Every k Cell is Compact
📂MetricSpaceEvery k Cell is Compact
Definition
For ai,bi∈R(1≤i≤k), the set I=[a1,b1]×[a2,b2]×⋯×[ak,bk] is called a k-cell. Here, × represents the Cartesian product of sets.
Theorem 1
Let’s assume a sequence of closed intervals on R, {In}, satisfies In⊃In+1 (n=1,2,⋯). Then, the following holds true.
i=1⋂∞In=∅
Proof
Let’s denote In=[an,bn]. Also, let E={an:n=1,2,⋯}. Then, E=∅ and is upper bounded by b1. Now, let’s define x=supE. For any two positive numbers m and n,
an≤am+n≤bm+n≤bm
holds true, thus for all n, x≤bn. Moreover, since x is the upper bound of E, it’s trivial that for all n, an≤x. Therefore, for all n, an≤x≤bn, which implies x∈In ∀n. Thus,
x∈i=1⋂nIn
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Theorem 2
Let {In} be a sequence of k-cells satisfying In⊃In+1(n=1,2,⋯). Then, ⋂i=1nIn=∅.
Theorem 2 is an extension of Theorem 1 to Rk.
Proof
Let’s represent In as follows.
In={x=(x1,⋯,xk):an,j≤xj≤bnj,(1≤j≤k; n=1,2,⋯)}
That is, In=In,1×⋯×In,k (In,j=[an,j,bn,j]). By Theorem 1, for each In,j, there exists xj∗∈In,j (an,j≤xj∗≤bn,j). Therefore,
x∗=(x1∗,⋯,xk∗)∈In,(n=1,2,⋯)
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Theorem 3
Every k-cell is compact.
Proof
Let’s consider an arbitrary k-cell I as follows.
I=I1×⋯×Ik=[a1,b1]×⋯×[ak,bk]
And let’s define as follows.
x=(x1,⋯,xk)andaj≤xj≤bj(1≤j≤k)
Now, let’s consider δ as follows.
δ=(j=1∑k(bj)−aj)2)21=∣b−a∣
Here, a=(a1,⋯,an), b=(b1,⋯,bn). Then, δ is the same as the distance between b and a. Therefore,
∣x−y∣≤δ∀x,y∈I
is valid. Now the proof begins in earnest, using a proof by contradiction. That is, assume that a k-cell is not compact. Then, by the definition of compactness, it’s the same as assuming that some open cover {Oα} of I does not have a finite subcover. Let’s denote cj=(aj+bj)/2. Then, each Ij can be divided into [aj,cj], [cj,bj] using cj, creating 2k 1-cells. Their union is naturally I, and by assumption, at least one of them cannot be covered by any finite subcover of {Oα}. Let’s call this cell I1. Then, by choosing intervals in the same way as I1 was chosen from I, we can obtain a sequence {In} satisfying the following three rules.
(i) I⊃I1⊃I2⊃⋯
(ii) Each In cannot be covered by any finite subcover of {Oα}.
(iii) ∣x−y∣≤2−nδ,∀x,y∈In
Then, by (i) and Theorem 2, there exists x∗∈In for all n. Since {Oα} is an open cover of I, there is some α for which x∗∈Oα. As Oα is an open set, there exists r>0 such that ∣x∗−y∣<r⟹y∈Oα. On the other hand, n can be sufficiently large so that 2−nδ<r. Then, by (iii), In⊂Oα. However, this contradicts (ii), so the assumption is wrong. Therefore, every k-cell is compact.
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From the above facts, we can prove the following useful theorems.
Equivalent Conditions for Compactness in Euclidean Space
For a subset E⊂Rk(or Ck) of the real (or complex) space, the following three propositions are equivalent.
(a) E is closed and bounded.
(b) E is compact.
(c) Every infinite subset
of E has an accumulation point p∈E.
Here, the equivalence of (a) and (b) is known as the Heine-Borel theorem. An E satisfying (c) is said to be ‘compact with respect to accumulation points’ or ‘having the Bolzano-Weierstrass property’. The equivalence of (b) and (c) holds in metric spaces but is not generally true in topological spaces.
Proof
(a) ⟹ (b)
Assuming (a), there exists a k-cell I such that E⊂I. Since I is compact, and a closed subset of a compact set is compact, E is compact.
(b) ⟹ (c)
This is proved by contradiction.
Let S be an infinite subset of a compact set E. Assume that S has no accumulation point. Then, every p∈E has at most one point of S in its neighborhood Np. When p∈S, that one point is p itself. And this implies that the open cover {Np} does not have a finite subcover for S. Since S⊂E, similarly, there’s no finite subcover for E either, contradicting the assumption that E is compact. Hence, S has an accumulation point p∈E.
(c) ⟹ (a)
This is proved by contradiction.
part 1. E is bounded
Let’s assume E is not bounded. Then, E contains points xn satisfying the following inequality.
∣xn∣>n(n=1,2,⋯)
Let’s denote S={xn:n=1,2,⋯}. S is infinite and obviously does not have an accumulation point in Rk,contradicting(c).Thus,E$ is bounded.
part 2. E is closed
Let’s assume E is not closed. Then, by definition, there exists an accumulation point x0 of E that is not included in E.Now,forn=1,2,\cdots,let’sconsider\mathbf{x}_{n} \in E$ satisfying the following conditions.
∣xn−x0∣<n1
Let’s denote the set of such xn as S. S is infinite and has x0 as an accumulation point. If x0 is the only accumulation point of S, then x0∈/E contradicts (c), proving E is closed. Now, consider y=x0 in Rk. Then,
∣xn−y∣≥∣x0−y∣−∣xn−x0∣≥∣x0−y∣−n1
For sufficiently large n, the following holds true.
∣xn−y∣≥∣x0−y∣−n1≥21∣x0−y∣
Furthermore, as n increases, xn gets closer to x0. This fact, along with (eq1), implies that we can find a neighborhood of y that contains no point other than y as n increases. Thus, y is not an accumulation point of S, proving x0 is the only accumulation point of S.Thiscontradicts(c),provingE$ is closed.
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Every bounded infinite subset of Rk has an accumulation point p∈Rk.
Proof
Let E be a bounded infinite subset of Rk. Since E is bounded, there exists a k-cell I such that E⊂I.Ask−cellsarecompact,Iiscompact.Then,bytheequivalentconditionforcompactnessin\mathbb{R}^{k}(b)\implies (c),Ehasanaccumulationpointp \in I \subset \mathbb{R}^{k}$.
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See Also
Specialization of Riesz’s Theorem
Riesz’s Theorem in normed spaces indicates the compactness of the closed unit ball B(0;1) as an equivalent condition of finite dimension. The k-cell [0,1]k in Euclidean space is compact, and since there is a homeomorphism with the closed unit ball, Riesz’s theorem can be seen as a generalization of the compactness of the k-cell.