and absorbing set but it is not necessarily convex. The balanced hull of a closed set need not be closed. Take for instance the graph of in The next example shows that the balanced hull of a convex set may fail to be convex (however, the convex hull of a balanced set is always balanced). For an example, let the convex subset be which is a horizontal closed line segment lying above the axis in The balanced hull is a non-convex subset that is "hour glass shaped" and equal to the union of two closed and filled isosceles triangles and where and is the filled triangle whose vertices are the origin together with the endpoints of (said differently, is the convex hull of while is the convex hull of ). Sufficient conditions A set is balanced if and only if it is equal to its balanced hull or to its balanced core in which case all three of these sets are equal: The Cartesian product of a family of balanced sets is balanced in the product space of the corresponding vector spaces (over the same field ). The balanced hull of a compact (respectively, totally bounded, bounded) set has the same property. The convex hull of a balanced set is convex and balanced (that is, it is absolutely convex). However, the balanced hull of a convex set may fail to be convex (a counter-example is given above). Arbitrary unions of balanced sets are balanced, and the same is true of arbitrary intersections of balanced sets. Scalar multiples and (finite) Minkowski sums of balanced sets are again balanced. Images and preimages of balanced sets under linear maps are again balanced. Explicitly, if is a linear map and and are balanced sets, then and are balanced sets. Balanced neighborhoods In any topological vector space, the closure of a balanced set is balanced. The union of and the topological interior of a balanced set is balanced. Therefore, the topological interior of a balanced neighborhood of the origin is balanced. However, is a balanced subset of that contains the origin but whose (nonempty) topological interior does not contain the origin and is therefore not a balanced set. Every neighborhood (respectively, convex neighborhood) of the origin in a topological vector space contains a balanced (respectively, convex and balanced) open neighborhood of the origin. In fact, the following construction produces such balanced sets. Given the symmetric