X is Compact Fx is Compact F is Continuous

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 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}. 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 <strong>Example 12.</strong> Prove that intersection of infinite collection of compact sets in\mathbb{R}is compact.  <strong>Solution:</strong> Let\sigma = \{C\}be an infinite collection of compact sets in\mathbb{R}.  LetS =\ \bigcap_{C\in\sigma} C. Since everyCbelonging to\sigmais compact, it is closed and henceSis a closed subset orR, which implies\mathbb{R}-Sis open.  Letg' = \{G : \alpha \in \Lambda\}be a family of open sets in\mathbb{R}which coversS, i.e.S \subset \ \bigcap_{\alpha\in\land} G_\alpha. Supposegis not an open cover ofCfor anyC \in \sigma.  Letg'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). SinceC \subset \mathbb{R},C \subset \left\lbrace \bigcup_{\alpha\in\land} G_\alpha\right\rbrace \cup \left(\mathbb{R} – S\right)for everyC \in \sigma. Theng'is an open cover ofCfor everyC \in \sigma. AsCis compact for everyC \in \sigma, there exists a sub collectiong"ofg'such thatg"also coversCfor everyC \in \sigma.  Letf" = \{G_{\alpha_1},\ G_{\alpha_2}, \ldots,\ G_{\alpha_m},\ \mathbb{R}\ – S\}.g"must contain\mathbb{R} – S, for otherwiseC \subset \bigcup_{i=1}^{m}G_{\alpha_i}which contradicts our assumptiongis not and open cover ofCfor anyC \in \sigma.  ThereforeC \subset \left\lbrace\bigcup_{i=1}^{m}G_{\alpha_i}\right\rbrace \cup (\mathbb{R} – S)for everyC \in \sigma.  SinceS \subset C,S \subset \bigcup_{i=1}^{m}G_{\alpha_i}. Letg"' = \{G_{\alpha_1},\ G_{\alpha_2}, \ldots,\ G_{\alpha_m}\}.  Theng"'is a finite sub collection ofgthat coversS. Henceghas a finite sub cover, which proves thatSis component.  <strong>Example 13.</strong> Using definition of compact set prove that the set(0, 1]is not a compact set in\mathbb{R}.  <strong>Solution:</strong> LetI_n = \left(\frac{1}{n+1},\ \frac{n+1}{n}\right), n \in \mathbb{N}and\sigma = \{I_n : n \in \mathbb{N}\}.\sigmais a collection of open sets in\mathbb{R}.  Letx \in (0, 1]. Ifx = 1thex \in I_nfor alln \in \mathbb{N}.  If0< x < 1, then, by the Archimedean property of\mathbb{R}, there exists a natural numbermsuch thatm \le\ \frac{1}{x} < m + 1

\Rightarrow \frac{1}{m+1} < x \le\ \frac{1}{m} <\ \frac{m+1}{m} \Rightarrow x \in I_mfor somem \in \mathbb{N}.  Hence(0, 1] \subset \bigcup_{n\in\mathbb{N}} I_nand therefore,\sigmais 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, wherer_1, r_2, \ldots, r_mare natural numbers greater than 1, such that(0, 1] \subset \bigcup_{k=1}^{m}I_{r_k}Letu = \max \{r_1 + 1, r_2 + 1,\ldots, r_m + 1\} \Rightarrow u \geq r_k + 1for allk = 1, 2, \ldots, mandv = \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 allk = 1, 2, \ldots, mThen0 < \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 allk = 1, 2,\ldots, m\Rightarrow (0, 1] \subset  \left(\frac{1}{u},\ \frac{1}{v}\right). Since0 < \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\sigmacover(0, 1], proving that(0, 1]is not a compact set.  <strong>Example 14.</strong> Leta < x < band\Gamma = \{(x – \varepsilon, x + \varepsilon), \varepsilon > 0\}. Is\Gammaan open cover of[a, b]having a finite sub cover? Justify your answer.  <strong>Solution:</strong> LetI_x = (x – \varepsilon, x + \varepsilon),\ \varepsilon > 0anda < x < b. Then\Gamma = \{I_x : a < x < b\}. Clearly every elementxof(a, b)is contained inI_x. By the density property of\mathbb{R}, there exist elementsx_1, x_2 \in (a, b)such thata < x_1 < a + \varepsilonandb – \varepsilon < x_2 < b. Thenx_1 – \varepsilon < a,\ x_1 + \varepsilon > aandb < x_2 + \varepsilon, x_2 – \varepsilon < b.  Thusa \in I_{x_1}andb \in I_{x_2}. Hence it is shown that[a, b] \subset \bigcap_{a<x<b} I_x \Rightarrow \Gammais 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,\Gammahas a finite sub cover. <h2><strong><u>Theorem 6</u></strong><strong>. (Lindelof's Covering Theorem).</strong></h2> <strong>Statement:</strong> - LetS \subset\mathbb{R}. Then every open cover ofShas a countable sub-cover.  <strong><em>Proof.</em></strong>  LetC = \{A : \alpha \in \Lambda\}Be a collection of open setsA_\alphain\mathbb{R},\Lambdabeing the index set, which coversS. ThenS \subset \bigcup_{\alpha\in\land} A_\alpha.  Letx \in S. Then there exists\alpha(x) \in \Lambdasuch thatx \in A_{\alpha(x)}. SinceA_{\alpha(x)}is open,xis its interior point. There exists a neighborhoodI(x)(which is an open interval) ofxsuch thatx \in I(x) \subset A_{\alpha(x)}. SinceI(x)is an open interval, we can always choose and open intervalJ(x) \subset I(x)such thatx \in J(x)andj(x)has rational and points. HenceS \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 coversS. 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 asC' = \{j_1, j_2, j_3, \ldots\}, which is an open cover ofS.  For eachJ_n \in C', there exists an elementx_n \in Ssuch thatx_n \in J_n \subset I(x_n) \subset A_{\alpha_n}, say.  Thus for eachJ_n \in C'there corresponds to an open setA_{\alpha_n} \in C.  IfC" = \{A_{\alpha_1},\ A_{\alpha_2}, \ldots,\ A_{\alpha_n}, \ldots\}thenC"is a countable sub-collection ofC, which also coversS. Hence the proof.  <strong><u>Example 15</u></strong><strong>. The set</strong>\bm{\{0\} \cup \left\{ }\frac{\bm{1}}{\bm{n}}\bm{:n\in\ \mathbb{N}\right\rbrace}<strong>is a compact set.</strong>  <strong><u>Solution:</u></strong>  LetS =\left\lbrace 0,\ 1,\ \frac{1}{2},\ \frac{1}{3}, \ldots\right\rbrace. LetTbe any infinite subset ofS.  Then eitherT = \left\lbrace 0,\ \frac{1}{m}\ :m\ \in\ N_1\right\rbraceorT = \left\lbrace\frac{1}{m}\ :m\ \in\ N_1\right\rbracewhereN_1is any infinite subset of\mathbb{N}. In any case0is the only limit point ofTand0 \in S. Then every infinite subset ofShas a limit point inSso (by Theorem 4.)Sis closed and bounded and hence by Heine-Borel theorem,Sis compact.  <strong><u>Example 16</u></strong><strong>.</strong> Construct a compact set in\mathbb{R}whose accumulation point from a countably infinite set in\mathbb{R}.  <strong>Solution:</strong>  LetA = \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, andC = \{0\}. ThenS = A \cup B \cup Cis a bounded subject of\mathbb{R}, sinceS \subset [0, 2].  Now derived sets ofA,\ B,\ Care respectivelyA' =\left\lbrace 0,\ \frac{1}{m}\ :m\ \in\ \mathbb{N}\right\rbrace,B' = \{0\},C' = \emptyset.  