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\newcommand{\COR}{\vspace{0.05in} \noindent \fbox{\large\bf Corollary:}  }
\newcommand{\PROP}{\vspace{0.05in} \noindent \fbox{\large\bf Proposition:}  }
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\newcommand{\THM}{\vspace{0.05in} \noindent \fbox{\large\bf Theorem:}  }
\newcommand{\EX}{\vspace{0.05in} \noindent \fbox{\large\bf Example:}  }
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% Inspired by a weird conversion with Quinn and Jeff. Draft typed up circa October 2023.

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\noindent
{\bf Weird Example}
\hfill {\Large \bf Space of Continuous Functions} \hfill 
{\bf Linear Algebra}

\vspace{0.15in}

\noindent {\it Note:} This example is less about linear algebra and more about cardinal numbers. Don't take it too seriously.

\vspace{0.1in}

\THM The dimension of the space of continuous function is continuum.

\vspace{0.1in}

\bproof
First, some notation. Let $\mathcal{C}=\mathcal{C}^0=\mathcal{C}(\mathbb{R}) = \{ f:\mathbb{R}\to\mathbb{R} \;|\; f \text{ is continuous}\}$ be the set of continuous functions (from the real numbers to themselves). This is a vector space over $\mathbb{R}$ since adding and scaling continuous functions yields continuous functions. Next, let $\mathcal{C}^k = \mathcal{C}^k(\mathbb{R}) = \{ f :\mathbb{R} \to \mathbb{R} \;|\; f^{(k)} \text{ exists and is continuous}\}$. This is the space of $k$-times continuously differentiable functions. Also, let $\mathcal{C}^\infty = \mathcal{C}^\infty(\mathbb{R}) = \{ f:\mathbb{R} \to \mathbb{R} \;|\; f^{(k)} \text{ exists for all }k \geq 0\}$. This is the space of smooth functions. 

\vspace{0.05in}

By definition, $f^{(k+1)}$ (the $(k+1)^\text{st}$ derivative of $f$) exists only if $f^{(k)}$ (the $k^\text{th}$ derivative of $f$) exists. Moreover, a basic result from Calculus tells us that differentiable functions must be continuous functions. In particular, if $f^{(k+1)}$ exists, by definition $f^{(k)}$ is differentiable and thus must be continuous. Therefore, for any positive integer $k$, $\mathcal{C}^\infty \subseteq \cdots \subseteq \mathcal{C}^k \subseteq \cdots \subseteq \mathcal{C}^1 \subseteq \mathcal{C}^0$ ($=\mathcal{C}$). These are non-empty since the constant function $0$ is smooth (and thus belongs to all of these sets). In addition, if $f^{(k)}$ and $g^{(k)}$ are continuous, then $(f+g)^{(k)}$ and $cf^{(k)}$ (for any $c \in \mathbb{R}$) are continuous as well. Therefore, these sets are (nested) subspaces of the space of all functions from the reals to the reals.

\vspace{0.05in}

Now we consider cardinalities. We denote countable infinity by $\aleph_0$ and continuum infinity by $\mathfrak{c}=2^{\aleph_0}$. A basic result from analysis says the rational numbers, $\mathbb{Q}$, are dense in the reals, $\mathbb{R}$. Another basic result tells us that a continuous function is determined by its values on any dense subset of its domain. Thus continuous functions are determined by their rational inputs. Therefore, there are at most $|\{f:\mathbb{Q}\to\mathbb{R}\}| = |\mathbb{R}|^{|\mathbb{Q}|} = (2^{\aleph_0})^{\aleph_0} = 2^{\aleph_0 \cdot \aleph_0} = 2^{\aleph_0} = \mathfrak{c}$ (i.e., continuum cardinality) continuous functions. On the other hand, constant functions: $f(t)=c$ for any fixed $c \in \mathbb{R}$ are infinitely differentiable (e.g., $f^{(k)}(t)=0$ for $k>0$), so that there are at least $|\mathbb{R}|=\mathfrak{c}$ smooth functions. 
This means $\mathfrak{c} \leq |\mathcal{C}^\infty| \leq \cdots \leq |\mathcal{C}^k| \leq \cdots \leq |\mathcal{C}| \leq \mathfrak{c}$. 
 Therefore, all of the above spaces (i.e., $\mathcal{C}$, $\mathcal{C}^\infty$, $\mathcal{C}^k$ for $k=1,2,\dots$) have cardinality $\mathfrak{c}$. Since vector spaces span themselves and every spanning set contains a basis: $\dim(V) \leq |V|$. In particular, the dimensions of these spaces are bounded above by their cardinality (i.e., $\mathfrak{c}$).

