logo

Every k Cell is Compact 📂MetricSpace

Every k Cell is Compact

Definition

For ai,biR(1ik)a_{i},b_{i} \in \mathbb{R} (1\le i \le k), the set I=[a1,b1]×[a2,b2]××[ak,bk]I=[a_{1},b_{1}] \times [a_{2},b_{2}]\times \cdots \times [a_{k},b_{k}] is called a kk-cell. Here, ×\times represents the Cartesian product of sets.

Theorem 1

Let’s assume a sequence of closed intervals on R\mathbb{R}, {In}\left\{ I_{n} \right\}, satisfies InIn+1 (n=1,2,)I_{n}\supset I_{n+1}\ (n=1,2,\cdots). Then, the following holds true.

i=1In \bigcap_{i=1}^{\infty}I_{n}\ne \varnothing

Proof

Let’s denote In=[an,bn]I_{n}=[a_{n},b_{n}]. Also, let E={an:n=1,2,}E=\left\{ a_{n} : n=1,2,\cdots \right\}. Then, EE\ne \varnothing and is upper bounded by b1b_{1}1. Now, let’s define x=supEx=\sup E. For any two positive numbers mm and nn,

anam+nbm+nbm a_{n} \le a_{m+n} \le b_{m+n} \le b_{m}

holds true, thus for all nn, xbnx\le b_{n}. Moreover, since xx is the upper bound of EE, it’s trivial that for all nn, anxa_{n} \le x. Therefore, for all nn, anxbna_{n}\le x \le b_{n}, which implies xIn nx\in I_{n}\ \forall n. Thus,

xi=1nIn x\in \bigcap _{i=1}^{n}I_{n}

Theorem 2

Let {In}\left\{ I_{n} \right\} be a sequence of kk-cells satisfying InIn+1(n=1,2,)I_{n}\supset I_{n+1}(n=1,2,\cdots). Then, i=1nIn\bigcap _{i=1}^{n}I_{n}\ne\varnothing.


Theorem 2 is an extension of Theorem 1 to Rk\mathbb{R}^{k}.

Proof

Let’s represent InI_{n} as follows.

In={x=(x1,,xk):an,jxjbnj,(1jk; n=1,2,)} I_{n}=\left\{ \mathbf{x}=(x_{1},\cdots,x_{k}) : a_{n,j} \le x_{j} \le b_{nj},\quad(1\le j \le k;\ n=1,2,\cdots) \right\}

That is, In=In,1××In,k (In,j=[an,j,bn,j])I_{n}=I_{n,1}\times \cdots\times I_{n,k}\ (I_{n,j}=[a_{n,j},b_{n,j}]). By Theorem 1, for each In,jI_{n,j}, there exists xjIn,j (an,jxjbn,j)x_{j}^{\ast}\in I_{n,j} \ (a_{n,j} \le x_{j}^{\ast} \le b_{n,j}). Therefore,

x=(x1,,xk)In,(n=1,2,) \mathbf{x^{\ast}} =(x_{1}^{\ast},\cdots ,x_{k}^{\ast})\in I_{n} ,\quad (n=1,2,\cdots)

Theorem 3

Every kk-cell is compact.

Proof

Let’s consider an arbitrary kk-cell II as follows.

I=I1××Ik=[a1,b1]××[ak,bk] I=I^{1}\times \cdots \times I^{k}=[a_{1},b_{1}]\times \cdots \times [a_{k},b_{k}]

And let’s define as follows.

x=(x1,,xk)andajxjbj(1jk) \mathbf{x}=(x_{1},\cdots,x_{k}) \quad \text{and} \quad a_{j} \le x_{j} \le b_{j}(1\le j \le k)

Now, let’s consider δ\delta as follows.

