eIn general topology, a compact set K is a set for which every open cover of K contains a finite subcover of K. This is the definition of a compact set. In metric spaces, a compact space (glorified set) K is compact if it is complete and totally bounded. Completeness means that every Cauchy sequence in K converges to a point that exists in K. The rationals Q are not complete, because we can form a sequence that converges to the irrational 2–√2, which does not exist within Q. Note that complete also implies closed, because a complete space necessarily contains all of its limit points, which coincides with the properties of a closed set in metric spaces. Totally bounded means that for any fixed ε>0, K can be covered by balls of radius ε. This is different than the definition of a bounded set / space: every totally bounded set is bounded, but not every bounded set is totally bounded. Typical counter examples happen in infinite dimensions with the unit ball. We note that every metric space induces a topology because of access to the distance function, permitting a natural definition of open sets. Furthermore, every compact space still satisfies the definition of a compact set as found in general topology. In real analysis, the real numbers R form the only complete, ordered field up to isomorphism. Furthermore R induces a metric space with its own natural distance function of subtraction (accounting for signs). We gain additional properties that implies the compactness of sets with R. From Heine-Borel, a finite dimensional set \(K \subseteq \Bbb R^n\) is compact if and only if it is closed and bounded. All previous properties of compact sets in general topology and metric spaces are still true for R. We just happen to gain convenient structure along the way. Examples: Every set F with finitely many elements is compact: leveraging the definition from general topology, for every open cover of F, it suffices to take a subcover, such that no two sets in the subcover overlap on the same points in F. This forces a finite subcover of any cover of F.
By Heine-Borel, all closed intervals of form [a,b] for some a≤b∈R are compact.
The open interval (a,b) for any a<b∈R is not compact, because it is not the case that every open cover of (a,b) has a finite subcover.
Consider the first part of the union. Is it bounded? Can you think about option 1)?