Suppose $X$ is a normed linear space. If $\mathcal {B}$ consists of $\phi$, $X$, all open balls centered at origin, and all open annulus centered at origin, then it is clear that $\mathcal {B}$ is a base for some topology $\mathcal {T}(\mathcal {B})$.
My Question is:
Is the addition and scalar multiplication continuous with respect to this topology $\mathcal {T}(\mathcal {B})$ ?
i.e. Does $X$ become a topological vector space with respect to the topology $\mathcal {T}(\mathcal {B})$ ?
Thanks in Advance.
So, we identify everything with the same norm, and apply the usual topology on $\mathbb{R}$ to those equivalence classes.
Scalar multiplication? That's continuous on $\mathbb{R}$, so it's continuous here.
Addition? That doesn't respect the identification, so it won't work. The preimage of $0$ under addition includes all pairs $(x,-x)$ but not the pairs $(x,x)$.
If $U$ is an open set containing $0$ but not $y$, then the preimage of $U$ under addition contains $(\frac y2,-\frac y2)$ but not $(\frac y2,\frac y2)$. On the other hand, every open set containing $-\frac y2$ also contains $\frac y2$, so a product of open sets that contains $(\frac y2,-\frac y2)$ also contains $(\frac y2,\frac y2)$. As such, the preimage of $U$ can't be open in the product topology, and addition is discontinuous.
Some versions of the definition of a topological vector space also include a separation axiom (such as $T_1$). This example, of course, would also fail that test.