\documentclass{article}
\title{The Impact of Functions on Disconnected Sets in Real Analysis}
\author{Bharath Krishnan}
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\begin{document}
\maketitle
\subsection{Introduction}
Mathematicians have analyzed specific cases of functions of disconnected functions (functions on disconnected sets) but seldom analyzed such functions in general. With careful analysis, one would find such functions can be continuous, differentiable and integrable. The concept is implied in definitions of analysis textbooks but there are no examples to reinforce these implications. Moreover, usually given examples of disconnected functions are "no-where continuous" Dirichlet function (the characteristic function of rationals), defined $1$ on rationals and $0$ on irrationals, and Thomae's function defined $1/q$ on $x=p/q$ ($\gcd(p,q)=1$) and $0$ on irrationals.
Such functions can be useful in textbooks for students to thoroughly understand definitions of continuous functions, derivatives and integrals. Moreover, they clear misconceptions students have on these definitions. Such functions are also interesting for people who see functions as a formula that is ambiguous.
Throughout this article, we will explain the rigorous definitions of limits, continuous functions and derivatives and see how these definitions apply to disconnected functions. Then we will observe the impact these issues have on concepts of differentiation and integration theorems.
\subsection{Using ${f(x)}^{g(x)}$}
We could nicely constructed example to understand this general concept. Rather than using a piece-wise function, we will apply this concept to "naturally-made functions" such as ${f(x)}^{g(x)}$. The function would be useful in textbooks and is interesting to analyze in the real and complex plane. The complex function $z^z$ was already analyzed by Mark D. Meyerson in "The $X^X$ spindle". Instead the focus of this article will be strictly on ${f(x)}^{g(x)}$ in the real plane.
Although analyzing ${f(x)}^{g(x)}$ on the cartesian plane is straight-forward, mathematicians are either unaware or dispute the notion that defined points exist at points outside of integers when $f(x)<0$. They would convert the following into a complex multi-valued function.
\subsection{Domain of ${f(x)}^{g(x)}$}
Generally when $a$ is negative, $a^b$ is thought to make sense only if $b$ is an integer. One reason $a^b$ is not continuous on the real plane in terms of the complex plane is that $\lim_{\epsilon \to 0}{(a+\epsilon i)}^b$, where $b \not \in \mathbb{Z}$, does not approach $a^{b}$. Also mathematicians find it crucial that $a^{b}=e^{b\log(a)}$, for all of $x$.
One solution is to convert $f(x)^{g(x)}$ into a complex function and find the point where each branch of ${f(z)}^{g(z)}$ intersects the real plane. The function ${f(x)}^{g(x)}$, when $f(x)<0$, extends to the complex plane as $e^{g(x)(\text{log}|f(x)|+(2n+1)i\pi)}$, where $n\in \mathbb{N}$ represents all branches of the complex logarithm. The complex function is equivelent to $|f(x)|^{g(x)}(\cos((2n+1)\pi g(x))+i\sin((2n+1)\pi g(x)))$.
In order to get the point where $|f(x)|^{g(x)}(\cos((2n+1)\pi g(x))+i\sin((2n+1)\pi g(x)))$ intersects the real plane plane for each $n\in\mathbb{N}$, $((2n+1)\pi g(x))$ must equal $k\pi$ for $k\in\mathbb{Z}$ so the imaginary component is zero.
Say $g(x)\in{\mathbb{R} \backslash \mathbb{Q}}$, then we need $(2n+1)\pi(g(x))=k\pi$ but since $g(x)$ is irrational $(2n+1)\pi(g(x)){\neq}k\pi$. Hence the $x$ values that make the output of $g(x)$ irrational would make $f(x)^{g(x)}$ undefined in the real plane.