δ=(j=1k(bj)aj)2)12=ba \delta =\left( \sum \limits_{j=1}^{k}(b_{j})-a_{j})^{2} \right)^{{\textstyle \frac{1}{2}}}=|\mathbf{b}-\mathbf{a}|

Here, a=(a1,,an)\mathbf{a}=(a_{1},\cdots,a_{n}), b=(b1,,bn)\mathbf{b}=(b_{1},\cdots,b_{n}). Then, δ\delta is the same as the distance between b\mathbf{b} and a\mathbf{a}. Therefore,

xyδx,yI |\mathbf{x}-\mathbf{y}| \le \delta \quad \forall \mathbf{x},\mathbf{y}\in I

is valid. Now the proof begins in earnest, using a proof by contradiction. That is, assume that a kk-cell is not compact. Then, by the definition of compactness, it’s the same as assuming that some open cover {Oα}\left\{ O_{\alpha} \right\} of II does not have a finite subcover. Let’s denote cj=(aj+bj)/2c_{j}=(a_{j}+b_{j})/2. Then, each IjI^{j} can be divided into [aj,cj][a_{j},c_{j}], [cj,bj][c_{j},b_{j}] using cjc_{j}, creating 2k2^{k} 1-cells. Their union is naturally II, and by assumption, at least one of them cannot be covered by any finite subcover of {Oα}\left\{ O_{\alpha} \right\}. Let’s call this cell I1I_{1}. Then, by choosing intervals in the same way as I1I_{1} was chosen from II, we can obtain a sequence {In}\left\{ I_{n} \right\} satisfying the following three rules.

  • (i)(\mathrm{i}) II1I2I\supset I_{1} \supset I_{2}\supset \cdots

  • (ii)(\mathrm{ii}) Each InI_{n} cannot be covered by any finite subcover of {Oα}\left\{ O_{\alpha} \right\}.

  • (iii)(\mathrm{iii}) xy2nδ,x,yIn|\mathbf{x}-\mathbf{y}|\le 2^{-n}\delta,\quad \forall \mathbf{x},\mathbf{y}\in I_{n}

Then, by (i)(\mathrm{i}) and Theorem 2, there exists xIn\mathbf{x}^{\ast}\in I_{n} for all nn. Since {Oα}\left\{ O_{\alpha} \right\} is an open cover of II, there is some α\alpha for which xOα\mathbf{x}^{\ast}\in O_{\alpha}. As OαO_{\alpha} is an open set, there exists r>0r>0 such that xy<r    yOα|\mathbf{x}^{\ast}-\mathbf{y}|<r \implies \mathbf{y}\in O_{\alpha}. On the other hand, nn can be sufficiently large so that 2nδ<r2^{-n}\delta<r. Then, by (iii)(\mathrm{iii}), InOαI_{n}\subset O_{\alpha}. However, this contradicts (ii)(\mathrm{ii}), so the assumption is wrong. Therefore, every kk-cell is compact.

From the above facts, we can prove the following useful theorems.

Equivalent Conditions for Compactness in Euclidean Space

For a subset ERk(or Ck)E\subset \mathbb{R}^{k}(\mathrm{or}\ \mathbb{C}^{k}) of the real (or complex) space, the following three propositions are equivalent.

(a) EE is closed and bounded.

(b) EE is compact.

(c) Every infinite subset

of EE has an accumulation point pEp \in E.


Here, the equivalence of (a) and (b) is known as the Heine-Borel theorem. An EE 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)     \implies (b)

    Assuming (a), there exists a kk-cell II such that EIE \subset I. Since II is compact, and a closed subset of a compact set is compact, EE is compact.

  • (b)     \implies (c)

    This is proved by contradiction.

    Let SS be an infinite subset of a compact set EE. Assume that SS has no accumulation point. Then, every pEp\in E has at most one point of SS in its neighborhood NpN_{p}. When pSp \in S, that one point is pp itself. And this implies that the open cover {Np}\left\{ N_{p} \right\} does not have a finite subcover for SS. Since SES \subset E, similarly, there’s no finite subcover for EE either, contradicting the assumption that EE is compact. Hence, SS has an accumulation point pEp \in E.

