is open then it is empty or contains $(1,1)$, while our preimage is nonempty and doesn't contain $(1,1)$.
A: Take $\mathbb R$ with the topology generated by the intervals of the form $[a,b)$.
A: Let $\phi\colon \mathbb R\to\mathbb R$ be some crazy bijection. For specificity, you could take $\phi(x)=x$ for $x\not\in\{0,1\}$ and $\phi(1)=0$, $\phi(0)=1$. We give $\mathbb R$ a new topology (and call it $\tilde{\mathbb R}$) so that $\phi\colon\tilde{\mathbb R}\to \mathbb R$ is a homeomorphism to the standard topology on the reals. The vector operations on $\tilde{\mathbb R}$ are not continuous. 
In other words, the open sets of $\tilde {\mathbb R}$ are the open sets of $\mathbb R$ with $1$ and $0$ interchanged. So $(-1/2,1)\cup(1,1/2)\cup\{0\}$ is an open set in $\tilde {\mathbb R}$. 
