state and prove Heine borel theorem
Answers
If a set is compact, then it must be closed.
Let S be a subset of Rn. Observe first the following: if a is a limit point of S, then any finite collection C of open sets, such that each open set U ∈ C is disjoint from some neighborhood VU of a, fails to be a cover of S. Indeed, the intersection of the finite family of sets VU is a neighborhood W of a in Rn. Since a is a limit point of S, W must contain a point x in S. This x ∈ S is not covered by the family C, because every U in C is disjoint from VU and hence disjoint from W, which contains x.
If S is compact but not closed, then it has a limit point a not in S. Consider a collection C ′ consisting of an open neighborhood N(x) for each x ∈ S, chosen small enough to not intersect some neighborhood Vx of a. Then C ′ is an open cover of S, but any finite subcollection of C ′ has the form of C discussed previously, and thus cannot be an open subcover of S. This contradicts the compactness of S. Hence, every limit point of S is in S, so S is closed.
The proof above applies with almost no change to showing that any compact subset S of a Hausdorff topological space X is closed in X.
If a set is compact, then it is bounded.
Let {\displaystyle S}S be a compact set in {\displaystyle \mathbf {R} ^{n}}\mathbf {R} ^{n}, and {\displaystyle U_{x}}U_x a ball of radius 1 centered at {\displaystyle x\in \mathbf {R} ^{n}}x\in\mathbf{R}^n. Then the set of all such balls centered at {\displaystyle x\in S}x\in S is clearly an open cover of {\displaystyle S}S, since {\displaystyle \cup _{x\in S}U_{x}}{\displaystyle \cup _{x\in S}U_{x}} contains all of {\displaystyle S}S. Since {\displaystyle S}S is compact, take a finite subcover of this cover. This subcover is the finite union of balls of radius 1. Consider all pairs of centers of these (finitely many) balls (of radius 1) and let {\displaystyle M}M be the maximum of the distances between them. Then if {\displaystyle C_{p}}C_{p} and {\displaystyle C_{q}}C_q are the centers (respectively) of unit balls containing arbitrary {\displaystyle p,q\in S}{\displaystyle p,q\in S}, the triangle inequality says: {\displaystyle d(p,q)\leq d(p,C_{p})+d(C_{p},C_{q})+d(C_{q},q)\leq 1+M+1=M+2.}{\displaystyle d(p,q)\leq d(p,C_{p})+d(C_{p},C_{q})+d(C_{q},q)\leq 1+M+1=M+2.} So the diameter of {\displaystyle S}S is bounded by {\displaystyle M+2}{\displaystyle M+2}.
A closed subset of a compact set is compact.
Let K be a closed subset of a compact set T in Rn and let CK be an open cover of K. Then U = Rn \ K is an open set and
{\displaystyle C_{T}=C_{K}\cup \{U\}}C_{T}=C_{K}\cup \{U\}
is an open cover of T. Since T is compact, then CT has a finite subcover {\displaystyle C_{T}',}C_{T}', that also covers the smaller set K. Since U does not contain any point of K, the set K is already covered by {\displaystyle C_{K}'=C_{T}'\setminus \{U\},}C_{K}'=C_{T}'\setminus \{U\}, that is a finite subcollection of the original collection CK. It is thus possible to extract from any open cover CK of K a finite subcover.
If a set is closed and bounded, then it is compact.
If a set S in Rn is bounded, then it can be enclosed within an n-box
{\displaystyle T_{0}=[-a,a]^{n}}T_{0}=[-a,a]^{n}
where a > 0. By the property above, it is enough to show that T0 is compact.
Assume, by way of contradiction, that T0 is not compact. Then there exists an infinite open cover C of T0 that does not admit any finite subcover. Through bisection of each of the sides of T0, the box T0 can be broken up into 2n sub n-boxes, each of which has diameter equal to half the diameter of T0. Then at least one of the 2n sections of T0 must require an infinite subcover of C, otherwise C itself would have a finite subcover, by uniting together the finite covers of the sections. Call this section T1.
Likewise, the sides of T1 can be bisected, yielding 2n sections of T1, at least one of which must require an infinite subcover of C. Continuing in like manner yields a decreasing sequence of nested n-boxes:
{\displaystyle T_{0}\supset T_{1}\supset T_{2}\supset \ldots \supset T_{k}\supset \ldots }T_{0}\supset T_{1}\supset T_{2}\supset \ldots \supset T_{k}\supset \ldots
where the side length of Tk is (2 a) / 2k, which tends to 0 as k tends to infinity. Let us define a sequence (xk) such that each xk is in Tk. This sequence is Cauchy, so it must converge to some limit L. Since each Tk is closed, and for each k the sequence (xk) is eventually always inside Tk, we see that L ∈ Tk for each k.
Since C covers T0, then it has some member U ∈ C such that L ∈ U. Since U is open, there is an n-ball B(L) ⊆ U. For large enough k, one has Tk ⊆ B(L) ⊆ U, but then the infinite number of members of C needed to cover Tk can be replaced by just one: U, a contradiction.
Thus, T0 is compact. Since S is closed and a subset of the compact set T0, then S is also compact (see above).
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