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Compact Sets and Continuous Functions on Compact Sets

by | Feb 23, 2022 | Math Learning, Math Lecture

INTRODUTION

In advance analysis, the notion of ‘Compact set’ is of paramount importance. In \mathbb{R}, Heine-Borel theorem provides a very simple characterization of compact sets. The definition and techniques used in connection with compactness of sets in \mathbb{R} are extremely important. In fact, the real line sets the platform to initiate the idea of compactness for the first time and the notion of compactness plays its important role in topological spaces.

The definition of compactness of sets in \mathbb{R} uses the notation of open cover of sets in \mathbb{R}. For this propose we need some definitions and illustrative examples to clear the meaning of cover of a set in \mathbb{R}.

COVER, OPEN COVER, SUB COVER

Definition (Cover): Let h be subset of \mathbb{R} and \sigma=\left\lbrace A_\alpha:\ \alpha\in\Lambda\right\rbrace be a collection of sub sets of \mathbb{R}. \sigma is said to cover H or, in other words, \sigma is said to be a covering of H if  H\subset\bigcup_{\alpha\in\Lambda} A_\alpha

i.e. if x\in H\Longrightarrow x\in A_\alpha for some \alpha\in\Lambda.

If A_\alpha, for each \alpha\in\Lambda, is an open set and H\subset\bigcup_{\alpha\in\Lambda} A_\alpha then \sigma is said to be an open cover of H.

For example

  • (i) The family \sigma=\left\{\left(-1,\ 1\right),\ \left(\frac{1}{2},\ \frac{3}{2}\right),\ \left(\frac{5}{4},\frac{5}{2}\right)\right\} is an open cover of [0, 2], since \left[0,\ 2\right]\subset\left(-1,1\ \right)\cup\left(\frac{1}{2},\frac{3}{2}\right)\cup\left(\frac{5}{4},\frac{5}{2}\right) and the sets in family \sigma are open sets.
  • (ii) The family \sigma=\left\{\left(\frac{1}{n+1},\frac{1}{n}\right):n\in\mathbb{N}\right\} is a collection of open sets in \mathbb{R}, but \sigma is not a cover of [0, 1], since 0 and 1 do not belong to \bigcup_{n\in\mathbb{N}}\left(\frac{1}{n+1},\frac{1}{n}\right).
  • (iii) The family \sigma=\left\{\left(\frac{1}{n},\frac{n+1}{n}\right):n\in\mathbb{N}\right\} is a collection of open sets in \mathbb{R} and x\in(0,\ 1]\Rightarrow x\in \left(\dfrac{1}{m},\dfrac{m+1}{m}\right) for some m\in\mathbb{N}. Hence \sigma is an open cover of (0, 1].

Note: If C=\left\lbrace I_n:n\in\mathbb{N}\right\rbrace be a collection of open intervals I_n in \mathbb{R} such that H\subset\bigcup_{n\in\mathbb{N}} I_n then C is also an open cover of H.

Definition (Sub-Cover): Let H\subset\mathbb{R} and \sigma be a collection of sets in \mathbb{R} which covers H. If \sigma^\prime  be a sub-collection of such that \sigma^\prime  itself is a cover of H then \sigma^\prime  is said to be a sub cover of \sigma. If \sigma^\prime  is a finite sub collection of \sigma such that \sigma^\prime  is a cover of H then \sigma^\prime  is said to be a finite sub cover of \sigma.

For example if H=(0,\ \infty) then \sigma=\left\{\left(\frac{1}{n},\ 2n\right):n\in\mathbb{N}\right\} is an open cover of H.

If \sigma^\prime=\left\{\left(\frac{1}{3n},\ 6n\right):n\in\mathbb{N}\right\} then \sigma^\prime\subset\sigma and H\subset\bigcup_{n\in\mathbb{N}}\left(\frac{1}{3n},\ 6n\right); this implies \sigma^\prime is an open sub cover of \sigma.

Definition (Countable set): A set X in \mathbb{R} is said to be a countable set if either it is finite or if it is infinite, it is enumerable i.e. there exists a bijective mapping from \mathbb{N} to X.

For example

  • (i) every finite set is countable,
  • (ii) \mathbb{N},\ \mathbb{Z},\ \mathbb{Q} are all countable sets,
  • (iii) the sets [0, 1],\ (0, 1),\ \mathbb{R} are not countable sets.

Definition (Countable Sub cover): Let H\subset\mathbb{R} and \sigma be a collection of sets in \mathbb{R} such that \sigma covers H. If \sigma^\prime be a countable sub-collection of \sigma such that \sigma^\prime covers H then \sigma^\prime is said to be a countable sub cover of \sigma.

