I stumbled across the following exercise in Wilansky - Modern methods in topological vector spaces:
Let $(X, \mathcal{T})$ be a TVS in which every bounded set is finite-dimensional. Show that every absorbing set is a bornivore.
So far I had one idea of how to approach the problem, but this doesn't seem to work: Let $B \subseteq X$ be a bounded set in the sense of the von Neumann Bornology such that for every neighbourhood of zero $U$, there exists a $\epsilon > 0$ with $t B \subseteq U$ for every $|t| \leq \epsilon$ and let $A \subseteq X$ be a absorbent set.
For neighborhoods of zero, this statement is clear.
Since $B$ is finite dimensional there exist $n \in \mathbb{N}$ and $b_1, \ldots, b_n \in X$ such that $\mbox{span}(\lbrace b_1, \ldots, b_n \rbrace) = B$ which is a closed linear subspace of $X$. Also $A$ is absorbent, so for each $i = 1, \ldots, n$ there exists $r_i > 0$ with $r_i b_i \in A$ or $b_i \in \frac{1}{r_i} A$ and $\mbox{span}(b_i) \in \mathbb{C}\frac{1}{r_i} A$. But from this on I can't seem to find any $\epsilon > 0$ such that $A$ absorbs $B$.
Edit: I noticed that the definition of a absorbent set differs from our definition. In the book the definition of an absorbent set is: For every $x \in X$ there exists a $r > 0$ such that $t x \in A$ for every $|t| < r$. Where the definition that I use is: For every $x \in X$ there exist a $r > 0$ such that $r x \in A$.
Both definitions are equivalent if $A$ is balanced or convex.
I would be thankful for every hint.
Hint: reconsider what you mean by "finite dimensional bounded sets". When is a subspace $V \subseteq X$ bounded?
Hint2: let $B \subseteq V$ be bounded. By assumption, it is contained in a finite dimensional subspace $\operatorname{span}(v_1, \ldots, v_n) \subseteq V$. Assume that each of the subspaces $\operatorname{span}(v_i)$ is unbounded. Argue that there is a $c > 0$ such that we have $|\lambda^i| < c$ for every $v \in B$ with basis decomposition $v = \sum_{i=1}^n \lambda^i v_i$.