Every k Cell is Compact 📂MetricSpace

Every k Cell is Compact

Definition

For $a_{i},b_{i} \in \mathbb{R} (1\le i \le k)$, the set $I=[a_{1},b_{1}] \times [a_{2},b_{2}]\times \cdots \times [a_{k},b_{k}]$ is called a $k$-cell. Here, $\times$ represents the Cartesian product of sets.

Theorem 1

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

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

Proof

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

$$ a_{n} \le a_{m+n} \le b_{m+n} \le b_{m} $$

holds true, thus for all $n$, $x\le b_{n}$. Moreover, since $x$ is the upper bound of $E$, it’s trivial that for all $n$, $a_{n} \le x$. Therefore, for all $n$, $a_{n}\le x \le b_{n}$, which implies $x\in I_{n}\ \forall n$. Thus,

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

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Theorem 2

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

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

Proof

Let’s represent $I_{n}$ as follows.

$$ I_{n}=\left\{ \mathbf{x}=(x_{1},\cdots,x_{k}) : a_{n,j} \le x_{j} \le b_{nj},\quad(1\le j \le k;\ n=1,2,\cdots) \right\} $$

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

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

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Theorem 3

Every $k$-cell is compact.

Proof

Let’s consider an arbitrary $k$-cell $I$ as follows.

$$ I=I^{1}\times \cdots \times I^{k}=[a_{1},b_{1}]\times \cdots \times [a_{k},b_{k}] $$

And let’s define as follows.

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

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

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

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

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

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

$(\mathrm{i})$ $I\supset I_{1} \supset I_{2}\supset \cdots$
$(\mathrm{ii})$ Each $I_{n}$ cannot be covered by any finite subcover of $\left\{ O_{\alpha} \right\}$.
$(\mathrm{iii})$ $|\mathbf{x}-\mathbf{y}|\le 2^{-n}\delta,\quad \forall \mathbf{x},\mathbf{y}\in I_{n}$

Then, by $(\mathrm{i})$ and Theorem 2, there exists $\mathbf{x}^{\ast}\in I_{n}$ for all $n$. Since $\left\{ O_{\alpha} \right\}$ is an open cover of $I$, there is some $\alpha$ for which $\mathbf{x}^{\ast}\in O_{\alpha}$. As $O_{\alpha}$ is an open set, there exists $r>0$ such that $|\mathbf{x}^{\ast}-\mathbf{y}|<r \implies \mathbf{y}\in O_{\alpha}$. On the other hand, $n$ can be sufficiently large so that $2^{-n}\delta<r$. Then, by $(\mathrm{iii})$, $I_{n}\subset O_{\alpha}$. However, this contradicts $(\mathrm{ii})$, 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 $E\subset \mathbb{R}^{k}(\mathrm{or}\ \mathbb{C}^{k})$ 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 $p \in E$.

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) $\implies$ (b)
Assuming (a), there exists a $k$-cell $I$ such that $E \subset I$. Since $I$ is compact, and a closed subset of a compact set is compact, $E$ is compact.
(b) $\implies$ (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 $p\in E$ has at most one point of $S$ in its neighborhood $N_{p}$. When $p \in S$, that one point is $p$ itself. And this implies that the open cover $\left\{ N_{p} \right\}$ does not have a finite subcover for $S$. Since $S \subset E$, similarly, there’s no finite subcover for $E$ either, contradicting the assumption that $E$ is compact. Hence, $S$ has an accumulation point $p \in E$.
(c) $\implies$ (a)
This is proved by contradiction.
- part 1. $E$ is bounded
  Let’s assume $E$ is not bounded. Then, $E$ contains points $\mathbf{x}_{n}$ satisfying the following inequality.
  $$ |\mathbf{x}_{n}| >n\quad (n=1,2,\cdots) $$
  Let’s denote $S=\left\{ \mathbf{x}_{n} :n=1,2,\cdots\right\}$. $S$ is infinite and obviously does not have an accumulation point in $\mathbb{R}^{k}, 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 $\mathbf{x}_{0}$ of $E$ that is not included in $E. Now, for $n=1,2,\cdots$, let’s consider $\mathbf{x}_{n} \in E$ satisfying the following conditions.
  $$ \left|\mathbf{x}_{n}-\mathbf{x}_{0} \right| < {\textstyle \frac{1}{n}} $$
  Let’s denote the set of such $\mathbf{x}_{n}$ as $S$. $S$ is infinite and has $\mathbf{x}_{0}$ as an accumulation point. If $\mathbf{x}_{0}$ is the only accumulation point of $S$, then $\mathbf{x}_{0}\notin E$ contradicts (c), proving $E$ is closed. Now, consider $\mathbf{y} \ne\mathbf{x}_{0}$ in $\mathbb{R}^{k}$. Then,
  $$ \begin{align*} \left| \mathbf{x}_{n} - \mathbf{y} \right| & \ge \left|\mathbf{x}_{0} - \mathbf{y} \right| - \left|\mathbf{x}_{n}-\mathbf{x}_{0} \right| \\ & \ge \left| \mathbf{x}_{0} - \mathbf{y} \right| -\frac{1}{n} \end{align*} $$
  For sufficiently large $n$, the following holds true.
  $$ \begin{equation} \left| \mathbf{x}_{n} - \mathbf{y} \right| \ge \left| \mathbf{x}_{0}- \mathbf{y} \right|-\frac{1}{n} \ge \frac{1}{2}\left|\mathbf{x}_{0}-\mathbf{y} \right| \label{eq1} \end{equation} $$
  Furthermore, as $n$ increases, $\mathbf{x}_{n}$ gets closer to $\mathbf{x}_{0}$. This fact, along with $\eqref{eq1}$, implies that we can find a neighborhood of $\mathbf{y}$ that contains no point other than $\mathbf{y}$ as $n$ increases. Thus, $\mathbf{y}$ is not an accumulation point of $S$, proving $\mathbf{x}_{0}$ is the only accumulation point of $S. This contradicts (c), proving $E$ is closed.

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Bolzano-Weierstrass Theorem

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

Proof

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

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Every k Cell is Compact

Definition

Theorem 1

Proof

Theorem 2

Proof

Theorem 3

Proof

Equivalent Conditions for Compactness in Euclidean Space

Proof

Bolzano-Weierstrass Theorem

Proof

See Also

Specialization of Riesz’s Theorem