For example, if H=[1,\ \infty) then \sigma=\left\{\left(r-1,\ r+1\right):r\in\mathbb{Q},\ r>0\right\} is an open cover of H and \sigma^\prime=\left\{\left(0,\ n\right):n\in\mathbb{N}\right\} is a countable sub-collection of \sigma, since if I_n=\left(0,\ n\right),\ n\in\mathbb{N}, the set \left\{I_1,\ I_2,\ I_3,\ \ldots\right\} has one-to-one correspondence with \mathbb{N}. \sigma^\prime also covers H. Hence \sigma^\prime is a countable sub cover of H. Note that there are infinitely many countable sub covers of \sigma, since \mathbb{Q} is a countable set and \sigma is the set of open intervals with rational end points and hence \sigma itself is a countable family of open sets so that every sub cover of \sigma is countable.

We now give some examples of open cover of a set in \mathbb{R} which has no finite sub cover.

Example 1. Let H=[0,\ \infty) and \sigma=\{I_n:n\in\mathbb{N}\} where I_n=\left(-1,\ n\right),\ n\in\mathbb{N}. Show that \sigma is an open cover of H but it has no finite sub cover.

Solution: Let c\in H. Then c\geq0 and by the Archimedean property of \mathbb{R}, there exists a natural number p such that p>c\Longrightarrow c\in\left(-1,\ p\right)=I_p for some  p\in\mathbb{N}. Hence  H\subset\bigcup_{n\in\mathbb{N}} I_n. This shows that \sigma=\{I_n:n\in\mathbb{N}\} is a collection of open sets in \mathbb{R} which covers H i.e \sigma is an open cover of H.

If possible, let \sigma^\prime=\{I_{r_1},\ I_{r_2},\ \ldots I_{r_m}\} where r_1,\ r_2,\ ...,\  r_m are natural numbers such that H\subset\bigcup_{k=1}^{m}I_{r_k} i.e. \sigma^\prime covers H.

Let p^\prime=\max{\ \left\{r_1,\ r_2,\ \ldots,\ r_m\right\}} then p^\prime\in\mathbb{N} and for k=1,\ 2,\ \ldots,\ m

I_{r_k}\subset I_{p^\prime}\Longrightarrow H\subset I_{p^\prime}.

Since p^\prime\in\mathbb{N},\ p^\prime\in H but  p^\prime\notin I_{p^\prime}, so we have a contradiction. Hence \sigma^\prime is not a cover of H.

Thus there exists no finite subcollection of \sigma that will cover H.

Example 2. Let  I_n=\left(-n,\ n\right),\ n\in\mathbb{N} and \sigma=\{I_n:n\in\mathbb{N}\}. Show that \sigma is an open cover of \mathbb{R}, but it has no finite sub cover.

Solution: Let x\in\mathbb{R}. Then \vert x\vert\geq0. By the Archimedean property of \mathbb{R} there exists a natural number p such that p>\left|x\right|\Longrightarrow-p<x<p\Longrightarrow x\in I_p for some  p\in\mathbb{N}\ \Longrightarrow\mathbb{R}\subset\bigcup_{n\in\mathbb{N}} I_n, which shows that \sigma=\{I_n:n\in\mathbb{N}\} is an open cover of \mathbb{R} (\sigma is also a countable cover of \mathbb{R}).

If possible, let \sigma^\prime=\{I_{r_1},\ I_{r_2}, \ldots I_{r_m}\} where r_1,\ r_2, ...,  r_m are natural numbers such that \mathbb{R}\subset\bigcup_{k=1}^{m}I_{r_k} i.e. \sigma' is a sub cover of \sigma.

Let  q=\max{\ \left\{r_1,\ r_2,\ \ldots,\ r_m\right\}}. Then q\in\mathbb{N} and I_{r_k}\subset\left(-q,\ q\right)=I_q for all  k=1,\ 2,\ \ldots,\ m.

Thus \mathbb{R}\subset\bigcup_{k=1}^{m}I_{r_k}\subset I_q,\ q\in\mathbb{N}\Longrightarrow q\in\mathbb{R} but q\notin I_q, a contradiction.

Thus it is proved that no finite subcollection of \sigma can cover \mathbb{R}.

Example 3. Let H=(0,\ 1) and  I_n=\left(\frac{1}{n},\frac{2}{n}\right),\ n=2,\ 3,\ 4,\ \ldots Let \sigma=\left\{I_n:n=2,\ 3,\ 4,\ \ldots\right\}. Show that \sigma is an open cover of H but no finite subcollection of \sigma can cover H.

Solution: Let c\in H. Then 0<c<1. For \frac{1}{2}<c<1,c\in I_2. Let 0<c\le\frac{1}{2}. Then \frac{1}{c}\geq2. By the Archimedean property of \mathbb{R}, there exists a natural number p>2 such that p-1\le\frac{1}{c}<p. Since p>2, \frac{p}{2}<p-1\le\frac{1}{c}<p\Longrightarrow\frac{1}{p}<c<\frac{2}{p} for p>2\Longrightarrow c\in I_p for some natural number p>2.