ThusS' = A' \cup B' \cup C' = \{0\} \cup \left\lbrace\frac{1}{m}\ :m\ \in\ \mathbb{N}\right\rbrace  = B \cup C \subset S.  HenceSis closed.  ThusSis a closed and bounded set in\mathbb{R}and henceSis a compact set in\mathbb{R}.  Now the mappingf : \mathbb{N} \to Bdefined byf(m) = \frac{1}{m},\ m \in\ \mathbb{N}, is clearly a bijective mapping and henceBis enumerable.  ThusS'is the union of a finite set and an enumerable set, which impliesS'is a countably infinite set in\mathbb{R}. <h2><strong>Theorem 7. (Cantor's Theorem).</strong></h2> <strong>Statement:</strong> - If\{F_n\}_nis a nesting sequence of non-empty compact sets in\mathbb{R}then\bigcap_{n\in\mathbb{N}} F_nis non-empty.  <strong>Proof :</strong> Since\{F_n\}_nis a nesting sequence of non-empty compact set,F_n \supset F_{n+1}for  alln \in \mathbb{N}, andF_nis closed and bounded for alln \in \mathbb{N}.  AlsoF_n \neq \emptyset, for everyn \in \mathbb{N}. We choose the fixed pointx_n \in F_n. Thus we have a sequence\{x_n\}_nsuch thatx_n \in F_n \subset F_mform \leq n.  If Xbe the range of the sequence\{x_n\}_nthen eitherXis finite or infinite.  <strong>Case-1:</strong>Xis finite. Then there is a point\alpha \in Xsuch thatx_n = \alphafor infinite values ofnin\mathbb{N}. For a fixedp \in \mathbb{N}, there is a natural numberk \geq psuch thatx_k = \alphaand hence\alpha = x_k \in F_k \subset F_p. This result is true for everyp \in \mathbb{N}.  Hence\alpha \in \bigcap_{n\in\mathbb{N}} F_n \Rightarrow \bigcap_{n\in\mathbb{N}} F_n \neq \emptyset.  <strong>Case-2:</strong>Xis infinite. Since eachF_nis bounded,F_1 \supset Xand henceXis bounded. By Bolzano-Weierstrass theorem,Xhas a limit point, say\beta.  Letmbe fixed positive integer. Since\betais a limit point ofX, any neighbourhood of\beta,N(\beta)contains infinite number of points ofXand henceN(\beta)contains infinite number of distinct pointsx_n's. Asx_n \in F_mforn \geq m, it follows thatN(\beta) \cap F_mis an infinite set. Hence\betais a limit point ofF_m. SinceF_m  is closed,\beta \in F_m. This result is true for everym \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}_nis non-empty.  <strong>Example 16.</strong> LetE = \{r \in \mathbb{Q} : \sqrt{2} < r < \sqrt{3} \}. We are going to show thatEis not compact in\mathbb{Q}, butEis Closed and bounded in\mathbb{Q}. Explain why this does not contradict the Heine-Borel theorem.  <strong>Solution:</strong> By the Density property of\mathbb{R}, there exist rational numbersx_1andy_1inE(\subset \mathbb{Q}) such that\sqrt{2} < x_1 < y_1 < \sqrt{3}. Again by the same property there exist rational numbersx_2andy_2inEsuch that\sqrt{2} < x_2 < x_1 < y_1 < x_2 < \sqrt3. Continuing this way we have sequences of rational points inE,\{x_n\}_nand\{y_n\}_nsuch that both are bounded;\{x_n\}_nis monotone decreasing which converges to\sqrt{2}and\{y_n\}_nis monotone increasing which converges to\sqrt3. Thus we have an infinite set of open intervals\sigma = \{I_n : n \in \mathbb{N}\}whereI_n = (x_n,\ y_n)such that\sqrt2 < x_n < y_n < \sqrt3for alln \in \mathbb{N}andx_n \in \mathbb{Q},y_n \in \mathbb{Q}for allm \in \mathbb{N}.  Letx \in E. Thenx \in I_mfor somem \in \mathbb{N}.  ThereforeE \subset\bigcup_{n\in\mathbb{N}} I_n. So\sigmais an open cover ofE.  