Stephen Tashi said:In the axioms for a topology, we assume the whole space is an open set. Its closure wouldn't be larger.
and the whole space is closed (because it's the complement of the empty set).Suppose A is an open set and not closed.
CompuChip said:But TS said
and the whole space is closed (because it's the complement of the empty set).
As you've worded it, your statement is true. Suppose that cl(A) is not strictly bigger. By definition, it is not smaller either (it's the smallest closed set that contains A), so it must be exactly A.
The closure of an open set A is the smallest closed set that contains all the points in A. This means that it includes all the boundary points of A, as well as the points within A.
The closure of an open set A may contain additional points, such as boundary points, that are not included in A itself. A is considered an open set because it does not contain its boundary points, while the closure of A does.
The closure of an open set A is important because it helps us to define the boundary of A and determine which points are included in A. It also allows us to extend the definition of A to include its boundary points, which can be useful in certain mathematical proofs and applications.
Yes, it is possible for the closure of an open set A to be equal to A itself. This would occur when A is already a closed set, meaning it contains all of its boundary points, and therefore does not require any additional points to be added to its closure.
The closure of an open set A can be calculated using the closure operator, denoted as "cl(A)". This operator takes the set A and adds all of its boundary points to create the closure of A. In mathematical notation, cl(A) = A ∪ Bd(A), where Bd(A) represents the boundary of A.