Hence H\subset\bigcup_{n=2}^{\infty}I_n\Longrightarrow\sigma is an open cover of H.

If possible, let \sigma^\prime=\{I_{r_1},\ I_{r_2},\ \ldots I_{r_m}\} be a finite subcollection of \sigma, where r_1,\ r_2, ..., r_m are natural numbers > 1, such that H\subset\bigcup_{k=1}^{m}I_{r_k}.

Let u=\max{\ \left\{r_1,\ r_2,\ \ldots,\ r_m\right\}} and v=\min{\ \left\{\frac{r_1}{2},\frac{r_2}{2},\ \ldots,\frac{r_m}{2}\right\}} then \frac{1}{u}\le\frac{1}{r_k}<\frac{2}{r_k}\le\frac{1}{v}\Longrightarrow I_{r_k}\subset\left(\frac{1}{u},\frac{1}{v}\right) for natural numbers r_k\left(k=1,\ \ldots,\ m\right)>1,\ u\geq2,\ v\geq1.

Thus H\subset\left(\frac{1}{u},\frac{1}{v}\right). Since u\geq2,\ 0<\frac{1}{u}\le\frac{1}{2}. Hence \frac{1}{u}\in H but \frac{1}{u}\notin\left(\frac{1}{u},\frac{1}{v}\right), which is a contradiction. Hence no finite subcollection of \sigma covers H.

Example 4. Let H = (0, 1) and x \in H. Let \sigma = \{Ix : x \in H\} where Ix = \left(\frac{1}{2}x,\frac{1}{2}(x+1)\right). Show that \sigma is an cover of H but it has no finite sub cover.

Solution: Let c \in H. Then 0 < c < 1 \Rightarrow \frac{1}{2} c < c < \frac{1}{2} (c + 1) \Rightarrow c \in Ic.

Hence H \subset \bigcup_{x\in H} I_x which implies \sigma is an open cover of H.

If possible, let \sigma^\prime = \left\{I\right._{x_1,}I_{x_2}\ldots.I_{x_m,}\left.\right\} where xi \in H for I = 1, 2… m and H \subset\bigcup_{i=1}^{m}I_x ,.

Let p = min\left\lbrace \dfrac{1}{2}x_{1}, \dfrac{1}{2}x_{2},\cdots,\dfrac{1}{2}x_{m}\right\rbrace

And q = max \left\lbrace \dfrac{1}{2}(x_{1}+1), \dfrac{1}{2}(x_{2}+2),\cdots,\dfrac{1}{2}(x_{m}+1)\right\rbrace.

Then

O < p \leq \left(\frac{1}{2}x_1< \frac{1}{2}{(x}_1+1)\right) \leq q < 1 \;\;\;\;\;\;\; (i = 1, 2,\ldots, m)

\Rightarrow I_{x_1} = \left(\frac{1}{2}x_1, \frac{1}{2}{(x}_1+1)\right) \subset (p, q) \;\;\;\;\;\;\;\;\;\;\ (i = 1, 2,\ldots, m)

\Rightarrow H \subset (p, q). Since p, q \in H but they do not belong to (p, q), we have a contradiction. Hence \sigma has no finite sub-collection that can cover H, i.e. \sigma has no finite sub cover.

Example 5. The collection \sigma = \{(a, b): a, b \in \mathbb{R} \} of open intervals is an uncountable cover of \mathbb{R} but  * = \{(n, n + 2) : n \in \mathbb{Z} where \mathbb{Z} is the set of integers, is countable sub cover of \mathbb{R}.

Solution: Let c \in\mathbb{R}. Since \mathbb{R} is both unbounded above and unbounded below, thee always exist two real numbers a and b such that a < c < b \Rightarrow\text{ there exists an open interval }(a, b)\text{ in }\mathbb{R} such that \in (a, b) in \mathbb{R} such that \in (a, b). Hence c \in \mathbb{R} \Rightarrow c \in (a, b).

Thus \mathbb{R} \subset U \{(a, b): a, b \in \mathbb{R} \}. As open interval (a, b) is an uncountable subset of \mathbb{R}, so \sigma is an uncountable cover of \mathbb{R}.

Thus c \in \mathbb{R}. Then by the Archimendeon property of real numbers, there exists an integer n such that n + 1 \leq c < n + 2. This implies c \in (n, n + 2) for some n \in \mathbb{Z}. Hence \mathbb{R} \subset U \{(n, n + 2): n \in \mathbb{Z}\}. Thus \sigma^{*} is a countable sub cover of \mathbb{R}.