If possible, let\sigma' = \{I_{r_1},\ I_{r_2}, \ldots,\ I_{r_m}\}be a finite sub collection of\sigmasuch that\sigma'also coversE, wherer_1, r_2,\ldots, r_mare natural numbers, i.e.E \subset \bigcup_{i=1}^{m}I_{r_i}.  Letp = \max \{r_1, r_2,\ldots, r_m\}. Thenp \in \mathbb{N}andI_{r_i} \subset I_pforI = 1, 2, \ldots, m. ThusE \subset I_p = (x_p,\ y_p).  Now\sqrt2 < y_p < \sqrt3 \Rightarrow y_p \in Ebuty_p \notin I_p, which is a contradiction. Therefore\sigmahas no infinite sub cover. HenceEis not compact in\mathbb{Q}.Eis bounded in\mathbb{Q}, sincex \in E \Rightarrow 1 < \sqrt2 < x < \sqrt3 < 2.  Thus rational numbers1and2are respectively a lower bound and an upper bound ofEin\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 thatEis closed in\mathbb{Q}, letp < \sqrt2. By the density property of\mathbb{R}, there exists a rationalrsuch thatp < r < \sqrt2. We consider a\delta-neighborhood ofp, for\delta = r – p > 0,(p – \delta, p + \delta)which contains no point ofE, sincep + \delta = r < \sqrt2. Hence(p – \delta, p + \delta) \cap E = \emptyset \Rightarrow pis not a limit point ofE.  Letq > \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 ofq,(p – \delta', p + \delta')contains no point ofE, sinceq – \delta' = s > \sqrt3.  Thus(p – \delta', p + \delta') \cap E = \emptyset \Rightarrow qis not a limit point ofE. Hence no rational number outsideEis a limit point ofE, i.e. limit points (obviously rational number) ofE, if any, will belong toE. HenceEis 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><strong>Theorem 8. (Alternative Proof of Heine-Borel Theorem).</strong></h2> Using Lindelof's covering theorem we are going to show that <strong>a closed and bounded subset in</strong>\bm{\mathbb{R}}<strong>is compact</strong>.  <strong>Proof.</strong> LetEbe a closed and bounded subset of\mathbb{R}.  Suppose\sigmabe a family of open sets in\mathbb{R}which coversE. By Lindelof's covering theorem,\sigmahas 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'coversE,E \subset\bigcup_{n\in\mathbb{N}} I_n.  LetA_n = E \cap \left(\bigcup_{k=1}^{n}I_k\right)^c = E – \bigcup_{k=1}^{n}I_k (n = 1, 2, 3,\ldots.).  ThenA_n \supset A_{n+1}for alln \in \mathbb{N}. Since\bigcup_{k=1}^{n}I_kis open andEis closed and bounded in\mathbb{R}, so for everyn \in \mathbb{N},A_nis closed and bounded in\mathbb{R}and\{A_n\}is a nesting sequence (also called contracting sequence).  SupposeA_n \neq \emptysetfor alln \in \mathbb{N}. Then by Cantor's theorem\bigcap_{n\in\mathbb{N}} A_n \neq \emptyset \Rightarrowthere is a pointc \in \bigcap_{n\in\mathbb{N}} A_n \Rightarrow c \in Ebutc \notin I_nfor alln \in \mathbb{N}i.e.c \notin \bigcup_{n\in\mathbb{N}} I_nwhich contradicts our assumption that\sigma'CoversE. Hence we can conclude that there is a natural number m such thatA_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 coversE. Therefore it is proved that every open cover ofEhas a finite sub cover and henceEis compact. <h1>CONTINUOUS FUNCTIONS ON COMPACT SETS</h1> <strong>Definition (Continuous Function):</strong> Letf : S \to \mathbb{R}, whereS \subset \mathbb{R}, be a function onS. Letc \in S.fis said to be continuous atcif for any\varepsilon> 0, there is exists a(\varepsilon, 0) > 0such thatf(c)\ – \varepsilon < f(x) < f(c) + \varepsilon,whenc – \delta < x < c + \deltaandx \in Si.e.f(x)\ \in\ N(f(c),\ \varepsilon)for allx \in N(c, \delta) \cap S.fis said to be continuous onSif it continuous at every point ofS. <h2><strong>Theorem 9.