\vspace{0.05in}

Finally, consider distinct real numbers $r_1,\dots,r_\ell \in \mathbb{R}$ and let $f_i(t) = e^{r_it}$ for $i=1,\dots,\ell$. Obviously, $f_i \in \mathcal{C}^\infty$. In fact, $f_i^{(k)}(t) = r_i^k e^{r_it}$. Suppose $c_1f_1+\cdots+c_\ell f_\ell=0$. Then $c_1f_1^{(k)}+\cdots+c_\ell f_\ell^{(k)}=0$ for any non-negative integer $k$. Evaluating at $t=0$, we get $c_1r_1^ke^{r_10}+\cdots+c_\ell r_\ell^ke^{r_\ell 0}=0$ and thus $c_1r_1^k+\cdots+c_\ell r_\ell^k=0$. In particular, using the first $k$ equations, we have:
$$\begin{bmatrix} 1 & 1 & \cdots & 1 \\ r_1 & r_2 & \cdots & r_\ell \\  \vdots & \vdots & \ddots & \vdots \\ r_1^{k-1} & r_2^{k-1} & \cdots & r_\ell^{k-1} \end{bmatrix} \begin{bmatrix} c_1 \\ c_2 \\ \vdots \\ c_\ell \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ \vdots \\ 0 \end{bmatrix}$$
The above coefficient matrix is called a Vandermonde matrix. It is well known that its determinant is $\prod\limits_{i<j}^\ell (r_j-r_i) \not= 0$. Since our coefficient matrix is non-singular, the only solution to this system is the trivial solution: $c_1=c_2=\cdots=c_\ell=0$. Consequently our functions are linearly independent. 

\vspace{0.05in}

We have shown that $S = \{ e^{rt} \;|\; r \in \mathbb{R}\}$ is a linearly independent set of smooth functions. Since linearly independent sets can be extended to bases, $\mathfrak{c}=|S| \leq \dim(\mathcal{C}^\infty) \leq \cdots \leq \dim(\mathcal{C}^k) \leq \cdots \leq \dim(\mathcal{C})$. However, we already know these dimensions are bounded above by $\mathfrak{c}$. Therefore, $\dim(\mathcal{C}^\infty) = \dim(\mathcal{C}^k) = \dim(\mathcal{C})=\mathfrak{c}$ (i.e., all of the function spaces under consideration are continuum-dimensional vector spaces over the real numbers).
\eproof

\vspace{0.1in}

\noindent
{\it Analytic Functions:} If we let $\mathcal{C}^\omega = \{ f:\mathbb{R} \to \mathbb{R} \;|\; f \text{ is analytic }\}$ denote all real analytic functions (these are functions which can be expanded into Taylor series). Then one easily checks that $\mathcal{C}^\omega$ is a subspace of $\mathcal{C}^\infty$. Since the constant functions are analytic and $e^{rt}$ is analytic for any $r \in \mathbb{R}$, the argument above shows that $\dim(\mathcal{C}^\omega)=\mathfrak{c}$ as well. In other words, the space of power series (= real analytic functions) is continuum dimensional. On the other hand, $\{1,t,t^2,\dots\}$ forms a basis for real polynomial functions so that it is only countable (i.e., $\aleph_0$) dimensional.