Now we could focus on the values of $g(x)$ that gives rationals. Say we find that $g(x)=\left\{\left.\frac{2p+1}{2q}\right|p,q \in \mathbb{Z}\right\}$ or $x=\left\{\left.g^{-1}\left(\frac{2p+1}{2q}\right)\right|p,q \in \mathbb{Z}\right\}$. Then if $(2n+1)\pi\left(\frac{2p+1}{2q}\right)=k\pi$, $2n+1=m(2q)$ for any $m\in{\mathbb{N}}$ in order for $(2n+1)\pi\left(\frac{2p+1}{2q}\right)$ to cancel into $k\pi$. So if $m=\frac{2n+1}{2q}$, then $m\not\in{\mathbb{N}}$. However, this is a contradiction since $m$ must be an integer and hence ${f(z)}^{g(z)}$ has no branches that intersect the real plane $x=\left\{\left.g^{-1}\left(\frac{2p+1}{2q}\right)\right|p,q \in \mathbb{Z}\right\}$.
If $g(x)=\left\{\left.\frac{2p+1}{2q+1}\right|p,q \in \mathbb{Z}\right\}$ where $p\text{,}q \in \mathbb{N}$ than $(2n+1)\pi\left(\frac{2p+1}{2q+1}\right)=k\pi$. In order for to happen $2n+1=m(2q+1)$ where $m\in \mathbb{N}$. Since $m=\frac{2n+1}{2q+1}$, $m$ must be an odd integer. Thus we can rewrite $\left\{\left. m \right|m\in{2\mathbb{N}+1} \right\}$ as $\left\{\left. {2m+1}\right|m \in \mathbb{N}\right\}$ and get $2n+1=(2m+1)(2q+1)$ which by substitution gives $(2m+1)(2q+1)\pi\left(\frac{2p+1}{2q+1}\right)$. As a result we get $\left|f\left(g^{-1}\left(\frac{2p+1}{2q+1}\right)\right)\right|^{\frac{2p+1}{2q+1}}\left(\cos((2m+1)(2p+1)\pi)+\right.$ $\left.i\sin((2m+1)(2p+1)\pi)\right)$. Since $(2m+1)(2p+1)$ is always odd we end up with ${f(x)}^{g(x)}=-{|f(x)|}^{g(x)}$ for $x=\left\{\left.g^{-1}\left(-\frac{2p+1}{2q+1}\right)\right|p,q \in \mathbb{Z}\right\}$ when $f\left(g^{-1}\left(\frac{2p+1}{2q+1}\right)\right)<0$.
However, if $g(x)$ is equal to $\frac{2p}{2q+1}$ where $p,q \in \mathbb{N}$ than $(2n+1)\pi\left(\frac{2p}{2q+1}\right)=k\pi$. In order for this to happen $2n+1=m(2q+1)$. Since $m=\frac{2n+1}{2q+1}$, $m$ must be odd. Hence we replace odd $\left\{\left.m\right| m\in 2\mathbb{N}+1\right\}$ with $\left\{\left.2m+1\right|m\in\mathbb{N}\right\}$ to get $2n+1=(2m+1)(2q+1)$. Using substitution, one gets $(2m+1)(2q+1)\pi(\frac{2p}{2q+1})$. Thus we end up with $|f(x)|^{g(x)}\left(\cos((2m+1)(2p)\pi)+i\sin((2m+1)(2p)\pi)\right)$. Since $(2m+1)(2p)$ is always even, we end up with $|f(x)|^{g(x)}$ for $x=\left\{\left.g^{-1}\left(\frac{2p}{2q+1}\right)\right|p,q \in \mathbb{Z}\right\}$ when $f\left(g^{-1}\left(\frac{2p+1}{2q+1}\right)\right)<0$.
Hence for $S(x)={f(x)}^{g(x)}$ we must find values where $g(x)=\left\{\left.\frac{p}{2q+1}\right|p,q\in\mathbb{Z}\right\}$ where $p$ is an odd or even integer and $q$ is an odd integer. Depending on if $p$ is odd or even it may end up in the rule $|f(x)|^{g(x)}$ or $-|f(x)|^{g(x)}$.