COMPACT SETS IN \mathbb{R}.

Definition (Compact set):  A set S (\subset\mathbb{R}) is said to be a compact set in \mathbb{R} if every open cover of S has a finite sub cover. More explicitly, S is said to be compact if for any open cover \sigma = \{A: \alpha\in \Lambda\} of S, there is a finite sub collection \sigma' =\left\{A_{\alpha_1}, A_{\alpha_2},\ldots, A_{\alpha_m} \right\} of \sigma such that S \subset \bigcup_{i=1}^{m}A_{\alpha_i} i.e. \sigma' is a finite sub-cover of \sigma.

NOTE: To prove that a set S is compact in \mathbb{R}, we must examine an arbitrary collection of open sets whose union contains S, and show that S is contained in the union of some finite number of sets in the given collection, i.e. we must have to show that any open cover of S has a finite sub-cover. But to prove that a set K is not compact, it is sufficient to choose one particular open cover \sigma has no finite sub-cover, i.e. union of any finite number of sets in \sigma fails to contain K.

Theorem 1 (Heine-Borel Theorem) :

Statement:A close and bounded subset of \mathbb{R} is a compact set in \mathbb{R}, or in other words every open cover of a closed and bounded subset of \mathbb{R} has a finite sub cover.

Proof, Let H be a closed and bounded subset of \mathbb{R}.

Let \sigma = \{A: \lambda \in \Lambda\} be an open cover of H. We assume that \sigma has no finite sub-cover. Then H is not a subject of the union of finite number of open sets in \sigma.

Since H is a bounded subject of \mathbb{R}, there exist real number a_1, b_1 (> a_1) such that H \subset [a_1 + b_1].

Let I_1 = [a_1 + b_1]. If c_1 = \frac{1}{2} (a_1 + b_1) then at least one of the two subsets H \cap [a_1 , c_1] and H \cap [c_1 , b_1] are subset of the union of finite number of open sets in \sigma, for otherwise both H \cap [a_1 , c_1] and H \cap [c_1 , b_1] are subsets of the union of finite number of open sets in \sigma contains H, contradicting our assumption that \sigma has no finite sub-cover.

We call I_2 = [a_1 , c_1] or [c_1 , b_1] according as H \cap [a_1 , c_1] \neq \emptyset and it is not a subset of the union of finite number of open sets in \sigma or H \cap [a_1 , b_1] \neq \emptyset and it is not a subset of the union of finite number of open sets in \sigma.

Let I_2 = [a_2, b_2] and c_2 = \frac{1}{2} (a_2 + b_2). The at least one of the subsets H \cap [a_2 , c_2] and H \cap [c_2 , b_2] is non-empty and it is not a subset of the union of finite number of open sets in \sigma. If the first subset is non-empty and it is not a subset of the union of finite number of open sets in \sigma, we call I_3 = [a_2 , c_2], otherwise we call I_3 = [a_2 , b_2].

Let I_3 = [a_3 , b_3], and C_3 = \frac{1}{2} (a_3 + b_3).

Continuing this process of bisection of intervals, we have a family of close and bounded intervals   \bm{\{I_n: n \in \mathbb{N}\}} such that

I_{n+1} \subset I_n, for all n \in \mathbb{N},

For all n \in \mathbb{N},\ H \cap I_n is non-empty and it is not a subset of the union of finite number of open sets in \sigma.

\bm{\vert I_n\vert = }\frac{\bm{1}}{\bm{2^{n-1}}}\bm{(b_1 - a_1)} such That \bm{\vert I_n\vert \rightarrow 0} as \bm{n \rightarrow \infty}.

Then by Nested Interval Theorem, \bm{\bigcap_{n\in\mathbb{N}}I_{n} = \{\alpha\}}, a singleton set. We shall show that \alpha \in H.

Since \bm{\lim_{n\rightarrow\infty}{\vert I_{n}\vert} = 0}, for any positive \epsilon, there exists a natural number p such that \vert I_p\vert < \epsilon i.e. b_p - a_p < \epsilon and \alpha \in I_p \Rightarrow a_p < \alpha < b_p \Rightarrow \alpha - \epsilon < b_p - \epsilon < a_p < \alpha < b_p < a_p + < \alpha + \epsilon. Hence \alpha \in I_p \Rightarrow I_p \subset (\alpha - \epsilon, \alpha + \epsilon). Since H \cap I_p \neq \emptyset and it is not a subset of the union of finite number of open sets in \sigma, I_p contains infinite number of elements of H \Rightarrow \alpha is a limit point of H. Since H is closed, \alpha \subset H.

