Unravel the Mysteries of Mathematics: 3 Brain-Twisting Challenges to Put Your Skills to the Test

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16. Find the sum of the series
* \(\sum_{k=1}^{\infty}\frac{\ln\left(\frac{k^{k+1}}{(k+1)^{k}} \right)}{k(k+1)}\).

17. Suppose the ratio test applies to \(\sum_{n=1}^{\infty}a_{n}\), i.e., \(r=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_{n}}\right|\) exists. Show that \(\limsup_{n\to\infty}|a_{n}|^{1/n}=r\).

### 1.5 Topology of the Real Line

Given a real number \(a\) and an \(\epsilon>0\), the \(\epsilon\)-_neighborhood_ of the point \(a\) is the set

\[V_{\epsilon}(a)=\{x\in\mathbb{R}:\quad|x-a|<\epsilon\}.\]

In other words \(V_{\epsilon}(a)=(a-\epsilon,a+\epsilon)\) is an open interval centered at \(a\) of radius \(\epsilon\).

**Definition 18**.: The set \(O\subset\mathbb{R}\) is _open_ if for all points \(x\in O\) there exists an \(\epsilon\)-neighborhood \(V_{\epsilon}(x)\subset O\).

**Example 31**.:
* Any open interval of the form \((a,b)=\{x\in\mathbb{R}:\quad a

The union of two open intervals is again an open set, in fact we have the following theorem:

Figure 1.20: The \(\epsilon\)-neighborhood of the point \(x\)

**Theorem 26**.:
* _The union of an arbitrary collection of open sets is open._
* _The intersection of a finite collection of open sets is open._

Proof

:
* Let \(\{O_{i}\}_{i\in I}\) be a collection of open sets in \(\mathbb{R}\), and let \(O=\bigcup_{i\in I}O_{i}\). Let \(x\) be an arbitrary element of \(O\). Since \(O\) is the union of \(O_{i}\)’s, \(x\in O\) implies that there is at least one particular \(O_{i}\), where \(x\in O_{i}\). Because we are assuming \(O_{i}\) is open, there is an \(\epsilon\)-neighborhood \(V_{\epsilon}(x)\) of \(x\) such that \(V_{\epsilon}(x)\subset O_{i}\). The fact \(O_{i}\subset O\) implies that \(V_{\epsilon}(x)\subset O\) as well.
* Let \(\{O_{i}\}_{i=1}^{N}\) be a finite collection of open subsets of \(\mathbb{R}\). Take \(x\in\bigcap_{i=1}^{N}O_{i}\), then \(x\in O_{i}\) for each \(1\leq i\leq N\). By the definition of open set, we know that for each \(1\leq i\leq N\), there exists an \(\epsilon_{i}\)-neighborhood of \(x\) such that \(V_{\epsilon_{i}}(x)\subset O_{i}\). Now we seek a single \(\epsilon\)-neighborhood of \(x\) contained in every \(O_{i}\). The natural candidate for \(\epsilon\) is \(\epsilon=\min\{\epsilon_{1},\epsilon_{2},\ldots,\epsilon_{N}\}\). It then follows that \[V_{\epsilon}(x)\subseteq\bigcap_{i=1}^{N}O_{i}\] as claimed.

**Remark 16**.: Statement b) above is not true if arbitrary collections are used in place of finite collections, for example, for \(i\in\mathbb{N}\), the intervals \(\left(\frac{-1}{i},\frac{1}{i}\right)\) are all open sets, but \(\bigcap_{i\in\mathbb{N}}\left(\frac{-1}{i},\frac{1}{i}\right)=\{0\}\) and \(\{0\}\) is not open. We will see that it indeed is a closed set.

We now give a description of the structure of open sets in \(\mathbb{R}\).

