Boundedness of numerical sets, their exact boundaries, and limit points of numerical sets.
References: Collection of Problems in Mathematics. Part 1. Edited by A.V. Efimov, B.P. Demidovich.
Let $X$ be an arbitrary non-empty set of real numbers. The number $M=\max X$ is called the greatest (maximum) element of the set $X$ if $M\in X$ and for every $x\in X$, the inequality $x\leq M$ holds. Similarly, the concept of the smallest (minimum) element $m=\min X$ of the set $X$ is defined.
The set $X$ is called bounded above if there exists a real number $a$ such that $x\leq a$ for all $x\in X$. Any number possessing this property is called an upper bound of the set $X$. For a given set $X$ bounded above, the set of all its upper bounds has a least element, which is called the least upper bound (supremum) of the set $X$ and is denoted by $\sup X$. Obviously, $\sup X=\max X$ if and only if $\sup X\in X$.
Similarly, the concepts of a set bounded below, lower bound, and greatest lower bound (infimum) of the set $X$ are defined. The latter is denoted by $\inf X$.
A set $X$, bounded both above and below, is called bounded.
Let $x\subset \mathbb{R}$. A number $x_0\in \mathbb{R}$ is called a limit point of the set $X$ if every neighborhood of the point $x_0$ contains a point from the set $X$ other than $x_0$, that is, for $$\forall\varepsilon>0\,\,\exists y\in X, y\neq x_0: |y-x_0|<\varepsilon.$$
The point $x_0$ itself may or may not belong to the set $X$.
Examples.
1.73. Let $X=\left\{1, \frac{1}{2}, \frac{1}{3}, \frac{1}{n}\right\}.$
a) Identify the smallest and largest elements of this set, if they exist.
b) Determine the sets of upper and lower bounds for the set $X$. Find $\sup X$ and $\inf X$.
Solution.
a) The given set has the greatest element $M=1$ since for all elements $x\in X$, the inequality $x\leq 1$ holds, and $1\in X$.
The set does not have a smallest element because for any element $x_n=\frac{1}{n}\in X$, there always exists an element $x_{n+1}=\frac{1}{n+1}\in X$ such that $x_{n+1}\leq x_n$.
b) Since for all elements $x$ of the set $X$, the inequality $x\leq 1$ holds, and $1\in X$, the set of upper bounds for the set $X$ is the set $[1, +\infty)$ with the least element equal to $1$. Thus, $\sup X=1$.
The set $X$ does not have a smallest element. Obviously, for all elements $x$ of the set $X$, $x\geq 0$, meaning the set $X$ is bounded below. Let's show that $0$ is a limit point of the set $X$. Indeed, for any $\varepsilon>0$, we can find a natural number $n>\frac{1}{\varepsilon}$, which implies $\frac{1}{n}<\varepsilon$, and $\frac{1}{n}\in X$. Thus, the set of lower bounds for $X$ is the set $(-\infty, 0]$ with the greatest element equal to $0$. Therefore, we find $\inf X=0$.
Answer: $M=1$, the smallest element does not exist, $[1, +\infty)$, $(-\infty, 0]$, $\sup X=1$, $\inf X=0$.
1.74. For the set $X=\left\{x\in R|\,\, x=\frac{1}{2^n},\,\, n\in N\right\},$ find $\max X$, $\min X$, $\sup X$, and $\inf X$ if they exist.
Solution.
Let's express the set $X$ as:
$$X=\left\{x\in R|\,\, x=\frac{1}{2^n},\,\, n\in N\right\}=\left\{\frac{1}{2},\frac{1}{4}, \frac{1}{8}, .., \frac{1}{2^n},...\right\}$$
This set has the greatest element $M=\frac{1}{2}$ since for all elements $x\in X$, the inequality $x\leq \frac{1}{2}$ holds. Additionally, $\frac{1}{2}\in X$.
The set does not have a smallest element because for any element $x_n=\frac{1}{2^n}\in X$, there always exists an element $x_{n+1}=\frac{1}{2^{n+1}}\in X$ such that $x_{n+1}\leq x_n$.
Since for all elements $x$ of the set $X$, the inequality $x\leq \frac{1}{2}$ holds, and $\frac{1}{2}\in X$, the set of upper bounds for the set $X$ is the set $\left[\frac{1}{2}, +\infty\right)$ with the least element equal to $\frac{1}{2}$. Thus, $\sup X=\frac{1}{2}$.
The set $X$ does not have a smallest element. Obviously, for all elements $x$ of the set $X$, $x\geq 0$, meaning the set $X$ is bounded below. Let's show that $0$ is a limit point of the set $X$. Indeed, for any $\varepsilon>0$, we can find a natural number $$n>\log_2\frac{1}{\varepsilon}\,\,\Rightarrow 2^n>\frac{1}{\varepsilon}\Rightarrow\,\,\frac{1}{2^n}<\varepsilon,\quad\frac{1}{2^n}\in X.$$Thus, the set of lower bounds for $X$ is the set $(-\infty, 0]$ with the greatest element equal to $0$. Therefore, we find $\inf X=0$.
Answer: $M=\frac{1}{2}$, the smallest element does not exist, $\sup X=\frac{1}{2}$, $\inf X=0$.
1.80. Let $X\subset \mathbb{R}$ be an arbitrary bounded set. Prove that the set $-X={x,|, -x\in X}$ is also bounded, and the following equalities hold:
$$\sup (-X)=-\inf X,\qquad \inf (-X)=-\sup X.$$
Proof:
Since the set $X$ is bounded, it is bounded above and below. Therefore, there exist numbers $a$ and $b$ such that $\forall x\in X, ,, a\leq x\leq b. $ Solving the inequality, it is evident that for elements $-x$, the inequality $-b\leq -x\leq -a$ holds. Thus, the set $-X={x,|, -x\in X}$ is also bounded.
Let $a=\inf X$. Then from the inequality $-x\leq -a$, we obtain $-x\leq -\inf X.$
If $a\in X$, then $-a\in -X$. In this case, it is obvious that $$-a=\sup(-X)\Rightarrow \sup(-X)=-\inf X.$$
If $a \notin X$, then $-a \notin -X$. Let's demonstrate that $-a$ is the smallest element belonging to the set of upper bounds. Indeed, suppose there exists an element $c \neq a$ such that $-c \notin -X$, but for all $-x \in -X$, $-x\leq -c\leq -a$. Then $c \notin X$, and the inequality $a\leq c\leq x$ holds. Therefore, $a\neq \inf X$, which leads to a contradiction. Thus,$$-a=\sup(-X)\Rightarrow \sup(-X)=-\inf X.$$
Similarly, it can be proven that $\inf (-X)=-\sup X$.
Which completes the proof.
Homework:
1.75. For the set $X=[-1, , 1]$, find $\max X$, $\min X$, $\sup X$, and $\inf X$ if they exist.
Answer: $1$, $-1$, $1$, $-1$.
1.76. For the set$X=\left\{x\in Z|\,\, -5\leq x <0\right\}$, find $\max X$, $\min X$, $\sup X$, and $\inf X$ if they exist.
Answer: Does not exist, $-5$, $0$, $-5$.
1.81. Let $X,,, Y\subset \mathbb{R}$ be arbitrary bounded sets. Prove that the set $X+Y={z\in \mathbb{R},|, z=x+y,,, x\in X,, y\in Y}$ is bounded above, and the following equality holds: $$\sup (X+Y)=\sup X+\sup Y.$$
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