Now \alpha \in H \Rightarrow \alpha \in -A_{\lambda} for some \lambda \in \Lambda. A_{\lambda} is an open set, hence there exists a positive \delta and hence I_k \subset (\alpha - \delta, \alpha + \delta) \Rightarrow I_k \subset A_{\lambda}. Since \sigma is an open cover of H, H \cap I_k \subset A_{\lambda} for some \lambda \in \Lambda, which goes against the construction of (I_n : n \in \mathbb{N}\}.

Hence our assumption that H is not a subset of the union of finite number of sets in \sigma is wrong and it is established that if H is closed and bounded, any open cover \sigma of H has a finite sub cover so that H is a compact set in \mathbb{R}.

Remark:  In the Heine-Borel theorem neither of the two conditions (i) H is closed (ii) H is bounded can be dropped. The theorem fails if one of the two conditions is withdrawn – this is evident if we go through the example 1.2.3 and the example 1.2.1. In example 1.2.3, H is closed but bounded and in example 1.2.1, H is closed but no bounded.

Thus both the conditions (i) and (ii) are necessary for a set in \mathbb{R} to be compact. Next we shall show that these two conditions are also sufficient for a set H to be compact in \mathbb{R}.

Theorem 2 (Converse of Heine-Borel Theorem):

Statement:A compact subset of \mathbb{R} is closed and bounded in \mathbb{R}.

Proof. Let H be a compact in \mathbb{R}. First we shall prove that H is a closed set in \mathbb{R}.

Let x \in H and y \in \mathbb{R} - H. Then exist two positive numbers \epsilon_x and \delta_x such that N(x,\ \delta_x) \cap N(y,\ \epsilon_x) = \emptyset.

Let \sigma = \{N(x,\ \delta_x): x \in H\text{ and }N(x,\ \delta_x) \cap N(y, \epsilon_x) = \emptyset\}.

Then \sigma is an open cover of H is compact, \sigma has a finite sub cover i.e. there exist elements x_1, x_2,\ldots, x_m of H and positive numbers \delta_{x_1}, \delta_{x_1}, \ldots, \delta_{x_m} such that H \subset \bigcup_{i=1}^{m}{N\ (x_i,\ \delta_{x_i}}). For each \delta_{x_i} (I = 1, 2,\ldots, m) there exists positive numbers \epsilon_{x_i} (i = 1, 2,\ldots, m) such that N(x_i, \delta_{x_i}) \cap N (y_i, \epsilon_{x_i}) = \emptyset (i = 1, 2,\ldots, m).

Let \epsilon ' = \min \{ \epsilon_{x_1},\epsilon_{x_2},\ldots, \epsilon_{x_m}\} > 0.

Then N (x_i,\delta_{x_i}) \cap N (x_i, \epsilon ') = \emptyset\text{ for } i = 1, 2,\ldots, m \text{ and }N (y, \epsilon ') \subset \mathbb{R} - H. Therefore y is an interior point of \mathbb{R} - H.

Since y is arbitrary point of \mathbb{R} - H, \mathbb{R} - H is open. Hence H is closed.

Nest we shall prove that H is bounded.

Let \delta be a fixed positive number. Then C = \{N(x, \delta) : x \in H\} is an open cover of H. Since His compact, C has a finite sub-cover. Then there exist points x_1, x_2,\ldots,x_m of H such that C' = \{N(x_1, \delta), N(x_2, \delta),\ldots, N(x_m, \delta)\} is a finite cover of H. If p = \min\{x_1, x_2,\ldots x_p\} and q = \max\{x_1, x_2, \ldots, x_m\} then H \subset \bigcup_{i=1}^{m}N(x_i, \delta) \subset [p - \delta, q + \delta] \Rightarrow H is bounded.

Hence it is proved that if H is a compact set in \mathbb{R}, it is closed and bounded in \mathbb{R}. This completes the proof.

Combining the theorems 1 and 2 we have the following theorem which gives a complete characterization of compact sets in \mathbb{R}.

Note:A set in \mathbb{R} is compact if and only if is closed and bounded in \mathbb{R}.

 

Definition (Heine-Borel Property): A set S (\subset\mathbb{R}) is said to possess Heine-Borel property if every open cover of S has a finite sub cover.

A set is said to be compacted if it has the Heine-Borel property.

Example 6. Using the definition of compact set, prove that the set [0, \infty) is not compact although it is a closed set in \mathbb{R}.

Solution: In example 1.2.1, it is shown that \sigma = \{I_{n} : n \in \mathbb{N}\}, where I_n = (-1, n), n \in \mathbb{N}, is an open cover of H = [0, \infty) and \sigma has no finite sub cover. Hence from definition H is not compact.

H is a closed set in \mathbb{R}, since \mathbb{R} - H is open.

Note: In the example 1.3.1, H does not satisfy Heine-Borel property, since H is not bounded in \mathbb{R}.