In piece-wise notation, we can express ${f(x)}^{g(x)}$ as the following.
\begin{equation}
{f(x)}^{g(x)}=\begin{cases} {|f(x)|}^{g(x)} & \left\{x= g^{-1}\left(\frac{2p}{2q+1} \right) |\ p, q \in \mathbb{Z}\right\}\land \left\{f(x)<0\right\}\\ -|f(x)|^{g(x)} & \left\{x=g^{-1} \left( \frac{2p+1}{2q+1} \right)| p, q \in \mathbb{Z}\right\}\land \left\{f(x)<0\right\}\\ {f(x)}^{g(x)} & f(x)>0\\ \end{cases}
\end{equation}
Here we find when $f(x)>0$, the function is defined on connected sets but when $f(x)<0$, the function is defined on disconnected sets that forms a totally disconnected space.
This can also be generalized to all disconnected functions.
\begin{equation}
S(x)=\begin{cases} I_1(x)& x\in A_1\\ I_2(x) & x\in A_2\\ I_3(x)& x\in A_3\\ {}\quad \text{.} & \text{}\quad{.}\\{}\quad \text{.}& {}\quad\text{.}\\I_n(x) & x\in A_n \end{cases}
\end{equation}
Where $I_1(x),I_2(x),..,I_n(x)$ are "rules" which are continuous almost everywhere and $A_1,A_2,A_3,..,A_n$ ($B_n=\bigcup_{i=1}^{n}A_i$ are disjointed. (Note that two or more rules may be equal to one another)
If we were to set $S(x)={f(x)}^{g(x)}$ then $I_1(x)={|f(x)|}^{g(x)}$, $I_2(x)={|f(x)|}^{g(x)}$, $A_1=\left\{\left.g^{-1} \left( \frac{2p+1}{2q+1}\right| p,q \in \mathbb{Z} \right)\right\}$ and $A_2=\left\{\left. g^{-1} \left( \frac{2p+1}{2q+1} \right)\right| p,q \in \mathbb{Z}\right\}$
\subsection{Limit Points and Continuity}
In standard Calculus, the definition of a limit is $\lim_{x\to a}f(x)=L$, which means that for every $\epsilon>0$ there is corresponding $\delta>0$ where $|x-a|<\delta$ such that $|f(x)-L|<\epsilon$.
This definition is true for a function defined on the interval. However, totally disconnected functions require specific sets of real numbers to approximate $a$. Hence a more precise definition is required.
"Bartles Elements of Real Analysis" states the following definition.
\subsubsection{Definition}
Let $B_n\subseteq\mathbb{R}$ ($B_n$) and let $a$ be a cluster point of B. For a function $f:B_n\to \mathbb{R}$ a real number $L$ is said to be the limit of $f$ at $a$ if, given any $\epsilon>0$ there exists $\delta>0$ such that if $x\in{B_n}$ and $0<|x-a|<\delta$, then $|f(x)-L|<\epsilon$.
According to definition, we can choose any $\epsilon$ and $\delta$ as long as $B$ has a cluster point at $a$. This means $B_n$ can closely approximate $a$. (Note that $a$ may not belong to $B_n$).
Inorder to determined the point where the limit exists we must find the values that are cluster points of $A_1$ and $A_2$.
We will begin by assuming that any $a\in\mathbb{R}$ in ${f(x)}^{g(x)}$ is a cluster point of $A_1$ and $A_2$. In order to prove this assumption, $A_1$ and $A_2$ must be dense in $\mathbb{R}$. If $g$ is continuous, analytic and bounded then if $\left\{\left.\frac{2p+1}{2q+1}\right|p,q\in \mathbb{Z}\right\}$ and $\left\{\left.\frac{2p}{2q+1}\right|p,q\in \mathbb{Z}\right\}$ are dense everywhere then so would $A_1=\left\{\left.g^{-1}\left(\frac{2p}{2q+1}\right)\right|p,q \in \mathbb{Z}\right\}$ and $A_2=\left\{\left.g^{-1}\left(\frac{2p+1}{2q+1}\right)\right|p,q \in \mathbb{Z}\right\}$.
