[Analysis] Neighborhoods and Limit points

Before we dive into real analysis in earnest, we will go over the important and basic properties of $\mathbb{R}$ in this post so as to understand about $\mathbb{R}$. One property that distinguishes the real numbers $\mathbb{R}$ with others such as $\mathbb{Q}$ and $\mathbb{N}$ is completeness. The completeness of set implies that there are no gaps in that set. For example, in case of rational numbers $\mathbb{Q}$, it has a gap at each irrational value.

There are several equivalent ways to express such a completeness of $\mathbb{R}$. For details, please refer to [1]. And one possible way that we will focus on is the least upper bound property. From this property, we will prove another important idea called Archimedean property of $\mathbb{R}$ and the fact that $\mathbb{Q}$ are dense in $\mathbb{R}$.

---

Supremum and Infimum

$\color{blue}{\mathbf{Definition.}}$ Boundedness
Let $S \subset \mathbb{R}$ be an non-empty set.

1. $S$ is bounded above if $\exists M \in \mathbb{R} \text{ s.t. } x \leq M \quad \forall x \in S$.
2. $S$ is bounded below if $\exists M \in \mathbb{R} \text{ s.t. } x \geq M \quad \forall x \in S$.
3. $S$ is bounded if $S$ is bounded above and below. $\blacksquare$

$\color{blue}{\mathbf{Definition.}}$ Supremum
Let $S \subset \mathbb{R}$ be an non-empty set that is bounded above.
We define the smallest upper bound as a supremum of $S$: $\mu = \text{ sup } S$ where

1. $x \leq \mu \quad \forall x \in S$
2. If $M$ is an upper bound for $S$, then $\mu \leq M$. $\blacksquare$

$\color{blue}{\mathbf{Definition.}}$ Infimum
Let $S \subset \mathbb{R}$ be an non-empty set that is bounded below.
We define the largest lower bound as a infimum of $S$: $\nu = \text{ inf } S$ where

1. $\nu \leq x \quad \forall x \in S$
2. If $m$ is a lower bound for $S$, then $\nu \geq m$. $\blacksquare$

And there are equivalent definitions for supremum and infimum:

$\color{red}{\mathbf{Thm\ 1.}}$ Equivalent definition of supremum
Let $S \subset \mathbb{R}$ be an non-empty set that is bounded above. Let $\mu \in \mathbb{R}$. Then, $\mu = \text{ sup } S$ if and only if for any $\varepsilon > 0$, $\exists \; x \in S$ such that $\mu - \varepsilon < x \leq \mu$.

$\color{red}{\mathbf{Proof.}}$

First of all, suppose $\mu = \text{ sup } S$.

Then, there exists $x \in (\mu - \varepsilon, \mu] \subset S$ for any $\varepsilon > 0$. If not, in other words if there are no points in $S$ in the interval, $\mu - \varepsilon$ will be supremum of $S$.

Conversely, let’s assume that for any $\varepsilon > 0$, $\exists \; x \in S$ such that $\mu - \varepsilon < x \leq \mu$.

Let $M$ be any upper bound for $S$ and assume that $M < \mu$. Then, contradiction occurs since $x < M < \mu$ for all $x \in S$ but if $\varepsilon = \mu - M > 0$, $M < x \leq \mu$.

$\color{red}{\mathbf{Thm\ 2.}}$ Equivalent definition of infimum
Let $S \subset \mathbb{R}$ be an non-empty set that is bounded below. Let $\nu \in \mathbb{R}$. Then, $\nu = \text{ inf } S$ if and only if for any $\varepsilon > 0$, $\exists \; x \in S$ such that $\nu \geq x < \nu + \varepsilon$.

$\color{red}{\mathbf{Proof.}}$

Similar with the previous theorem.

Completeness of $\mathbb{R}$

For now, let’s accept the fact that every nonempty set of $\mathbb{R}$ that is bounded from above has a supremum, and every nonempty set of $\mathbb{R}$ that is bounded from below has an infimum. This property is called least upper bound property.