**Theorem 27**.: _Every open set \(O\) in \(\mathbb{R}\) can be written uniquely as a countable union of disjoint open intervals._

Proof

: Let \(O\) be an open set in \(\mathbb{R}\) and \(x\in O\). Since \(O\) is an open set, there is a non-trivial open interval (we called this a “neighborhood” of \(x\) previously) \(x\) is contained in. Next define two points \(a_{x}\) and \(b_{x}\) as \(a_{x}=\inf\{ax:(x,b)\subset O\}\), and denote the interval containing \(x\) as \(I_{x}=(a_{x},b_{x})\). This way every \(x\in O\) is in \(I_{x}\) and moreover \(I_{x}\subset O\). Thus \(O=\bigcup_{x\in O}I_{x}\). We still need to show this union of intervals \(\{I_{x}\}\) is disjoint and countable. Suppose it is not disjoint. Then we have two open intervals \(I_{x}\) and \(I_{y}\) such that \(I_{x}\cap I_{y}\neq\emptyset\). Consider \(I_{x}\cup I_{y}\), which is an open set contained in \(O\) containing \(x\). However by construction \(I_{x}\) is the maximal one satisfying thisproperty, which forces \((I_{x}\cup I_{y})\subset I_{x}\), similarly \((I_{x}\cup I_{y})\subset I_{y}\), which in turn implies that \(I_{x}=I_{y}\), meaning any distinct intervals in the collection must be disjoint. To prove that the union \(O=\bigcup_{x\in O}I_{x}\) is countable, we need to recall that the set of rational numbers is countable and every interval \(I_{x}\) contains a rational number. Since different intervals are disjoint, they must contain distinct rationals, and thus the above union is a countable union.

**Remark 17**.: Because the representation of \(O=\bigcup_{x\in O}I_{i}\) is unique and the union is disjoint, the above theorem enables us to “measure” the open set \(O\) in terms of the sum of the length of the intervals \(I_{i}\), i.e., the measure of \(O=\sum_{i=1}^{\infty}|I_{i}|\). This is also the beginning of the concept of “measure.” However this fact is about open sets, it does not extend to other sets in \(\mathbb{R}\), and there is no direct analog of this theorem for subsets of \(\mathbb{R}^{n}\).

**Definition 19**.: Let \(A\subset\mathbb{R}\). A point \(x\in A\) is called an _interior_ point of \(A\) if there is an open set \(U\) such that \(x\in U\subset A\). The interior of \(A\) is the collection of all interior points of \(A\) and is denoted by \(\mathring{A}\). This set might be empty.

The condition \(x\in\mathring{A}\) is equivalent to the following: There is \(\epsilon>0\) such that \((x-\epsilon,x+\epsilon)\subset A\). If \(A=\{a\}\) is a single point set, then \(\mathring{A}=\emptyset\), and if \(A=[a,b]\), then \(\mathring{A}=(a,b)\). The interior of a set can also be described as the union of all open subsets of \(A\); thus \(\mathring{A}\) is the largest open subset of \(A\). It is clear that \(A\) is open if and only if \(A=\mathring{A}\).

**Definition 20**.: Let \(S\) be a subset of \(\mathbb{R}\). We say \(S\) is a _closed_ set in \(\mathbb{R}\) if the complement of \(S\) is an open set in \(\mathbb{R}\).

**Example 32**.:
* The empty set \(\emptyset\) and \(\mathbb{R}\) are both open and closed subsets of \(\mathbb{R}\).
* A single point set \(\{a\}\) is closed because its complement is the union of two open sets, i.e., its complement is \((-\infty,a)\cup(a,+\infty)\).
* An interval of the form \([a,b]\) is a closed set.

