What is shown below is a reference from the text Functional Analysis by Walter Rudin.
If $X$ s a vector space over the field $\Bbb F$ then the following notation well be used $$ x+A=\{x+a:a\in A\}\\ x-A:=\{x-a:a\in A\}\\ A+B:=\{a+b:a\in A\wedge b\in B\}\\ \lambda A:=\{\lambda a:a\in A\} $$ for any $A,B\subseteq X$, for any $x\in\ X$ and for any $\lambda\in\Bbb F$.
Definition
If $\tau$ is a topology on a vector space $X$ such that
- every point of $X$ is a closed set
- the vector space operation are continuous with respect to $\tau$
we say that $X$ is a topological vector space equipped with the vector topology $\tau$.
So we note that the condition $1$ imply that any topological vector space will be $T_1$ but many authors omitted the above condition from the definition of a topological vector space.
Proposition
Let be $X$ a topological vector space over the field $\Bbb F$ and for some $v\in X$ and $\lambda\in\Bbb F$ we define the $v$-traslation and $\lambda$-multiplication functions through the condition $$ T_v(x):=x+v\,\,\,\text{and}\,\,\,M_\lambda(x):=\lambda x $$ for any $x\in X$. So the above two defined functions are homeomorphisms.
Proof. Omitted.
One consequence of the above proposition is that every vector topology $\tau$ is translation-invariant, that is a set $E\subseteq X$ is open if and only if each of its translates $v+E$ is open and thus $\tau$ is completely determined by any local base so that the open sets of $X$ are then precisely those that are unions of translates of memebrs of any local base.
So I like to discuss the last statement: indeed since a neighborhood is not union of basic neighborhoods then the image of local base undera traslation is not a base so what means that a vector topology is completely determined by any local base? Could someone help me to understand this statement, please?
If $\mathscr{B}_x$ is a local base at $x$, $\mathscr{B}_y=\{B+(y-x):B\in\mathscr{B}\}$ is a local base at $y$, and $\bigcup_{y\in X}\mathscr{B}_y$ is a base for the topology of $X$. In other words, we can get a base for the topology and therefore the whole topology directly from that one local base $\mathscr{B}_x$.