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These are two examples in which both the subset and the whole set are infinite, and the subset has the same cardinality (the concept that corresponds to size, that is, the number of elements, of a finite set) as the whole; such cases can run counter to one's initial intuition.
If the base set is finite, then = ℘ since every subset of , and in particular every complement, is then finite.This case is sometimes excluded by definition or else called the improper filter on . [2] Allowing to be finite creates a single exception to the Fréchet filter’s being free and non-principal since a filter on a finite set cannot be free and a non-principal filter cannot contain ...
In reality, obtaining an unbiased sample can be difficult as many parameters (in this example, country, age, gender, and so on) may strongly bias the estimator and it must be ensured that none of these factors play a part in the selection process.
A measure in which all subsets of null sets are measurable is complete. Any non-complete measure can be completed to form a complete measure by asserting that subsets of null sets have measure zero. Lebesgue measure is an example of a complete measure; in some constructions, it is defined as the completion of a non-complete Borel measure.
If A is a subset of B, then one can also say that B is a superset of A, that A is contained in B, or that B contains A. In symbols, A ⊆ B means that A is a subset of B, and B ⊇ A means that B is a superset of A. Some authors use the symbols ⊂ and ⊃ for subsets, and others use these symbols only for proper subsets. For clarity, one can ...
A set is infinite if and only if for every natural number, the set has a subset whose cardinality is that natural number. [2] If the axiom of choice holds, then a set is infinite if and only if it includes a countable infinite subset. If a set of sets is infinite or contains an infinite element, then its union is infinite.
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Named examples. The singleton set = {} is called the indiscrete or trivial filter on . [25] [10] It is the unique minimal filter on because it is a subset of every filter on ; however, it need not be a subset of every prefilter on .