Question

Prove that every compact subset of set of real numbers is bounded and closed

Answer 1

Answer:

Step-by-step explanation:

One is based on sequences (every sequence has a convergent subsequence), and the other is based on open sets (every open cover has a finite subcover). We will prove that a subset of R is compact if and only if it is closed and bounded. For example, every closed, bounded interval [a, b] is compact.

Answer 2

Answer:

a set K ⊂ R which is both bounded and closed is compact. s = supC where C = {c ∈ [a,b]: O has a finite subcover for [a,c]}. It is clear that a ∈C, since O is a cover for [a,b]. So C is not empty, and s ∈ [a,b].

Step-by-step explanation:

Step 1: If a subset of Euclidean space is closed and bounded, it is said to be compact. This means that each infinite sequence from the set has a subsequence that converges to a point in the set according to the Bolzano-Weierstrass theorem.

Step 2:A set D ⊆ Rn is open if all its points are interior points: for any x ∈ D, there is some sufficiently small r > 0 such that the ball B(x,r) is contained in D. A set D ⊆ Rn is closed if it contains all its limit points: whenever we have a sequence x(0),x(1),x(2),.

Step 3: It remains to prove that a set K ⊂ R which is both bounded and closed is compact. s = supC where C = {c ∈ [a,b]: O has a finite subcover for [a,c]}. It is clear that a ∈C, since O is a cover for [a,b]. So C is not empty, and s ∈ [a,b].

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