Show That the Lower Level Sets of a Continuous Function Are Closed

INTRODUTION
COVER, OPEN COVER, SUB COVER
COMPACT SETS IN
- Theorem 1 (Heine-Borel Theorem) :
- Theorem 2 (Converse of Heine-Borel Theorem):
- Theorem 3.
- Theorem 4.
- Theorem 5.
  - - Bibliography

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

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

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 $H$ is 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 $n \in \mathbb{N}$ . Thus $\{A_n : n \in \mathbb{N}\}$ is an infinite collection of compact sets in $\mathbb{R}$ .

But $\bigcup_{n=1}^{\infty}{A_n} =(- \infty, \infty) = \mathbb{R}$ , which is not a compact set (see example 1.2.2). Hence union of infinite number of compact sets in $\mathbb{R}$ is not necessarily compact.

Example 11. Let $A$ be a closed subset of $\mathbb{R}$ and $B$ be a component subset of $\mathbb{R}$ . Prove that $A \cap B$ is component.

Solution: Since $B$ is compact, by converse of Heine-Borel theorem $B$ is closed. $A$ being closed, $F = A \cap B$ is a closed subset of compact set $B$ . Then following exactly similar arguments given in example 1.3.3 (replacing $Example 12. Prove that intersection of infinite collection of compact sets in$ \mathbb{R} $is compact. Solution: Let$ \sigma = \{C\} $be an infinite collection of compact sets in$ \mathbb{R} $. Let$ S =\ \bigcap_{C\in\sigma} C $. Since every$ C $belonging to$ \sigma $is compact, it is closed and hence$ S $is a closed subset or$ R $, which implies$ \mathbb{R}-S $is open. Let$ g' = \{G : \alpha \in \Lambda\} $be a family of open sets in$ \mathbb{R} $which covers$ S $, i.e.$ S \subset \ \bigcap_{\alpha\in\land} G_\alpha $. Suppose$ g $is not an open cover of$ C $for any$ C \in \sigma $. Let$ g' $be a family of open sets$ \{G : \alpha \in \Lambda\} $together with$ \mathbb{R} – S $. Then$ \mathbb{R} \subset \left\lbrace\bigcup_{\alpha\in\land} G_\alpha\right\rbrace \cup \left(\mathbb{R} – S\right) $. Since$ C \subset \mathbb{R} $,$ C \subset \left\lbrace \bigcup_{\alpha\in\land} G_\alpha\right\rbrace \cup \left(\mathbb{R} – S\right) $for every$ C \in \sigma $. Then$ g' $is an open cover of$ C $for every$ C \in \sigma $. As$ C $is compact for every$ C \in \sigma $, there exists a sub collection$ g" $of$ g' $such that$ g" $also covers$ C $for every$ C \in \sigma $. Let$ f" = \{G_{\alpha_1},\ G_{\alpha_2}, \ldots,\ G_{\alpha_m},\ \mathbb{R}\ – S\} $.$ g" $must contain$ \mathbb{R} – S $, for otherwise$ C \subset \bigcup_{i=1}^{m}G_{\alpha_i} $which contradicts our assumption$ g $is not and open cover of$ C $for any$ C \in \sigma $. Therefore$ C \subset \left\lbrace\bigcup_{i=1}^{m}G_{\alpha_i}\right\rbrace \cup (\mathbb{R} – S) $for every$ C \in \sigma $. Since$ S \subset C $,$ S \subset \bigcup_{i=1}^{m}G_{\alpha_i} $. Let$ g"' = \{G_{\alpha_1},\ G_{\alpha_2}, \ldots,\ G_{\alpha_m}\} $. Then$ g"' $is a finite sub collection of$ g $that covers$ S $. Hence$ g $has a finite sub cover, which proves that$ S $is component. Example 13. Using definition of compact set prove that the set$ (0, 1] $is not a compact set in$ \mathbb{R} $. Solution: Let$ I_n = \left(\frac{1}{n+1},\ \frac{n+1}{n}\right), n \in \mathbb{N} $and$ \sigma = \{I_n : n \in \mathbb{N}\} $.$ \sigma $is a collection of open sets in$ \mathbb{R} $. Let$ x \in (0, 1] $. If$ x = 1 $the$ x \in I_n $for all$ n \in \mathbb{N} $. If$ 0< x < 1 $, then, by the Archimedean property of$ \mathbb{R} $, there exists a natural number$ m $such that$ m \le\ \frac{1}{x} < m + 1