Example 7. Using definition of compact set show that a finite subset of \mathbb{R} is a compact set in \mathbb{R}.

Solution: Let S = \{x_1, x_2,\ldots, x_n\} be a finite subset of \mathbb{R}. Let \sigma = \{A: \lambda \in \Lambda\} be an open cover of S. Then each x_i is contained in some open set A_{\lambda_i} of \sigma for some \lambda_I \in \Lambda. Let \sigma' = \{A_{\lambda_1}, A_{\lambda_2},\ldots, A_{\lambda_n}\}. Then S \subset\bigcup_{i=1}^{n}A_{\lambda_1}. Thus \sigma' also covers S.

Hence \sigma' is a finite sub cover of \sigma. Therefore, by definition, S is a compact set in \mathbb{R}.

Theorem 3.

Statement:– If S be a compact subset of \mathbb{R}, then every infinite subset of S has a limit point belonging to S.

Let T be an infinite subset of the compact subset of S of \mathbb{R} such that T has no limit point belonging to S.

Let x \in S. Then x is not a limit point of T. There exists a positive \delta such that N'(x, \delta) \cap T = \emptyset where N'(x, \delta) = (x - \delta, x + \delta) - \{x\}, called deleted \delta-\text{ neighborhood of }x.

Let \sigma = \{N(x, \delta): x \in S, N'(x, \delta) \cap T = \emptyset\}, which is a collection of open sets in \mathbb{R}. Since S \subset \bigcup_{x\in S} N\left(x,\ \delta\right) so \sigma is an open cover of S.

Since S is compact, there exists a finite sub collection \sigma' = \{N(x_1, \delta), N(x_2, \delta),\ldots, N(x_m, \delta)\} of \sigma where x_i \in S and N'(x_i, \delta) \cap T = \emptyset,\ (i = 1, 2, \ldots, m) such that \sigma' covers S i.e. S \subset \bigcup_{i=1}^{m}{N(x_i,\ \delta)}

i.e. S \subset \left\lbrace \bigcup_{i=1}^{m}{N(x_i,\ \delta)}\right\rbrace\cup \{x_1, x_2,\ldots, x_m\}

\Rightarrow T \cap S \subset\bigcup_{i=1}^{m}{(T\cap N'(x_i,\ \delta))} \cup (T \cap \{x_1, x_2,\ldots, x_m\})

\Rightarrow T \subset T \cap \{x_1, x_2,\ldots, x_m\}

[Since T \subset S and T \cap N'(x_i,\ \delta) = \emptyset for I = 1, 2, …., m]

\Rightarrow T \subset \{x_1, x_2,\ldots, x_m\}

which shows that T is a finite subset of a compact set S in \mathbb{R} has a limit point in S.

Theorem 4.

Statement: – If S \subset \mathbb{R} be such that every infinite subset of S has a limit point in S then S is closed and bounded in \mathbb{R}.

Proof. First we shall prove that S is bounded. If possible, let S be unbounded above. Let x_1 be any point of S. Since S is unbounded above, there exists a point x_2 in S such that x_2 > x_1 + 1. By similar argument there exists a point x_3 in S such that x_3 > x_2 + 1 and so on. Continuing this process indefinitely we ultimately have an infinite subject \{x_1, x_2, x_3,\ldots\} of S, which being a discrete set, has no limit point in S is a bounded above subset of \mathbb{R}. Similarly, If S is unbounded below we can construct an infinite subset of S which has no limit point. Hence S is also bounded below so that S is a bounded subset of \mathbb{R}.

Next we shall prove that S is closed in \mathbb{R}.

Since S is an infinite and bounded subset of \mathbb{R}, by the Bolzano-Weierstrass theorem on set, S has a limit point in \mathbb{R}.

Let \alpha be a limit point of S. Then for any \epsilon > 0, N'(\alpha,\ \epsilon) \cap S is infinite.

For \epsilon =1, N' (\alpha,\ 1) \cap S is infinite. Let us take a point x_1 \in N'(\alpha, 1) \cap S.

For \epsilon =\frac{1}{2}, N' (\alpha, \frac{1}{2}) \cap S is infinite. Let us take a point x_2 \neq x_1,

such that x_2   N'(\alpha, \frac{1}{2}) \cap S, Continuing this process, we have an infinite subset

T = \{x_1, x_2,\ldots, x_n,\ldots\} of S such that

x_i \in N' (\alpha, \frac{1}{i}) for I = 1, 2,\ldots, n,\ldots. We shall show that T has a unique limit point which is \alpha.