**Theorem 28**.:
* _The intersection of an arbitrary collection of closed sets is closed._
* _The union of a finite collection of closed sets is closed._

Proof

:
* Let \(\{S_{i}\}_{i\in I}\) be any collection of closed subsets of \(\mathbb{R}\). The proof follows from De Morgan’s law: \[\left(\bigcap_{i\in I}S_{i}\right)^{c}=\bigcup_{i\in I}S_{i}^{c}\]and the first part of Theorem 26.
2. If \(\{S_{i}\}_{i=1}^{N}\) is a finite collection of closed subsets of \(\mathbb{R}\), then by De Morgan’s law we have \[\left(\bigcup_{i=1}^{N}S_{i}\right)^{c}=\bigcap_{i=1}^{N}S_{i}^{c}.\] Now apply the second part of Theorem 26 to complete the proof.

**Remark 18**.: Statement b) above is not true if arbitrary collections are used in place of finite collections; for example, if we take closed sets \(S_{k}=\left[\frac{1}{k+1},\frac{k}{k+1}\right]\), then \(\bigcup_{k\in\mathbb{N}}\left[\frac{1}{k+1},\frac{k}{k+1}\right]=(0,1)\).

Theorems 26 and 28 have many applications. In particular they allow us to define the largest open set contained in a given set \(S\) and the smallest closed set containing the given set \(S\) as in the following:

**Definition 21**.: Let \(S\) be a subset of \(\mathbb{R}\).

1. The _interior_ of \(S\) is the set \[\mathring{S}:=\bigcup\{O\subseteq S\text{ and }O\text{ is open in }\mathbb{R}\}.\]
2. The _closure_ of \(S\) is the set \[\overline{S}:=\bigcap\{F\supseteq S\text{ and }F\text{ is closed in }\mathbb{R}\}.\]

Note that every set \(S\) contains the empty set \(\emptyset\) and is contained in \(\mathbb{R}\); hence \(\overline{S}\) and \(\mathring{S}\) are well defined. Moreover, by Theorems 26 and 28 we immediately see that \(\overline{S}\) is a closed set and \(\mathring{S}\) is an open set. The following result claims more: \(\overline{S}\) is the smallest closed set which contains \(S\) and \(\mathring{S}\) is the largest open set contained in \(S\).

**Proposition 4**.: _Let \(S\) be a subset of \(\mathbb{R}\). Then_

1. \(\mathring{S}\subseteq S\subseteq\overline{S}\)_._
2. _If_ \(G\) _is open and_ \(G\subseteq S\)_, then_ \(G\subseteq\mathring{S}\)_._
3. _If_ \(K\) _is closed and_ \(K\supseteq S\)_, then_ \(K\supseteq\overline{S}\)_._

Proof

: The proof of part a) is clear. For the proof of parts b) and c), use the above definition, and observe that if \(O\) is an open set contained in \(S\), then \(O\subseteq\mathring{S}\) and \(K\) is a closed set containing \(S\), and then \(\overline{S}\subseteq K\)

**Remark 19**.: Parts b) and c) imply that

\[S=\hat{S}\quad\text{if and only if}\quad S\text{ is open}\]

and

\[S=\overline{S}\quad\text{if and only if}\quad S\text{ is closed}.\]

Furthermore, these concepts have a natural extension to the subsets of \(\mathbb{R}^{n}\) or to an arbitrary metric space \(M\) which we will define in Chapter 2. For a detailed discussion of the topology of \(\mathbb{R}^{n}\) or an arbitrary metric space, we refer the reader to [52], pp. 303-342.

**Definition 22**.: Let \(S\) be a subset of \(\mathbb{R}\). We say \(x\in\mathbb{R}\) is an _accumulation point_ of \(S\) if, for every \(\epsilon>0\), we have \(((x-\epsilon,x+\epsilon)\setminus\{x\})\cap S\neq\emptyset\).

Thus, \(x\) is an accumulation point of \(S\) if every interval around \(x\) contains points of \(S\) other than \(x\). A set need not have any accumulation points, for example, a set consisting of a single point or the set of integers in \(\mathbb{R}\) has no accumulation points. Moreover, the set of accumulation points of a set need not lie in that set. For example, if we set \(S=(0,1)\), then the set of accumulation points of S is the whole interval \([0,1]\). The following lemma clarifies the concept of accumulation point further.