  • (c)     \implies (a)

    This is proved by contradiction.

    • part 1. EE is bounded

      Let’s assume EE is not bounded. Then, EE contains points xn\mathbf{x}_{n} satisfying the following inequality.

      xn>n(n=1,2,) |\mathbf{x}_{n}| >n\quad (n=1,2,\cdots)

      Let’s denote S={xn:n=1,2,}S=\left\{ \mathbf{x}_{n} :n=1,2,\cdots\right\}. SS is infinite and obviously does not have an accumulation point in Rk,contradicting(c).Thus,\mathbb{R}^{k}, contradicting (c). Thus, E$ is bounded.

    • part 2. EE is closed

      Let’s assume EE is not closed. Then, by definition, there exists an accumulation point x0\mathbf{x}_{0} of EE that is not included in E.Now,forE. Now, for n=1,2,\cdots,letsconsider, let’s consider \mathbf{x}_{n} \in E$ satisfying the following conditions.

      xnx0<1n \left|\mathbf{x}_{n}-\mathbf{x}_{0} \right| < {\textstyle \frac{1}{n}}

      Let’s denote the set of such xn\mathbf{x}_{n} as SS. SS is infinite and has x0\mathbf{x}_{0} as an accumulation point. If x0\mathbf{x}_{0} is the only accumulation point of SS, then x0E\mathbf{x}_{0}\notin E contradicts (c), proving EE is closed. Now, consider yx0\mathbf{y} \ne\mathbf{x}_{0} in Rk\mathbb{R}^{k}. Then,

      xnyx0yxnx0x0y1n \begin{align*} \left| \mathbf{x}_{n} - \mathbf{y} \right| & \ge \left|\mathbf{x}_{0} - \mathbf{y} \right| - \left|\mathbf{x}_{n}-\mathbf{x}_{0} \right| \\ & \ge \left| \mathbf{x}_{0} - \mathbf{y} \right| -\frac{1}{n} \end{align*}

      For sufficiently large nn, the following holds true.

      xnyx0y1n12x0y \begin{equation} \left| \mathbf{x}_{n} - \mathbf{y} \right| \ge \left| \mathbf{x}_{0}- \mathbf{y} \right|-\frac{1}{n} \ge \frac{1}{2}\left|\mathbf{x}_{0}-\mathbf{y} \right| \label{eq1} \end{equation}

      Furthermore, as nn increases, xn\mathbf{x}_{n} gets closer to x0\mathbf{x}_{0}. This fact, along with (eq1)\eqref{eq1}, implies that we can find a neighborhood of y\mathbf{y} that contains no point other than y\mathbf{y} as nn increases. Thus, y\mathbf{y} is not an accumulation point of SS, proving x0\mathbf{x}_{0} is the only accumulation point of S.Thiscontradicts(c),provingS. This contradicts (c), proving E$ is closed.

Bolzano-Weierstrass Theorem

Every bounded infinite subset of Rk\mathbb{R}^{k} has an accumulation point pRkp \in \mathbb{R}^{k}.

Proof

Let EE be a bounded infinite subset of Rk\mathbb{R}^{k}. Since EE is bounded, there exists a kk-cell II such that EI.AsE \subset I. As kcellsarecompact,-cells are compact, Iiscompact.Then,bytheequivalentconditionforcompactnessin is compact. Then, by the equivalent condition for compactness in \mathbb{R}^{k} (b)\implies (c),, Ehasanaccumulationpoint has an accumulation point p \in I \subset \mathbb{R}^{k}$.

See Also

Specialization of Riesz’s Theorem

Riesz’s Theorem in normed spaces indicates the compactness of the closed unit ball B(0;1)\overline{B (0;1)} as an equivalent condition of finite dimension. The kk-cell [0,1]k[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 kk-cell.


  1. Any bnb_{n} will suffice. ↩︎