</strong></h2> LetS \subset \mathbb{R}andf : S \to \mathbb{R}be a function ofS. Letfbe continuous onS. IfSis compact in\mathbb{R}then the image setf(S)is also compact in\mathbb{R}andf\left(S\right)has both greatest and least elements.  <strong>Proof.</strong> Let\sigma = \{I : \alpha \in \Lambda\}be a family of open intervals such thatf(S)\ \subset\ \bigcup_{\alpha\in\Lambda} I_\alpha. Then\sigmais an open cover off(S).  Leta \in S. Thenf(a)\ \in\ f(S)and there exists an open intervalI_{\alpha'}in\sigmasuch thatf(S) \in I_{\alpha'}. SinceI_{\alpha'}is an open set,f(a)is its interior point. So there exists an\varepsilon_athere exists a positive\delta_asuch thatf(x)\ \in\ N(f(a),\ \varepsilon_a)\ \subset\ I_\alpha'for allx \in N(a,\delta_a) \cap S.  We consider the familyC = \{N(a, \delta_a) : a \in S\}which is a family of open sets and it coversS. SinceSis compact,Chas a finite sub collectionC' = \{N(a, \delta_a) : i = 1, 2, \ldots, m\}, wherea_i \in Sand\delta_{a_i} > 0fori = 1, 2, \ldots, m, such thatC'coversS, i.e.S \subset \bigcup_{i=1}^{m}{N(a_i,\ \delta_{a_i})}.  Any point off(S)isf(x)for somex \in S. Thisx \in N(a_p,\ \delta_{a_p})for somep = 1, 2, \ldots, mand hencex \in N(a_p, \varepsilon_{a_p}) \cap Sfor somep = 1, 2, \ldots, m.  Thereforef(x) \in\ N(f(a_p),\ \varepsilon_{a_p})\ \subset\ I_{{\alpha'}_p}for somep = 1, 2, \ldots, m. Thusf(x) \in\ f(S)\ \Rightarrow\ f(x) \in\ I_{{\alpha'}_p}for somep = 1, 2, \ldots, m. Hencef(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 coversf(S)implying\sigma'is an open sub cover of\sigma. Hencef(S)is compact.  Sincef(S)is a compact subset of\mathbb{R}, it is closed and bounded in\mathbb{R}. Hence\sup{f(S)}both exist and belong tof(S). Hencef(S)has a greatest and least element.  <strong>Notes :</strong> <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>Iff : S \to \mathbb{R}be a continuous function onS (\subset R)such thatf(S)is compact thenSis 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>LetS = \left\lbrace x \in (0, 1): x \neq \frac{1}{n}, n \in \mathbb{N}\right\rbraceand\sigma = \left\lbrace \left(\frac{1}{n+1},\ \frac{1}{n}\right) : n \in \mathbb{N}\right\rbrace. Shows that\sigmais an open cover ofS. Justify whether\sigmahas a finite sub cover forS.</li>  	<li>IfA = \left[\frac{1}{2},\ \frac{7}{2}\right]andB =\left(1,\ \frac{9}{2}\right), try to check whetherA \cup Bis compact or not.</li>  	<li>IfAbe a compact set in\mathbb{R}andBis a non-empty finite subset of\mathbb{R}, isA \cup Bis 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>IfAandBare two compact subset of\mathbb{R}then show thatA \cap Bis compact.</li>  	<li>Is intersection of finite number of compact subsets in\mathbb{R}is a compact set in\mathbb{R}$?

Bibliography

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

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