\vspace{0.1in}

\newpage

\noindent
{\bf Weird Example}
\hfill {\Large \bf Rational Functions} \hfill 
{\bf Linear Algebra}

\vspace{0.15in}

\noindent {\it Note:} Like the previous page, this example is weird in terms of our usual undertakings. Don't take it too seriously.

\vspace{0.1in}

Let $\mathbb{F}$ be a field. Recall that $\mathbb{F}[t] = \{ a_nt^n+\cdots+a_1t+a_0 \;|\; n \geq 0\; ;\; a_0,\dots,a_n \in \mathbb{F}\}$ is the set of all polynomials in indeterminate $t$ with coefficients in $\mathbb{F}$. This collection has the structure of a (commutative associative unital) {\it algebra} (over $\mathbb{F}$). Essentially this means one can add, scale (by $\mathbb{F}$), and multiply these polynomials. In terms of ring theory, $\mathbb{F}[t]$ is quite a nice of ring. It is a {\it Euclidean domain} (we can divide with remainder and the extended Euclidean algorithm works). This implies it is also a {\it principal ideal domain} (PID) and thus a {\it unique factorization domain} (UFD). Ignoring polynomial multiplication, $\mathbb{F}[t]$ is an infinite dimensional vector space over $\mathbb{F}$ with standard basis $\beta =\{1,t,t^2,\dots\}$. This means $\dim(\mathbb{F}[t])=\aleph_0$ (i.e., it is countably infinite dimensional). 

\vspace{0.05in}

Given an integral domain (a commutative ring where cancellation works), one can formally construct its {\it field of fractions}. For polynomials $\mathbb{F}[t]$, this would be $\mathbb{F}(t) = \{ f(t)/g(t) \;|\; f(t),g(t) \in \mathbb{F}[t] \text{ and } g(t)\not=0\}$. This ring $\mathbb{F}(t)$ is called the field of {\it rational functions} with coefficients in $\mathbb{F}$. Just like $\mathbb{F}[t]$, $\mathbb{F}(t)$ is a (commutative associative unital) algebra (over $\mathbb{F}$).
While it is (just like $\mathbb{F}[t]$) a Euclidean domain, PID, UFD, and vector space (over $\mathbb{F}$), it is even nicer. $\mathbb{F}(t)$ is a field. We ask, what is $\dim(\mathbb{F}(t)$ (when working over $\mathbb{F}$)?

\vspace{0.05in}

Let us sketch out how this works over a general field and then specialize to when $\mathbb{F}$ is the real or complex numbers. We recall $f(t) \in \mathbb{F}[t]$ is {\it monic} if its leading coefficient is $1$, and $f(t)$ is {\it irreducible} if $f(t)$ is non-constant and has no non-trivial factorizations (i.e., $f(t)=g(t)h(t)$ implies $g(t)$ or $h(t)$ is constant). Because $\mathbb{F}[t]$ is a unique factorization domain, every non-constant polynomial can be factored uniquely (up to order and associates) into irreducibles.