\Rightarrow \frac{1}{m+1} < x \le\ \frac{1}{m} <\ \frac{m+1}{m} \Rightarrow x \in I_m $for some$ m \in \mathbb{N} $. Hence$ (0, 1] \subset \bigcup_{n\in\mathbb{N}} I_n $and therefore,$ \sigma $is an open cover of$ (0, 1] $. If possible, let$ \sigma' = \{I_{r_1},\ I_{r_2}, \ldots,\ I_{r_m}\} $be a finite sub collection of$ \sigma $, where$ r_1, r_2, \ldots, r_m $are natural numbers greater than 1, such that$ (0, 1] \subset \bigcup_{k=1}^{m}I_{r_k} $Let$ u = \max \{r_1 + 1, r_2 + 1,\ldots, r_m + 1\} \Rightarrow u \geq r_k + 1 $for all$ k = 1, 2, \ldots, m $and$ v = \min \{\frac{r_1}{r_1+ 1},\ \frac{r_2}{r_2+ 1}, \ldots,\ \frac{r_m}{r_m+\ 1}\} \Rightarrow v \leq \frac{r_k}{r_k+ 1} $for all$ k = 1, 2, \ldots, m $Then$ 0 < \frac{1}{u} \leq \frac{1}{r_k+ 1} < \frac{r_k+ 1}{r_k} \leq \frac{1}{v},\ x = 1, 2,\ldots, m\Rightarrow I_{r_k} \subset \left(\frac{1}{u},\ \frac{1}{v}\right) $for all$ k = 1, 2,\ldots, m\Rightarrow (0, 1] \subset \left(\frac{1}{u},\ \frac{1}{v}\right) $. Since$ 0 < \frac{1}{u} < 1 $, so$ \frac{1}{u} \in (0, 1] $but$ \frac{1}{u} \left(\frac{1}{u},\ \frac{1}{v}\right) $, which is a contradiction. Hence no finite sub collection of$ \sigma $cover$ (0, 1] $, proving that$ (0, 1] $is not a compact set. Example 14. Let$ a < x 0\} $. Is$ \Gamma $an open cover of$ [a, b] $having a finite sub cover? Justify your answer. Solution: Let$ I_x = (x – \varepsilon, x + \varepsilon),\ \varepsilon > 0 $and$ a < x < b $. Then$ \Gamma = \{I_x : a < x < b\} $. Clearly every element$ x $of$ (a, b) $is contained in$ I_x $. By the density property of$ \mathbb{R} $, there exist elements$ x_1, x_2 \in (a, b) $such that$ a < x_1 < a + \varepsilon $and$ b – \varepsilon < x_2 a $and$ b < x_2 + \varepsilon, x_2 – \varepsilon < b $. Thus$ a \in I_{x_1} $and$ b \in I_{x_2} $. Hence it is shown that$ [a, b] \subset \bigcap_{a<x<b} I_x \Rightarrow \Gamma $is an open cover of$ [a, b] $. Since$ [a, b] $is closed and bounded it is compact by Heine-Borel theorem. Hence, by definition of compact set,$ \Gamma $has a finite sub cover. <h2>Theorem 6. (Lindelof's Covering Theorem).</h2> Statement: - Let$ S \subset\mathbb{R} $. Then every open cover of$ S $has a countable sub-cover. Proof. Let$ C = \{A : \alpha \in \Lambda\} $Be a collection of open sets$ A_\alpha $in$ \mathbb{R} $,$ \Lambda $being the index set, which covers$ S $. Then$ S \subset \bigcup_{\alpha\in\land} A_\alpha $. Let$ x \in S $. Then there exists$ \alpha(x) \in \Lambda $such that$ x \in A_{\alpha(x)} $. Since$ A_{\alpha(x)} $is open,$ x $is its interior point. There exists a neighborhood$ I(x) $(which is an open interval) of$ x $such that$ x \in I(x) \subset A_{\alpha(x)} $. Since$ I(x) $is an open interval, we can always choose and open interval$ J(x) \subset I(x) $such that$ x \in J(x) $and$ j(x) $has rational and points. Hence$ S \subset \bigcup_{x\in S} I\left(x\right) $so that$ \{J(x) : x \in S\} $is a collection of open intervals with rational end points, that covers$ S $. Since we know that a set of open intervals with rational end points is countable, the family$ \{J(x) : x \in S\} $can be enumerated as$ C' = \{j_1, j_2, j_3, \ldots\} $, which is an open cover of$ S $. For each$ J_n \in C' $, there exists an element$ x_n \in S $such that$ x_n \in J_n \subset I(x_n) \subset A_{\alpha_n} $, say. Thus for each$ J_n \in C' $there corresponds to an open set$ A_{\alpha_n} \in C $. If$ C" = \{A_{\alpha_1},\ A_{\alpha_2}, \ldots,\ A_{\alpha_n}, \ldots\} $then$ C" $is a countable sub-collection of$ C $, which also covers$ S $. Hence the proof. Example 15. The set$ \bm{\{0\} \cup \left\{ }\frac{\bm{1}}{\bm{n}}\bm{:n\in\ \mathbb{N}\right\rbrace} $is a compact set. Solution: Let$ S =\left\lbrace 0,\ 1,\ \frac{1}{2},\ \frac{1}{3}, \ldots\right\rbrace $. Let$ T $be any infinite subset of$ S $. Then either$ T = \left\lbrace 0,\ \frac{1}{m}\ :m\ \in\ N_1\right\rbrace $or$ T = \left\lbrace\frac{1}{m}\ :m\ \in\ N_1\right\rbrace $where$ N_1 $is any infinite subset of$ \mathbb{N} $. In any case$ 0 $is the only limit point of$ T $and$ 0 \in S $. Then every infinite subset of$ S $has a limit point in$ S $so (by Theorem 4.)$ S $is closed and bounded and hence by Heine-Borel theorem,$ S $is compact. Example 16. Construct a compact set in$ \mathbb{R} $whose accumulation point from a countably infinite set in$ \mathbb{R} $. Solution: Let$ A = \left\lbrace\frac{1}{m}+\frac{1}{n}\ :m\ \in\ \mathbb{N},\ n\ \in\mathbb{N}\right\rbrace $,$ B = \left\lbrace 0,\ \frac{1}{m}\ :m\ \in\ \mathbb{N}\right\rbrace $, and$ C = \{0\} $. Then$ S = A \cup B \cup C $is a bounded subject of$ \mathbb{R} $, since$ S \subset [0, 2] $. Now derived sets of$ A,\ B,\ C $are respectively$ A' =\left\lbrace 0,\ \frac{1}{m}\ :m\ \in\ \mathbb{N}\right\rbrace $,$ B' = \{0\} $,$ C' = \emptyset $. Thus$ S' = A' \cup B' \cup C' = \{0\} \cup \left\lbrace\frac{1}{m}\ :m\ \in\ \mathbb{N}\right\rbrace = B \cup C \subset S $. Hence$ S $is closed. Thus$ S $is a closed and bounded set in$ \mathbb{R} $and hence$ S $is a compact set in$ \mathbb{R} $. Now the mapping$ f : \mathbb{N} \to B $defined by$ f(m) = \frac{1}{m},\ m \in\ \mathbb{N} $, is clearly a bijective mapping and hence$ B $is enumerable. Thus$ S' $is the union of a finite set and an enumerable set, which implies$ S' $is a countably infinite set in$ \mathbb{R} $. <h2>Theorem 7. (Cantor's Theorem).</h2> Statement: - If$ \{F_n\}_n $is a nesting sequence of non-empty compact sets in$ \mathbb{R} $then$ \bigcap_{n\in\mathbb{N}} F_n $is non-empty. Proof : Since$ \{F_n\}_n $is a nesting sequence of non-empty compact set,$ F_n \supset F_{n+1} $for all$ n \in \mathbb{N} $, and$ F_n $is closed and bounded for all$ n \in \mathbb{N} $. Also$ F_n \neq \emptyset $, for every$ n \in \mathbb{N} $. We choose the fixed point$ x_n \in F_n $. Thus we have a sequence$ \{x_n\}_n $such that$ x_n \in F_n \subset F_m $for$ m \leq n $. If$ X $be the range of the sequence$ \{x_n\}_n $then either$ X $is finite