Question

Prove that a subspace of R is compact if and only if it is closed and bounded.

Answer 1

An unbounded subset of R has 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.
Similarly, if K is not closed then it has a boundary point x ∉ K. But then the collec-
tion of sets R\ [x −ε,x +ε] with ε > 0 is an open cover with no finite subcover.
We have shown that a set which is either unbounded or not closed is not compact.
It remains to prove that a set K ⊂ R which is both bounded and closed is compact.
First, we show that a closed and bounded interval [a,b] is compact.
Let O be an open cover of [a,b].
Define:
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].
Pick some U ∈ O with s ∈U, and let ε > 0 so that [s −ε,s +ε] ⊆U.
Pick some c ∈ C with c > s −ε (it exists by the definition of s). Then there is a finite
subcover Q of O for [a,c]. But then Q∪ {U}
is a finite subcover of O for [a,c]
where,
C^l = min(s + ε,b). Thus c^l ∈ C. If s < b then this contradicts the definition of s,
therefore s = b, and b = c^l ∈C, in other words there is a finite subcover for [a,b].

Finally, let K ⊂ R be closed and bounded, and let O be an open cover of K. Then
U = R \K is open, and O ∪ {U} is an open cover of [a,b], where K ⊆ [a,b] (which
can be so arranged because K is bounded). Since [a,b] is compact, it has a finite
subcover Q. Now Q\{U} ⊆ O is a finite cover of K, and the proof is complete.