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The Cantor set is an unusual closed set in the sense that it consists entirely of boundary points and is nowhere dense. Singleton points (and thus finite sets) are closed in T 1 spaces and Hausdorff spaces. The set of integers is an infinite and unbounded closed set in the real numbers.
These examples show that the closure of a set depends upon the topology of the underlying space. The last two examples are special cases of the following. In any discrete space, since every set is closed (and also open), every set is equal to its closure.
Conversely, if closed sets are given and every intersection of closed sets is closed, then one can define a closure operator C such that () is the intersection of the closed sets containing X. This equivalence remains true for partially ordered sets with the greatest-lower-bound property , if one replace "closed sets" by "closed elements" and ...
A subset of a space X is regular open if it equals the interior of its closure; dually, a regular closed set is equal to the closure of its interior. [21] An example of a non-regular open set is the set U = (0,1) ∪ (1,2) in R with its normal topology, since 1 is in the interior of the closure of U, but not in U.
As discussed above, given a topological space we may define the closure of any subset to be the set () = {|}, i.e. the intersection of all closed sets of which contain . The set c ( A ) {\displaystyle \mathbf {c} (A)} is the smallest closed set of X {\displaystyle X} containing A {\displaystyle A} , and the operator c : ℘ ( X ) → ℘ ( X ...
Closure operators are determined by their closed sets, i.e., by the sets of the form cl(X), since the closure cl(X) of a set X is the smallest closed set containing X. Such families of "closed sets" are sometimes called closure systems or "Moore families". [1] A set together with a closure operator on it is sometimes called a closure space.
For another example, consider the relative interior of a closed disk in . It is locally closed since it is an intersection of the closed disk and an open ball. On the other hand, { ( x , y ) ∈ R 2 ∣ x ≠ 0 } ∪ { ( 0 , 0 ) } {\displaystyle \{(x,y)\in \mathbb {R} ^{2}\mid x\neq 0\}\cup \{(0,0)\}} is not a locally closed subset of R 2 ...
By definition, the map : is a relatively closed map if and only if the surjection: is a strongly closed map. If in the open set definition of "continuous map" (which is the statement: "every preimage of an open set is open"), both instances of the word "open" are replaced with "closed" then the statement of results ("every preimage of a ...