Let \delta be any positive number. Then by the Archimedean property of \mathbb{R}, there exists a natural number m such that 0 < \frac{1}{m} < \delta \Rightarrow N (\alpha,\frac{1}{m}) \subset N (\alpha, \delta) and N(\alpha, \delta) contains infinite subset \{x_m, x_{m + 1},\ldots\} of T. Thus for every positive \delta, N (\alpha, \delta) \cap T is infinite which proves that \alpha is a limit point of T.

To prove uniqueness, let \beta (\neq \alpha) be a limit point of T. Let 2_{\varepsilon_0} = \vert\beta - \alpha\vert > 0. Then the neighborhoods N(\alpha, \varepsilon_0) and N(\beta, \varepsilon_0) are disjoint (since either \alpha - \varepsilon_0 = \beta + \varepsilon_0 or \alpha + \varepsilon_0 = \beta - \varepsilon_0). By the Archimedean property of \mathbb{R}, there exists a natural number p such that 0 < \frac{1}{p} < \varepsilon_0 \Rightarrow N (\alpha, \frac{1}{p}) \subset N(\alpha, \varepsilon_0). Since each of x_p, x_{p +1},\ldots belongs to N(\alpha, \frac{1}{p}), so N(\alpha, \varepsilon_0) contains all elements of T expect some finite number of elements and hence N(\beta, \varepsilon_0) can contain almost finite number of elements of T. This implies \beta is not a limit point of T. Hence \alpha is the only limit point of T. By the condition of the theorem \alpha \in S. Hence S is closed.

Thus it is proved that S is closed and bounded in \mathbb{R}.

Note:A subset S of a compact subset of \mathbb{R} is compact if and only if every infinite subset of S has a limit point belonging to S.

Theorem 5.

Statement: – A Subset S of \mathbb{R} is compact if and only if every sequence in S has a subsequence that converges to a point in S.

Proof. Let S be compact. Then S is closed and bounded.

Let \{x_n\}_n be a sequence of points in S. Since S is bounded, \{x_n\}_n is bounded. By the Bolzano-Weierstrass theorem on sequence, there exists a subsequence \{x_{r_n}\} of \{x_n\}_n that converges to a point, say x. Since S is closed, if x \notin S, x \in S^c and S^c is open. Then there exists a neighborhood N(x) of x which contains no point of S. This implies N(x) contains no element of the sequence \{x_{r_n}\} which contradicts that \lim x_{r_n} = x. Thus x \in S. Hence every sequence in S has a subsequence converging to a point of S.

Suppose S is not closed. Then S has a limit point, say \alpha which is not in S. Since \alpha is a limit point of S, there is a sequence \{x_n\}_n in S, where x_n \neq \alpha for all n \in N, such that \lim x_n = \alpha. Then every subsequence of \{x_n\}_n converges to \alpha. Since \alpha \notin S, there is no subsequence of \{x_n\}_n that converges to a point of S.

Suppose S is not bounded. Then there exists a sequence \{x_n\}_n in S such that x_n > n for all n \in \mathbb{N}. Then every subsequence of unbounded sequence \{x_n\}_n is unbounded and hence no subsequence of \{x_n\}_n converges to a point in S.

Hence, by contrapositive argument, it is proved that if every sequence in S has a subsequence that converges to a point of S the S is closed and bounded and hence by Heine-Borel theorem S is compact.

Note:  Following theorem 5., an alternative definition of compact set can be given in the from:

“A set S in \mathbb{R} is called a compact set in \mathbb{R} if every sequence in S has a subsequence that converges to a point of S.”

Theorem 5 and the Heine-Borel theorem together prove the equivalence of the two definitions.

Example 8. If F is a closed subset of a compact set S in \mathbb{R} then using definition of compact set, prove that F is compact.

Solution: F^c = \mathbb{R} - F is open, since F is closed.

Let \sigma = \{A : \lambda \in \Lambda\} be an open cover of F. Suppose \sigma is not an open cover of S. Let \sigma' = \{A : \lambda \in \Lambda\} \cup F^c. Then R \subset \left(\bigcup_{\lambda\in\land} A_\lambda\right) \cup F^c i.e., \sigma' is an open cover of \mathbb{R}. Since S \subset \mathbb{R}, \sigma' is also an open cover of S. S being compact, \sigma' has a finite sub collection \sigma'' = \{A_{\lambda_1}, A_{\lambda_2}, \ldots, A_{\lambda_m}, F^c\} such that S \subset \left\lbrace\bigcup_{i=1}^{m}A_{\lambda_i}\right\rbrace \cup F^c, where \lambda_i \in \Lambda. \sigma'' must contain F^c, otherwise S \subset \bigcup_{i=1}^{m}A_{\lambda_i} which implies \sigma is an open cover of S, which is contrary to our assumption.

Since F \subset S we have F \subset \bigcup_{i=1}^{m}A_{\lambda_i}.

