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For example, the union of three sets A, B, and C contains all elements of A, all elements of B, and all elements of C, and nothing else. Thus, x is an element of A ∪ B ∪ C if and only if x is in at least one of A, B, and C. A finite union is the union of a finite number of sets; the phrase does not imply that the union set is a finite set ...
The union of two intervals is an interval if and only if they have a non-empty intersection or an open end-point of one interval is a closed end-point of the other, for example (,) [,] = (,]. If R {\displaystyle \mathbb {R} } is viewed as a metric space , its open balls are the open bounded intervals ( c + r , c − r ) , and its closed balls ...
Now let be an infinite set under the discrete metric – that is, two points , have distance 1 if they're not the same point, and 0 otherwise. Under the resulting metric space, any singleton set is open; hence any set, being the union of single points, is open. Since any set is open, the complement of any set is open too, and therefore any set ...
The union is the join/supremum of and with respect to because: L ⊆ L ∪ R {\displaystyle L\subseteq L\cup R} and R ⊆ L ∪ R , {\displaystyle R\subseteq L\cup R,} and if Z {\displaystyle Z} is a set such that L ⊆ Z {\displaystyle L\subseteq Z} and R ⊆ Z {\displaystyle R\subseteq Z} then L ∪ R ⊆ Z . {\displaystyle L\cup R\subseteq Z.}
Locally connected does not imply connected, nor does locally path-connected imply path connected. A simple example of a locally connected (and locally path-connected) space that is not connected (or path-connected) is the union of two separated intervals in , such as (,) (,).
As another example, in a 2-interval hypergraph, the vertex set is the disjoint union of two real lines, and each edge is a union of two intervals: one in line #1 and one in line #2. The following two concepts are defined for d-interval hypergraphs just like for finite hypergraphs:
A disjoint union may mean one of two things. Most simply, it may mean the union of sets that are disjoint. [11] But if two or more sets are not already disjoint, their disjoint union may be formed by modifying the sets to make them disjoint before forming the union of the modified sets. [12] For instance two sets may be made disjoint by ...
Provided X has at least two elements, this is equivalent to saying that the open intervals (,) = {< <} together with the above rays form a base for the order topology. The open sets in X are the sets that are a union of (possibly infinitely many) such open intervals and rays.