or infinite. Case-1:$ X $is finite. Then there is a point$ \alpha \in X $such that$ x_n = \alpha $for infinite values of$ n $in$ \mathbb{N} $. For a fixed$ p \in \mathbb{N} $, there is a natural number$ k \geq p $such that$ x_k = \alpha $and hence$ \alpha = x_k \in F_k \subset F_p $. This result is true for every$ p \in \mathbb{N} $. Hence$ \alpha \in \bigcap_{n\in\mathbb{N}} F_n \Rightarrow \bigcap_{n\in\mathbb{N}} F_n \neq \emptyset $. Case-2:$ X $is infinite. Since each$ F_n $is bounded,$ F_1 \supset X $and hence$ X $is bounded. By Bolzano-Weierstrass theorem,$ X $has a limit point, say$ \beta $. Let$ m $be fixed positive integer. Since$ \beta $is a limit point of$ X $, any neighbourhood of$ \beta $,$ N(\beta) $contains infinite number of points of$ X $and hence$ N(\beta) $contains infinite number of distinct points$ x_n $'s. As$ x_n \in F_m $for$ n \geq m $, it follows that$ N(\beta) \cap F_m $is an infinite set. Hence$ \beta $is a limit point of$ F_m $. Since$ F_m $is closed,$ \beta \in F_m $. This result is true for every$ m \in \mathbb{N} $, hence$ \beta \in \bigcap_{n\in\mathbb{N}} F_n \Rightarrow \bigcap_{n\in\mathbb{N}}{F^\prime}_n \neq \emptyset $. Thus it is proved that, in either case,$ \bigcap_{n\in\mathbb{N}}{F^\prime}_n $is non-empty. Example 16. Let$ E = \{r \in \mathbb{Q} : \sqrt{2} < r < \sqrt{3} \} $. We are going to show that$ E $is not compact in$ \mathbb{Q} $, but$ E $is Closed and bounded in$ \mathbb{Q} $. Explain why this does not contradict the Heine-Borel theorem. Solution: By the Density property of$ \mathbb{R} $, there exist rational numbers$ x_1 $and$ y_1 $in$ E $($ \subset \mathbb{Q} $) such that$ \sqrt{2} < x_1 < y_1 < \sqrt{3} $. Again by the same property there exist rational numbers$ x_2 $and$ y_2 $in$ E $such that$ \sqrt{2} < x_2 < x_1 < y_1 < x_2 < \sqrt3 $. Continuing this way we have sequences of rational points in$ E $,$ \{x_n\}_n $and$ \{y_n\}_n $such that both are bounded;$ \{x_n\}_n $is monotone decreasing which converges to$ \sqrt{2} $and$ \{y_n\}_n $is monotone increasing which converges to$ \sqrt3 $. Thus we have an infinite set of open intervals$ \sigma = \{I_n : n \in \mathbb{N}\} $where$ I_n = (x_n,\ y_n) $such that$ \sqrt2 < x_n < y_n < \sqrt3 $for all$ n \in \mathbb{N} $and$ x_n \in \mathbb{Q} $,$ y_n \in \mathbb{Q} $for all$ m \in \mathbb{N} $. Let$ x \in E $. Then$ x \in I_m $for some$ m \in \mathbb{N} $. Therefore$ E \subset\bigcup_{n\in\mathbb{N}} I_n $. So$ \sigma $is an open cover of$ E $. If possible, let$ \sigma' = \{I_{r_1},\ I_{r_2}, \ldots,\ I_{r_m}\} $be a finite sub collection of$ \sigma $such that$ \sigma' $also covers$ E $, where$ r_1, r_2,\ldots, r_m $are natural numbers, i.e.$ E \subset \bigcup_{i=1}^{m}I_{r_i} $. Let$ p = \max \{r_1, r_2,\ldots, r_m\} $. Then$ p \in \mathbb{N} $and$ I_{r_i} \subset I_p $for$ I = 1, 2, \ldots, m $. Thus$ E \subset I_p = (x_p,\ y_p) $. Now$ \sqrt2 < y_p < \sqrt3 \Rightarrow y_p \in E $but$ y_p \notin I_p $, which is a contradiction. Therefore$ \sigma $has no infinite sub cover. Hence$ E $is not compact in$ \mathbb{Q} $.