A set S of real numbers is compact if and only if every open cover C of S can be reduced to a finite subcovering. Compact sets share many properties with finite sets. For example, if A and B are two non-empty sets with A B then A B # 0.

The set ℝ 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 ℝ but there is no finite subcover. The Cantor set is compact.

In and a set is compact if and only if it is closed and bounded. In general the answer is no. There exists metric spaces which have sets that are closed and bounded but aren’t compact. Theorem 2: There exists a metric space that has a closed and bounded set that is not compact.

Characterization of compact sets: A subset of R is compact if, and only if, it is closed and bounded. Proof. An unbounded subset of Rhas an open cover consisting of all bounded, open intervals. This has no finite subcover, since the union of a finite set of bounded intervals is bounded.

So the number line is not compact because we have found an open cover that does not have a finite subcover. A set does not have to be infinite in length or area to be non-compact. A closed interval and an open interval make a good case study for how we can think about compactness.

The empty set ∅ and R are both open and closed; they’re the only such sets. Most subsets of R are neither open nor closed (so, unlike doors, “not open” doesn’t mean “closed” and “not closed” doesn’t mean “open”).

The boundary of the empty set is the empty set, since it has no members for any contiguous open interval* of a boundary point to contain. Therefore all zero boundary points of the empty set are contained by the empty set, so the empty set is closed.

For example, one can prove that a certain set is not empty by proving that its cardinality is big, as in the proof that there exist transcendental numbers : The set of algebraic numbers is countable, but the set of real numbers is uncountable, so there is uncountably many transcendental numbers.

Let S be a set. Then S is said to be non-empty if and only if S has at least one element. By the Axiom of Extension, this may also be phrased as: S≠∅

2 Answers. Vector spaces can’t be empty, because they have to contain additive identity and therefore at least 1 element! The empty set isn’t (vector spaces must contain 0). However, {0} is indeed a subspace of every vector space.

1 Answer. The answer is no. The empty set is empty in the sense that it does not contain any elements. Thus the zero vector is not a member of the empty set.

It is impossible to have a common point between two parallel lines as parallel lines do not intersect. So the last set is also empty. Therefore option (A) is the only set which is an example of non-empty set.

Hence {x:x is a real number and x2+1=0} is an empty set.

This includes the class of groups, the class all fields, the class of all lattices, etc. Function classes into proper classes never form a set. For example, let Grp denote the class of all groups. Then the class of all functions N→Grp does not form a set.

A={x∣xϵI,x2=9} is a singleton set.