Why $ \mathbb R$ with topology generated by base of form [a,b) is not topological vector space?

Question

Why $ \mathbb R$ with topology generated by base of form [a,b) is not topological vector space?

Topological Vector space:Suppose $\tau$ is topology on vector space X such that

1) every singleton is closed

2) both vector space operation is continuous

I can show that every singleton of R with given topology is closed but how to not continous to conclude result

Any Help will be appreciated

Answer 1

The most direct way is to note that $\Bbb R_l$ is not even a topological group w.r.t. $+$, as the map $f(x)=-x$ is not continuous: $[0,1)$ is open but $f^{-1}[[0,1)) = (-1,0]$ is not open, as $0$ is not an interior point of it (all basic neighbourhoods of $0$, which are of the form $[0,r)$, "stick out"). This doesn't use the scalar multiplication, so it's irrelevant over what field we're working.

An indirect way is to note that $\Bbb R_l$ is not locally compact (all compact subsets are at most countable, even) but as a $1$-dimensional vector space (over $\Bbb R$) it would have to be locally compact.

Also, $\Bbb R_l$ is totally disconnected, while all real or complex vector spaces are even path-connected.