Which shows that,

\sigma''' = \{A_{\lambda_1}, A_{\lambda_2}, \ldots, A_{\lambda_m}\} is a finite sub collection of \sigma and \sigma''' covers F, which implies \sigma''' is a finite sub cover of \sigma. Therefore, by definition, F is compact.

Example 9. Every compact set in \mathbb{R} has greatest as well as least element.

Solution: Let S be any compact set in \mathbb{R}. If possible, let S have no greatest element. Then for each element p \in S. let A_p = \{x \in \mathbb{R} : x < p\} = (-\infty, p). Then A_p is an open set. Let \sigma = \{A_p : p \in S\}, a family of open sets in \mathbb{R}. Let y \in S. Since S has no greatest element, there exists an element q in S such that q > y \Rightarrow y \in A_q. Thus y \in S \Rightarrow y \in A_q. Hence \sigma is an open cover of S. S being compact, \sigma has a finite sub-cover, say \sigma'.

Let \sigma' = \{A_{p_1}, A_{p_2}, \ldots, A_{p_m}\} Then p_i \in S for I = 1, 2,\ldots, m and S \subset \bigcup_{i=1}^{m}A_{p_i}. Let p_0 = \max \{p_1, p_2,\ldots, p_m\}.

Then S \subset A_{p_0} and p_0 \in S. This leads to a contradiction, since p_0 \notin A_{p_0}. Hence our assumption is not tenable and S has greatest element.

To prove the next part, let, if possible, S has no least element. Then for each element u \in  S, let A_u is an open set. Let \sigma = \{A_u : u \in S\}, a family of open sets in \mathbb{R}. Let z \in S. Since S has no least element there exists an element v in S such that z > v. Then z \in A_v and z \in S \Rightarrow z \in A_v. Thus \sigma is an open cover of S. Since S is compact, \sigma has a finite sub collection \sigma' that covers S.

Let \sigma' = \{A_{u_1}, A_{u_2}, \ldots, A_{u_m}\} where u_i \in S (I = 1, 2, \ldots, n) and s \subset \bigcup_{i=1}^{m}A_{u_i}. If  u_0 = \min\{u_1, u_2, \ldots, u_n\} then A_{u_i} \subset A_{u_0} \Rightarrow S \subset A_{u_0}.

Now u_0 \in S but u_0 \notin A_{u_0} which contradicts our assumption. Hence S has least element.

 

Example 10.  If K and T are component sets in \mathbb{R}, show that K \cup T is also compact. Give an example to show that union of an infinite number of compact sets in \mathbb{R} is not necessarily a compact set in \mathbb{R}.

Solution: Let \sigma = \{A : \lambda \in \Lambda\} be a family of open sets in \mathbb{R} such that K \cup T \subset \bigcup_{\lambda\in\land} A_\lambda i.e. \sigma is an open cover of K \cup T.

Since K \subset K \cup T \subset \bigcup_{\lambda\in\land} A_\lambda and T \subset K \cup T \subset\bigcup_{\lambda\in\land} A_\lambda,

\sigma is an open cover of both K and T. Since K and T are both compact sets in \mathbb{R}, then there exist two finite sub collections

\sigma' = \left\{A_{m_1},\ A_{m_2}, \ldots,\ A_{m_p}\right\} and \sigma'' = \left\{A_{n_1},\ A_{n_2},\ldots,\ A_{n_q}\right\} of \sigma such that

k \subset \bigcup_{i=1}^{p}A_{m_i} and T \subset \bigcup_{j=1}^{q}A_{n_j} where m_i \in \Lambda (i = 1, 2,\ldots, p) and n_j \in \Lambda (j = 1, 2,\ldots, q).

Let \sigma''' = \sigma' \cup \sigma''. Then \sigma''' is a finite sub collection of \sigma such that

x \in K \cup T \Rightarrow x \in K or x \in T

\Rightarrow x \in A_{m_i} for some i = 1, 2,\ldots, p

Or x \in A_{n_j} for some j = 1, 2, \ldots, q

\Rightarrow x \in \left\lbrace\bigcup_{i=1}^{p}{A_{m_i}}\right\rbrace \cup \left\lbrace\bigcup_{j=1}^{q}{A_{n_j}}\right\rbrace

Hence K \cup T \subset \left\lbrace\bigcup_{i=1}^{p}{A_{m_r}}\right\rbrace \cup \left\lbrace\bigcup_{j=1}^{q}{A_{n_j}}\right\rbrace

\Rightarrow \sigma''' cover K \cup T. Hence, from definition, K \cup T is compact.

Second Part.

Let A_n = [- n, n], n \in \mathbb{N}. Then for each n \in \mathbb{N}, A_n is closed and bounded set in \mathbb{R} and by Heine-Borel Theorem, A_n is compact for every n \in \mathbb{N}.