Let $X$ be a topological vector space over $\mathbb K$ and $\mathfrak B$ be neighborhood basis of $0$ s.t all of the basis elements are balanced (such choice is possible). Call a subset $E\subseteq X$ precompact if for every $B\in\mathfrak B$ there is a finite subset $F\subseteq X$ such that $E \subseteq F+B$.
If $E\subseteq X$ is precompact, is it true that for every $\lambda\in\mathbb K$ also the set $\lambda E$ is precompact?
Minus the trivial case of $\lambda =0$, if $|\lambda|≤ 1$, then due to balancedness of $B\in\mathfrak B$ we get $$\lambda E\subseteq \lambda F +\lambda B\subseteq \lambda F+ B. $$ I can't for the life of me figure out what to do for $|\lambda|>1$. We do have precompactness for $\lambda ^{-1}E$, but can't think of anything else. This is something that supposedly is immediate from the definition :(
Let $B \in \mathfrak{B}$. Note that $\lambda^{-1} B$ is a neighborhood of $0$, so there is some other $U \in \mathfrak{B}$ with $U \subset \lambda^{-1} B$, which is to say $\lambda U \subset B$. Now cover $E$ with finitely many translates of $U$. Then $\lambda E$ is covered with finitely many translates of $\lambda U$, hence with finitely many translates of $B$.