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The word "bounded" makes no sense in a general topological space without a corresponding metric. Boundary is a distinct concept; for example, a circle (not to be confused with a disk) in isolation is a boundaryless bounded set, while the half plane is unbounded yet has a boundary. A bounded set is not necessarily a closed set and vice
The collection of all bounded sets on a topological vector space is called the von Neumann bornology or the (canonical) bornology of .. A base or fundamental system of bounded sets of is a set of bounded subsets of such that every bounded subset of is a subset of some . [1] The set of all bounded subsets of trivially forms a fundamental system of bounded sets of .
Every compact set is totally bounded, whenever the concept is defined. [clarification needed] Every totally bounded set is bounded. A subset of the real line, or more generally of finite-dimensional Euclidean space, is totally bounded if and only if it is bounded. [5] [3]
If X is any set and M is a metric space, then the set of all bounded functions: (i.e. those functions whose image is a bounded subset of ) can be turned into a metric space by defining the distance between two bounded functions f and g to be (,) = ((), ()).
A bounded operator: is not a bounded function in the sense of this page's definition (unless =), but has the weaker property of preserving boundedness; bounded sets are mapped to bounded sets (). This definition can be extended to any function f : X → Y {\displaystyle f:X\rightarrow Y} if X {\displaystyle X} and Y {\displaystyle Y} allow for ...
If is the real line, or -dimensional Euclidean space, then a function has compact support if and only if it has bounded support, since a subset of is compact if and only if it is closed and bounded. For example, the function f : R → R {\displaystyle f:\mathbb {R} \to \mathbb {R} } defined above is a continuous function with compact support ...
A real-valued or complex-valued function defined on some topological space is called a locally bounded functional if for any there exists a neighborhood of such that () is a bounded set. That is, for some number M > 0 {\displaystyle M>0} one has | f ( x ) | ≤ M for all x ∈ A . {\displaystyle |f(x)|\leq M\quad {\text{ for all }}x\in A.}
The set of all real numbers is not compact as there is a cover of open intervals that does not have a finite subcover. For example, intervals (n − 1, n + 1) , where n takes all integer values in Z, cover but there is no finite subcover.