$ E $is bounded in$ \mathbb{Q} $, since$ x \in E \Rightarrow 1 < \sqrt2 < x < \sqrt3 < 2 $. Thus rational numbers$ 1 $and$ 2 $are respectively a lower bound and an upper bound of$ E $in$ \mathbb{Q} $. <img class="aligncenter size-full wp-image-8697" src="https://www.mathacademytutoring.com/wp-content/uploads/2022/02/Compact-sets.png" alt="" width="563" height="82" /> To show that$ E $is closed in$ \mathbb{Q} $, let$ p < \sqrt2 $. By the density property of$ \mathbb{R} $, there exists a rational$ r $such that$ p < r < \sqrt2 $. We consider a$ \delta $-neighborhood of$ p $, for$ \delta = r – p > 0 $,$ (p – \delta, p + \delta) $which contains no point of$ E $, since$ p + \delta = r < \sqrt2 $. Hence$ (p – \delta, p + \delta) \cap E = \emptyset \Rightarrow p $is not a limit point of$ E $. Let$ q > \sqrt3 $. By the density property of$ \mathbb{R} $, there exists a rational s such that$ \sqrt3 <s < q $. Let$ \delta' = q – s > 0 $. Then$ \delta' $-neighborhood of$ q $,$ (p – \delta', p + \delta') $contains no point of$ E $, since$ q – \delta' = s > \sqrt3 $. Thus$ (p – \delta', p + \delta') \cap E = \emptyset \Rightarrow q $is not a limit point of$ E $. Hence no rational number outside$ E $is a limit point of$ E $, i.e. limit points (obviously rational number) of$ E $, if any, will belong to$ E $. Hence$ E $is closed in$ \mathbb{Q} $. This problem does not contradict Heine-Borel theorem, since this theorem is valid for the set$ \mathbb{R} $which is a complete metric space while$ \mathbb{Q} $is not so. <h2>Theorem 8. (Alternative Proof of Heine-Borel Theorem).</h2> Using Lindelof's covering theorem we are going to show that a closed and bounded subset in$ \bm{\mathbb{R}} $is compact. Proof. Let$ E $be a closed and bounded subset of$ \mathbb{R} $. Suppose$ \sigma $be a family of open sets in$ \mathbb{R} $which covers$ E $. By Lindelof's covering theorem,$ \sigma $has a countable sub cover, say$ \sigma' $. If$ \sigma' $is finite then the theorem is proved from the definition of compact set. Let$ \sigma' $be infinite. Then it is enumerable and we can write$ \sigma' = \{I_n : n \in \mathbb{N}\} $Since$ \sigma' $covers$ E $,$ E \subset\bigcup_{n\in\mathbb{N}} I_n $. Let$ A_n = E \cap \left(\bigcup_{k=1}^{n}I_k\right)^c = E – \bigcup_{k=1}^{n}I_k (n = 1, 2, 3,\ldots.) $. Then$ A_n \supset A_{n+1} $for all$ n \in \mathbb{N} $. Since$ \bigcup_{k=1}^{n}I_k $is open and$ E $is closed and bounded in$ \mathbb{R} $, so for every$ n \in \mathbb{N} $,$ A_n $is closed and bounded in$ \mathbb{R} $and$ \{A_n\} $is a nesting sequence (also called contracting sequence). Suppose$ A_n \neq \emptyset $for all$ n \in \mathbb{N} $. Then by Cantor's theorem$ \bigcap_{n\in\mathbb{N}} A_n \neq \emptyset \Rightarrow $there is a point$ c \in \bigcap_{n\in\mathbb{N}} A_n \Rightarrow c \in E $but$ c \notin I_n $for all$ n \in \mathbb{N} $i.e.