for instance, is not Hausdorff), in which case the pairing is not a dual pair. By the Hahn–Banach theorem, if is a Hausdorff locally convex space then separates points of and thus forms a dual pair. Properties If covers then the canonical map from into is well-defined. That is, for all the evaluation functional on meaning the map is continuous on If in addition separates points on then the canonical map of into is an injection. Suppose that is a continuous linear and that and are collections of bounded subsets of and respectively, that each satisfy axioms and Then the transpose of is continuous if for every there is some such that In particular, the transpose of is continuous if carries the (respectively, ) topology and carry any topology stronger than the topology (respectively, ). If is a locally convex Hausdorff TVS over the field and is a collection of bounded subsets of that satisfies axioms and then the bilinear map defined by is continuous if and only if is normable and the -topology on is the strong dual topology Suppose that is a Fréchet space and is a collection of bounded subsets of that satisfies axioms and If contains all compact subsets of then is complete. Polar topologies on the continuous dual space Throughout, will be a TVS over the field with continuous dual space and and will be associated with the canonical pairing. The table below defines many of the most common polar topologies on Notation: If denotes a polar topology then endowed with this topology will be denoted by (e.g. if then and so that denotes with endowed with ). If in addition, then this TVS may be denoted by (for example, ). The reason why some of the above collections (in the same row) induce the same polar topologies is due to some basic results. A closed subset of a complete TVS is complete and that a complete subset of a Hausdorff and complete TVS is closed. Furthermore, in every TVS, compact subsets are complete and the balanced hull of a compact (resp. totally bounded) subset is again compact (resp. totally bounded). Also, a Banach space can be complete without being weakly complete. If is bounded then is absorbing in (note that being absorbing is a necessary condition for to be a neighborhood of the origin in any TVS topology on ). If is a locally