**Lemma 6**.: _Let \(S\subset\mathbb{R}\). Then \(x\in\mathbb{R}\) is an accumulation point of \(S\) if and only if every neighborhood of \(x\) contains infinitely many points of \(S\)._

Proof

: Let \(x\) be an accumulation point of \(S\) and \(\epsilon>0\). By Definition 22 there is an \(\epsilon\)-neighborhood of \(x\), \((x-\epsilon,x+\epsilon)\) such that \(x_{1}\in S\cap(x-\epsilon,x+\epsilon)\) and \(x_{1}\neq x\). Let \(\epsilon_{1}=|x-x_{1}|>0\). Again using the definition of an accumulation point, we can find \(x_{2}\in S\cap(x-\epsilon_{1},x+\epsilon_{1})\) such that \(x_{2}\neq x\). Continuing this process will yield an infinite set of elements in \(S\cap(x-\epsilon,x+\epsilon)\).

Let \(S^{*}\) denote a set of all accumulation points of \(S\). If \(x\in S\) but \(x\notin S^{*}\), then \(x\) is called an _isolated_ point of \(S\). Note that an isolated point is always an element of the set, while an accumulation point does not necessarily belong to the set. Accumulation points are also referred to as “cluster points” or “limit points.”

**Proposition 5**.: _A point \(x\) is an accumulation point of a set \(S\) if and only if \(x=\lim\limits_{n\to\infty}x_{n}\) for some sequence \(\{x_{n}\}\) contained in \(S\) satisfying \(x_{n}\neq x\) for all \(n\in\mathbb{N}\)._

Proof

: Given a point \(s\in S\), where \(s\) is the limit of the constant sequence \(\{s,s,s,\dots\}\). To handle this uninteresting situation, the condition \(x_{n}\neq x\) is given in the statement of the theorem. Assume \(x\) to be an accumulation point of the set \(S\), and we want to produce a sequence \(\{x_{n}\}\) with \(x_{n}\to x\). Using the definition of accumulation point and by taking \(\epsilon_{n}=\dfrac{1}{n}\), we can find

\[x_{n}\in V_{\frac{1}{n}}(x)\cap S\quad\text{for each}\quad n\in\mathbb{N}\]with the condition that \(x_{n}\neq x\). Given arbitrary \(\epsilon>0\), choose \(N\) so that \(\frac{1}{N}<\epsilon\), and then we have \(|x_{n}-x|<\epsilon\) holds true for all \(n\geq N\). We leave the proof of the reverse implication to the reader.

The adjective _closed_ is often used in mathematics; it roughly means if an operation on the elements of a given set is performed, we still obtain an element of the same given set. For example, for the definition of a vector space, one can simply state a vector space is a set which is closed under addition and scalar multiplication. However, in analysis, the operation one considers is the limiting operation. The following proposition follows directly from the definition of a closed set.

**Proposition 6**.: _A set \(S\subseteq\mathbb{R}\) is closed if it contains all of its accumulation points._

**Example 33**.: To prove the closed interval \([a,b]=\{x\in\mathbb{R}:\quad a\leq x\leq b\}\) is a closed set, let \(x\) be an accumulation point of \([a,b]\). Then there exists a sequence \(\{x_{n}\}\) such that \(x_{n}\to x\). We need to show that \(x\in[a,b]\). Using the relationship between inequalities and the limit, we obtain

\[a\leq x_{n}\leq b\quad\text{implies that}\quad a\leq x\leq b,\]

proving \(x\in[a,b]\) and the set \([a,b]\) is closed.

**Definition 23**.: Given \(S\subseteq\mathbb{R}\) and \(S^{*}\) the set of all accumulation points of \(S\), the _closure_ of a set \(S\) is defined to be \(\overline{S}=S\cup S^{*}\).