There is a rigorous way of proving that $\left\{\left.\frac{2p+1}{2q+1}\right|p,q\in \mathbb{Z}\right\}$ and $\left\{\left.\frac{2p}{2q+1}\right|p,q\in \mathbb{Z}\right\}$ is dense everywhere; but I will give an informal proof.
For example to prove $\left\{\left.\frac{2p+1}{2q+1}\right|p,q \in \mathbb{Z}\right\}$ is dense. Show the set can approximate $a$ from the values lesser than or greater than $a$.
We could then solve for the integer $p$ to get
\begin{equation}
p=\frac{1}{2}\left\lfloor{a(2q+1)}\right\rfloor-\frac{1}{2}
\end{equation}
The floor function is used since $p$ must be an integer.
Then substitute $p$ into (3) to get which leads to an inequality
\begin{equation}
\frac{2\left\lfloor\frac{1}{2}{a(2q+1)}-\frac{1}{2}\right\rfloor+1}{2p+1}0$ making ${|f(x)|}^{g(x)}>0$ and $-{|f(x)|}^{g(x)}<0$. However ${f(x)}^{g(x)}$, ${|f(x)|}^{g(x)}$ and ${-|f(x)|^{g(x)}}$ can intersect when $f(x)=0$.
On the other hand, if we substitute ${f(x)}^{g(x)}$ into $P(x)$ such that $P\left(|f(x)|^{g(x)}\right)=P\left(-|f(x)|^{g(x)}\right)$ then the $\underset{x\to a}{\lim}{f(x)}^{g(x)}$ can exist for any $a\in\mathbb{R}$. The limit would be in the form of $P\left(|{f(a)|}^{g(a)}\right)$.
Mathematicians might argue one could extend $P(f(x)^{g(x)})$ into $P(|f(x)|^{g(x)})$. However, for the purpose of analysis we will see if continuity, differentiability or integrability can exist for this disconnected function.
Using the analysis definition of a limit we could apply a similar principal for continuity of ${f(x)}^{g(x)}$.
\subsubsection{Definition}
Let $B_n\subseteq\mathbb{R}$, let $f:B_n \to \mathbb{R}$ and let $a \in {B}$. We say that $f$ is continuous at $a$ if, given any number $\epsilon>0$ such that $x$ is any point of $B$ satisfying $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.
This rigorous definition states that as the $\epsilon$ band narrows and approaches $L$ the $\delta$ band should narrows and approaches $a$, $L$ should equal $f(a)$.
Informally, the limit at $a$ which is a cluster point of $A_1$ and $A_2$ must equal ${f(a)}^{g(a)}$. Hence the following is required.
\begin{equation}
\underset{x\in A_1}{\underset{x\to a}\lim}{f(x)^{g(x)}}=\underset{x\in A_2}{\underset{x\to a}\lim}{f(x)^{g(x)}}={f(a)}^{g(a)}
\end{equation}
We already know from (8), the limit can exist only at $f(x)=0$. If ${0}^{g(f(0))}$ is defined, then the point is continuous.
For $P({f(x)}^{g(x)})$ the limit can exist anywhere but continuous points exist where $P\left({f(a)}^{g(a)}\right)$ is defined. The continuous points exist in the domain of $P({f(x)}^{g(x)})$ and hence the function is continuous.
Some may have misconceptions that continuous function should not have discontinuous, undefined "breaks". However, these "breaks" are actually removable singularities: points where the limit is finite and undefined. Generally, continuities and discontinuities are determined in the domain making removable singularities neither continuities nor discontinuities. The only exceptions are at vertical asymptotical discontinuities, which splits a function into several parts.
Mathematician tend to replace removable singularities with continuous points as singularities are viewed as unnecessary; however, mathematical textbooks state functions need only be continuous in the domain. Here is an analogy, picture a steel wire made up of atoms which are made of particles that are made of even smaller particles and so on. Now matter how small the particles are there is always space between each one, making the steel wire disconnected. However, we see the steel wire as connected since we only see the "defined particles" instead of the "undefined space". Replacing the undefined space with a defined space is merely an illusion.
