subset of is a neighborhood of the origin. This is no longer guaranteed if the set is not convex (a counter-example exists even in ) or if is not of the second category in itself.InteriorIf and has non-empty interior then and The topological interior of a disk is not empty if and only if this interior contains the origin. More generally, if is a balanced set with non-empty interior in a TVS then will necessarily be balanced; consequently, will be balanced if and only if it contains the origin. For this (i.e. ) to be true, it suffices for to also be convex (in addition to being balanced and having non-empty interior).; The conclusion could be false if is not also convex; for example, in the interior of the closed and balanced set is If is convex and then Explicitly, this means that if is a convex subset of a TVS (not necessarily Hausdorff or locally convex), and then the open line segment joining and belongs to the interior of that is, If is any balanced neighborhood of the origin in then where is the set of all scalars such that If belongs to the interior of a convex set and then the half-open line segment and If is a balanced neighborhood of in and then by considering intersections of the form (which are convex symmetric neighborhoods of in the real TVS ) it follows that: and furthermore, if then and if then Non-Hausdorff spaces and the closure of the origin A topological vector space is Hausdorff if and only if is a closed subset of or equivalently, if and only if Because is a vector subspace of the same is true of its closure which is referred to as in This vector space satisfies so that in particular, every neighborhood of the origin in contains the vector space as a subset. The subspace topology on is always the trivial topology, which in particular implies that the topological vector space a compact space (even if its dimension is non-zero or even infinite) and consequently also a bounded subset of In fact, a vector subspace of a TVS is bounded if and only if it is contained in the closure of Every subset of also carries the trivial topology and so is itself a compact, and thus also complete, subspace (see footnote for a proof). In particular, if is not Hausdorff