Does every absorbing set of a Banach space contain a neighborhood of origin? Let $X$ be a Banach space and $A$ be any absorbing subset of $X$. Does $A$ contain a neighborhood of the origin?
 A: We should expect the answer to this question to be "no", since being absorbent depends only on the algebraic structure of $B$, and is thus unlikely to imply the topological property of containing a neighbourhood of $0$. In fact, the answer is even no in the finite-dimensional case, where the algebra determines the topology.
Suppose $A$ is an absorbing set in $B$, and $v$ is a vector. We know that there exists an $r$ such that $v \in sA$ for $s>r$. In other words, $\lambda v$ is in $A$ for all sufficiently small positive $\lambda$. Since this is true for every $v$, $A$ must contain a radial neighbourhood of $0$: that is, it must contain an open line segment in every direction. Conversely, it's not hard to show that if $A$ contains a radial neighbourhood of $0$ then $A$ is absorbing.
Thus the question is whether a radial neighbourhood of $0$ must be an actual neighbourhood. The answer is no: containing open segments in every straight line direction doesn't stop $A$ omitting a curve that goes to $0$. For example, in $\Bbb R^2$, the set $A$ consisting of the whole space with the nonzero points of the parabola $y=x^2$ removed is an absorbing set.
