countable family of bounded sets in a metrizable locally convex space, it is possible to scale each set by its own positive real so that they become uniformly bounded. Generalizations Uniformly bounded sets A family of sets of subsets of a topological vector space is said to be in if there exists some bounded subset of such that which happens if and only if its union is a bounded subset of In the case of a normed (or seminormed) space, a family is uniformly bounded if and only if its union is norm bounded, meaning that there exists some real such that for every or equivalently, if and only if A set of maps from to is said to be if the family is uniformly bounded in which by definition means that there exists some bounded subset of such that or equivalently, if and only if is a bounded subset of A set of linear maps between two normed (or seminormed) spaces and is uniformly bounded on some (or equivalently, every) open ball (and/or non-degenerate closed ball) in if and only if their operator norms are uniformly bounded; that is, if and only if Assume is equicontinuous and let be a neighborhood of the origin in Since is equicontinuous, there exists a neighborhood of the origin in such that for every Because is bounded in there exists some real such that if then So for every and every which implies that Thus is bounded in Q.E.D. Let be a balanced neighborhood of the origin in and let be a closed balanced neighborhood of the origin in such that Define which is a closed subset of (since is closed while every is continuous) that satisfies for every Note that for every non-zero scalar the set is closed in (since scalar multiplication by is a homeomorphism) and so every is closed in It will now be shown that from which follows. If then being bounded guarantees the existence of some positive integer such that where the linearity of every now implies thus and hence as desired. Thus expresses as a countable union of closed (in ) sets. Since is a nonmeager subset of itself (as it is a Baire space by the Baire category theorem), this is only possible if there is some integer such that has non-empty interior in Let be any point belonging to this open subset of Let be