In my lecture notes, I found the following exercise where I am not entirely sure how to proceed:
Let $E$ be a vector space. We are given a set $\mathfrak{U}:= \{U| U \text{ is absorbing, balanced}\}$ and for every $V \in \mathfrak{U}$ there exists $U \in \mathfrak{U}$ such that $U + U \subseteq V$.
The claim is that there exists a unique topology such that $\mathfrak{U}$ is a neighbourhoodsubbasis of zero and that the vector space structure is compatible with this topology.
I already read this, which gave me a rough idea for the topology. Since $\mathfrak{U}$ should be a subbasis, I define $\mathfrak{B}(0) = \{\bigcap_{i=1}^n U_i | n \in \mathbb{N}, U_i \in \mathfrak{U}\}$, for each $x \in E$, we set $\mathfrak{B}(x) = x + \mathfrak{B}(0)$.
Now I can define a topology, with $$ A \in \tau :\iff \forall x \in A \exists B_x \in \mathfrak{B}(x): B_x \subseteq A. $$ This is all in accordance with the link above.
Now I want to verify that $\mathfrak{B}(x)$ is a neighbourhoodbasis for the point $x$. In particular, every $B \in \mathfrak{B}(x)$ must be a neighbourhood of x. Let $B \in \mathfrak{B}(x)$ with a representation $B = x + \bigcap_{i=1}^n U_i$. It suffices to find a $V \subseteq X$ with $x \in V \subseteq B$ such that for every $y \in V$ there exists $B_y \in \mathfrak{B}(y)$ such that $B_y \subseteq V$ (i.e. V is open).
No matter how hard I try, I am not able to verify the last step, giving me reason to think that my construction for $\tau$ is not correct.
However, I already succeeded to proof that addition in my vector space is continuous (using the fact that $\mathfrak{B}(x)$ gives a neighbourhood basis without actually proofing it). This gives me reason to think, that it must be correct.
So, is my initial construction of $\tau$, using $\mathfrak{B}$ correct and if so, how I can show that $\mathfrak{B}(x)$ is a neighbourhood basis.
Thanks in advance!