$ c \notin \bigcup_{n\in\mathbb{N}} I_n $which contradicts our assumption that$ \sigma' $Covers$ E $. Hence we can conclude that there is a natural number m such that$ A_m = \emptyset \Rightarrow E \subset \bigcup_{k=1}^{m}I_k $. Hence there exists a finite sub collection$ \sigma" = \{I_1, I_2, \ldots, I_m\} $of$ \sigma' $and hence of$ \sigma $, which also covers$ E $. Therefore it is proved that every open cover of$ E $has a finite sub cover and hence$ E $is compact. <h1>CONTINUOUS FUNCTIONS ON COMPACT SETS</h1> Definition (Continuous Function): Let$ f : S \to \mathbb{R} $, where$ S \subset \mathbb{R} $, be a function on$ S $. Let$ c \in S $.$ f $is said to be continuous at$ c $if for any$ \varepsilon> 0 $, there is exists a$ (\varepsilon, 0) > 0 $such that$ f(c)\ – \varepsilon < f(x) < f(c) + \varepsilon $,when$ c – \delta < x < c + \delta $and$ x \in S $i.e.$ f(x)\ \in\ N(f(c),\ \varepsilon) $for all$ x \in N(c, \delta) \cap S $.$ f $is said to be continuous on$ S $if it continuous at every point of$ S $. <h2>Theorem 9.</h2> Let$ S \subset \mathbb{R} $and$ f : S \to \mathbb{R} $be a function of$ S $. Let$ f $be continuous on$ S $. If$ S $is compact in$ \mathbb{R} $then the image set$ f(S) $is also compact in$ \mathbb{R} $and$ f\left(S\right) $has both greatest and least elements. Proof. Let$ \sigma = \{I : \alpha \in \Lambda\} $be a family of open intervals such that$ f(S)\ \subset\ \bigcup_{\alpha\in\Lambda} I_\alpha $. Then$ \sigma $is an open cover of$ f(S) $. Let$ a \in S $. Then$ f(a)\ \in\ f(S) $and there exists an open interval$ I_{\alpha'} $in$ \sigma $such that$ f(S) \in I_{\alpha'} $. Since$ I_{\alpha'} $is an open set,$ f(a) $is its interior point. So there exists an$ \varepsilon_a $there exists a positive$ \delta_a $such that$ f(x)\ \in\ N(f(a),\ \varepsilon_a)\ \subset\ I_\alpha' $for all$ x \in N(a,\delta_a) \cap S $. We consider the family$ C = \{N(a, \delta_a) : a \in S\} $which is a family of open sets and it covers$ S $. Since$ S $is compact,$ C $has a finite sub collection$ C' = \{N(a, \delta_a) : i = 1, 2, \ldots, m\} $, where$ a_i \in S $and$ \delta_{a_i} > 0 $for$ i = 1, 2, \ldots, m $, such that$ C' $covers$ S $, i.e.$ S \subset \bigcup_{i=1}^{m}{N(a_i,\ \delta_{a_i})} $. Any point of$ f(S) $is$ f(x) $for some$ x \in S $. This$ x \in N(a_p,\ \delta_{a_p}) $for some$ p = 1, 2, \ldots, m $and hence$ x \in N(a_p, \varepsilon_{a_p}) \cap S $for some$ p = 1, 2, \ldots, m $. Therefore$ f(x) \in\ N(f(a_p),\ \varepsilon_{a_p})\ \subset\ I_{{\alpha'}_p} $for some$ p = 1, 2, \ldots, m $. Thus$ f(x) \in\ f(S)\ \Rightarrow\ f(x) \in\ I_{{\alpha'}_p} $for some$ p = 1, 2, \ldots, m $. Hence$ f(S) \subset\ \bigcup_{P=1}^{m}{\ I_{{\alpha'}_p}} $. Hence$ \sigma' = \{{I}_{{\alpha'}_1},\ I_{{\alpha'}_2}, \ldots,\ I_{{\alpha'}_m}\} $is a finite sub-collection of$ \sigma $, which covers$ f(S) $implying$ \sigma' $is an open sub cover of$ \sigma $. Hence$ f(S) $is compact. Since$ f(S) $is a compact subset of$ \mathbb{R} $, it is closed and bounded in$ \mathbb{R} $. Hence$ \sup{f(S)} $both exist and belong to$ f(S) $. Hence$ f(S) $has a greatest and least element. Notes : <ol> <li>The above theorem can be equivalently stated as: Continuous image of a compact set in$ \mathbb{R} $is a compact set in$ \mathbb{R} $.