Hence $P(f(x)^{g(x)})$ is continuous if there are no discontinuities at vertical asymptotes.
Generalizing the following case to $S(x)$, if $S(x)$ has sets defined by more than one rule, then $\underset{\left\{x\in{B_n}\right\}\to a} \lim S(x)=\bigcap_{i=1}^{n}I_{i}(a)$ and is continuous at $x=a_1$ where $\left\{a_1\right
\}$ is the set of values of $a$ where $\bigcup_{i=1}^{n}I_n(a)$.
If $S(x)$ is defined by the same rule, then $\underset{{\left\{x\in A_1\right\}}\to a} \lim S(x)={I_1}(a)=I_2(a)=...=I_n(a)$ and $S(x)$ is continuous at $x=B_n$.
\subsection{The Existence of a Derivative}
In this section we will show that the derivative of a disconnected function can exist.
If ones takes the newton-quotient of the derivative.
\begin{equation}
\underset{x\in B_n}{\underset{h\to0}{\lim}}\frac{S(x+h)-S(x)}{h}
\end{equation}
Since $S(x)$ is disconnected, we must restrict $h$ to values $f(x+h)$ is defined. Finding such an $h$ would be tedious.
In most real analysis textbooks, the derivative is not shown as a newton quotient. Instead they use the following notation.
\begin{equation}
\lim_{x\to a}\frac{S(x)-S(a)}{x-a}
\end{equation}
This is preferred since we can focus properties of $f(x)$ as opposed to $f(x+h)$.
In "Elementary Real Analysis", the author states a similar but rigorous definintion. He says, "Let $B_n\subset \mathbb{R}$ be an interval, let $S:B_n\to \mathbb{R}$, and let $a \in B_n$. We say that the real number $L$ is the derivative of $f$ at $a$ if any given $\epsilon>0$ there exists $\delta(\epsilon)>0$ such that if $x\in B_n$ satisfies $0<|x-a|<\delta(\epsilon)$.
\begin{equation}
\left|\frac{S(x)-S(a)}{x-a}-L\right|<\epsilon
\end{equation}
The definition states as the epsilon band $L-\epsilon<\frac{S(x)-S(a)}{x-a}S(x)$ and a minimum exist if $S(x_0)~~0$ and a disconnected set when $f(x)<0$.
3. A group of elementary functions that is not Riemman integrable or Lebesque integrable.
The summary of ${P\left({f(x)}^{g(x)}\right)}$
1. Has a limit that can exist everywhere.
2. Is "naturally" continuous and differentable on a domain that is a disconnected space(Otherwise discontinuous at vertical asymptotes)
3. Has no Riemman or Lebesque integral. But can have an "Extended Riemman integral".
\subsection{Propositions}
1. For the function to be a continuous, differentable or integratable function; the function must be continuous, differentable or integratable in the domain.
2. The Darboux sum should not have the requirement $f:[a,b]\to\mathbb{R}$
3. The Riemman integral should be extended to include functions that are not defined everywhere.
4. Integrals should exist for functions not defined everywhere as the defined points belong to set dense in $\mathbb{R}$.
\subsubsection{Conclusion on Properties of Disconnected functions}
1. A function can be continuous and differentiable but not defined almost everywhere.
2. The anti-derivative and indefinite integral are not the same. The "usual" Riemmen indefinite integral is different from the anti-derivative at removable or jump discontinuities. The extended Riemmen integral may differ at removable singularities.
3. If the sets $B_n$ are not defined everywhere then $S(x)$ is an example of a functions that is not Lebesque nor Riemman integrtable. A new or extended kind of an integral could be formed which compares defined points of one set to defined points of another set.
\subsection{Allowing an extended Riemman integral}
The extended Riemman integral is different from Indefinite integral at points that are not continuous (removable singularities and discontinuities)
3. (Proposition) The indefinite integral should not have the $C$ sign.
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