$\color{blue}{\mathbf{Axiom.}}$ Least Upper Bound Property
Every non-empty subset $S$ of $\mathbb{R}$ that is bounded above has a least upper bound, i.e. supremum. $\blacksquare$

Although the proof is omitted, but it can be easily proved with Dedekind cuts [3]. Based on this axiom, another important property of $\mathbb{R}$ is obtained:

Archimedean property of the real numbers
Let $\varepsilon, M \in \mathbb{R}$ be non-zero positive number. Then, $\exists k \in \mathbb{N} \text{ s.t. } M < \varepsilon \cdot k$. $\blacksquare$

$\color{red}{\mathbf{Proof.}}$

Suppose, for the sake of contradiction, that $M \in \mathbb{R}$ and $M > n$ for every $n \in \mathbb{N}$. Then $M$ is an upper bound for $\mathbb{N}$. It implies that $\mathbb{N}$ is a non-empty set that is bounded above, hence by the least upper bound property, $\mathbb{N}$ has a least upper bound, let $\lambda$.

But consider $\lambda - 1$. For $n \in \mathbb{N}$, $n + 1 \in \mathbb{N}$ too. Since $\lambda$ is an upper bound for $\mathbb{N}$, $\lambda \geq n + 1$. In other words, $\lambda - 1 \geq n$.

Since this is true for all $n \in \mathbb{N}$, $\lambda - 1$ is an upper bound for $\mathbb{N}$. But it contradicts the fact that $\lambda$ is the least upper bound for $\mathbb{N}$. This contradiction proves that cannot be bounded above, and that no such $M$ can exist.

Here are some examples of analyzing sets with boundedness (supremum / infimum) and Archimedean property of $\mathbb{R}$.

$\color{green}{\mathbf{Example\ 1.}}$ $S = \{ (-1)^k (1- \frac{1}{k}) \; | \; k \in \mathbb{N} \}$

As $-1 < (-1)^k (1- \frac{1}{k}) < 1$ for any natural number $k$, a set $S$ is clearly bounded. And we claim that $\text{sup } S = 1$.

Suppose not. Then, there exists an upper bound $0 < M < 1$, but it leads to contradiction that $M$ is upper bound of set $S$ since $M < (1 - \frac{1}{2p}) < 1$ for some $k = 2p \in \mathbb{N}$ by Archimedean property. Precisely, $1 - M < 1$ and by Archimedean property, there exists $2p > 0$ such that $2p \cdot (1 - M) > 1$.

In analogous way, we can prove that $\text{inf } S = -1$ easily.

$\color{green}{\mathbf{Example\ 2.}}$ $S = (a, b)$

We claim that $\text{sup } S = b$. Suppose it isn’t. Then, since $b$ is an upper bound of $S$, $\exists \; M < b$ and $M$ is an upper bound of $S$. But, it implies $\exists \; x$ where $M < x < b$. For example, $x = \frac{M + b}{2} \in S$. Thus contradiction occurs; $\text{sup } S = b$.

In analogous way, we can prove that $\text{inf } S = a$ easily.

$\color{green}{\mathbf{Example\ 3.}}$ Does $S = \{ x_k = \sum_{i=0}^k \frac{1}{i!} \}$ bounded above?

Yes.

v

Hence, $\text{sup } S$ will exist. And one can easily prove that $\text{sup } S = e$.

Notably, as it was mentioned above, Archimedean property leads to the fact that $\mathbb{Q}$ is dense in $\mathbb{R}$.

$\color{red}{\mathbf{Thm\ 3.}}$
Let $a, b \in \mathbb{R}$ and $a < b$. Then there exist $x \in \mathbb{Q} \subset \mathbb{R}$ such that $a < x < b$.

$\color{red}{\mathbf{Proof.}}$

Set $M = 1$, $\varepsilon = b - a > 0$. Then, by Archimedean property, $\exists \; M \in \mathbb{N}$ such that $1 < k \cdot ( b - a )$. Then,

\[\begin{aligned} k \cdot a \leq \ell \in \mathbb{N} < k \cdot a + 1 < k \cdot b \end{aligned}\]

Thus, $a < \frac{\ell}{k} \in \mathbb{Q} < b$.

Reference

[1] Douglass, Steven A. “An introduction to mathematical analysis.” (No Title) (1996).
[2] Wikipedia, Completeness of the real numbers
[3] Reading Guides for Math 331 from Foundations of Analysis, second edition, by David Belding and Kevin Mitchel