</li> <li>If$ f : S \to \mathbb{R} $be a continuous function on$ S (\subset R) $such that$ f(S) $is compact then$ S $is not necessarily compact in$ \mathbb{R} $. This is shown by the following example.</li> </ol> <h3>Some worksheets for readers</h3> <ul> <li>Obtain an open cover of the interval$ (2, 3] $that has no finite sub cover.</li> <li>Justify the statement: The set of all natural numbers is not a compact set in$ \mathbb{R} $.</li> <li>Let$ S = \left\lbrace x \in (0, 1): x \neq \frac{1}{n}, n \in \mathbb{N}\right\rbrace $and$ \sigma = \left\lbrace \left(\frac{1}{n+1},\ \frac{1}{n}\right) : n \in \mathbb{N}\right\rbrace $. Shows that$ \sigma $is an open cover of$ S $. Justify whether$ \sigma $has a finite sub cover for$ S $.</li> <li>If$ A = \left[\frac{1}{2},\ \frac{7}{2}\right] $and$ B =\left(1,\ \frac{9}{2}\right) $, try to check whether$ A \cup B $is compact or not.</li> <li>If$ A $be a compact set in$ \mathbb{R} $and$ B $is a non-empty finite subset of$ \mathbb{R} $, is$ A \cup B $is compact?</li> <li>Examine whether$ \mathbb{Q} $, the set of all rational numbers, is compact or not.</li> <li>Give an example, with proper justification, of a set of real numbers which is bounded but not compact.</li> <li>Is the set of all irrational numbers compact? Give reasons in support of your conclusion.</li> <li>If$ A $and$ B $are two compact subset of$ \mathbb{R} $then show that$ A \cap B $is compact.</li> <li>Is intersection of finite number of compact subsets in$ \mathbb{R} $is a compact set in$ \mathbb{R}$?

Bibliography

An introduction to functional analysis, By Charles Swartz.
Introduction to Topology, By Bert Mendelson.
The Elements of Real Analysis, R.G
Real Analysis, H.L. Royden

Thank you for reading this article. We will meet next time with a new article

Also, you may want to check our blog posts about Fourier Series, Riemann Integral, Lebesgue Outer Measure, and All The Logarithm Rules You Know and Don't Know About

And don't forget to join us on our Facebook page stay updated on any new articles and a lot more!!!!!

millerhumbecioned.blogspot.com

Source: https://www.mathacademytutoring.com/blog/compact-sets-and-continuous-functions-on-compact-sets

Show That the Lower Level Sets of a Continuous Function Are Closed

INTRODUTION

COVER, OPEN COVER, SUB COVER

COMPACT SETS IN $\mathbb{R}.$

Theorem 1 (Heine-Borel Theorem) :

Theorem 2 (Converse of Heine-Borel Theorem):

Theorem 3.

Theorem 4.

Theorem 5.

Bibliography

0 Response to "Show That the Lower Level Sets of a Continuous Function Are Closed"

Post a Comment

Iklan Atas Artikel

Iklan Tengah Artikel 1

Iklan Tengah Artikel 2

Iklan Bawah Artikel