\vspace{0.05in}

 Consider $\frac{f(t)}{g(t)} \in \mathbb{F}(t)$.  Suppose $g(t)=a(t)b(t)$ where $a(t),b(t) \in \mathbb{F}[t]$ are relatively prime. By the extended Euclidean algorithm, there exists $m(t),n(t) \in \mathbb{F}[t]$ such that $m(t)a(t)+n(t)b(t)=1$. Thus $\frac{f(t)}{g(t)}=\frac{f(t)(m(t)a(t)+n(t)b(t))}{a(t)b(t)} = \frac{f(t)m(t)}{b(t)}+\frac{f(t)n(t)}{a(t)}$. From this we learn if $g(t)=p_1(t)^{k_1}\cdots p_\ell(t)^{k_\ell}$ where $p_i(t)$ are non-associate (i.e., distinct and thus relatively prime) irreducibles and $k_i>0$, then we can find $h_1(t),\dots,h_\ell(t) \in \mathbb{F}[t]$ such that $\frac{f(t)}{g(t)} = \frac{h_1(t)}{p_1(t)^{k_1}}+\cdots+\frac{h_\ell(t)}{p_\ell(t)^{k_\ell}}$. Focus on a factor $\frac{h(t)}{p(t)^k}$. Because $\mathbb{F}[t]$ is a Euclidean domain, we have a division algorithm and can divide $h(t)$ by $p(t)$. We have unique $q(t),r(t) \in \mathbb{F}[t]$ such that $h(t)=p(t)q(t)+r(t)$ with $r(t)=0$ or $\deg(r(t)) < \deg(p(t))$. Thus $\frac{h(t)}{p(t)^k} = \frac{q(t)}{p(t)^{k-1}}+\frac{r(t)}{p(t)^k}$. Continuing in this fashion, we can write $\frac{h(t)}{p(t)^k} = q(t)+\frac{r_1(t)}{p(t)}+\frac{r_2(t)}{p(t)^2}+\cdots+\frac{r_k(t)}{p(t)^k}$ where $q(t),r_1(t),\dots,r_k(t) \in \mathbb{F}[t]$ and each $r_i(t)$ is either 0 or has degree at most $\deg(p(t))$.
 Putting all of this together, \fbox{$\frac{f(t)}{g(t)} = q(t)+\sum_{i=1}^\ell \sum_{j=1}^{k_i} \frac{r_{ij}(t)}{p_i(t)^j}$} for some $q(t),r_{ij}(t) \in \mathbb{F}[t]$ where $r_{ij}(t)=0$ or $\deg(r_{ij}(t))<\deg(p_i(t))$ for all $i,j$. The boxed in equation is called the {\it partial fraction decomposition} of $\frac{f(t)}{g(t)}$. 

\vspace{0.05in}

It can be shown (see \cite{parfrac}) this decomposition is {\it unique}. Thus if we let
$\mathcal{I} = \{ p(t) \in \mathbb{F}[t] \;|\; p(t) \text{ monic irreducible}\}$, then we have 

\vspace{-0.25in}

$$\beta = \{ t^k \;|\; k \in \mathbb{Z}_{\geq 0} \} \cup \left\{ \frac{t^k}{p(t)^\ell} \;\middle|\; p(t) \in \mathcal{I} \text{ and } k,\ell \in \mathbb{Z} \text{ where } 0 \leq k<\ell \right\}$$

\vspace{-0.05in}

\noindent
is a basis for $\mathbb{F}(t)$. One can show (roughly because finite subsets of an infinite set have the same cardinality as that set) the cardinality of $\mathbb{F}[t]$ is $\aleph_0 \cdot |\mathbb{F}|$ (which is $|\mathbb{F}|$ if for infinite fields and $\aleph_0$ for finite fields). Note $\mathcal{I}$ contains linear monics: $x-c$ for each $c \in \mathbb{F}$, so $|\mathcal{I}| \geq |\mathbb{F}|$. Also, it can be shown that finite fields have irreducibles of all positive degrees, thus $|\mathcal{I}|$ must always be infinite. Therefore, $|\mathcal{I}| \geq \aleph_0 \cdot |\mathbb{F}|$. Next, since $\mathcal{I} \subseteq \mathbb{F}[t]$, we have its cardinality is at most $|\mathbb{F}[t]|=\aleph_0 \cdot |\mathbb{F}|$, so $|\mathcal{I}|=\aleph_0 \cdot |\mathbb{F}|$. Consider $S=\{ (k,\ell) \in \mathbb{Z}_{\geq 0} \times \mathbb{Z}_{> 0} \;|\; k<\ell \}$. This set has at least $|\mathbb{Z}_{\geq 0}|=\aleph_0$ elements and at most $|\mathbb{Z}_{\geq 0}\times \mathbb{Z}_{>0}| = \aleph_0 \cdot \aleph_0 = \aleph_0$ elements, so $|S|=\aleph_0$. Putting this together, 

\vspace{-0.1in}

$$|\beta|=|\{t^k\;|\;k \in \mathbb{Z}_{\geq 0}\}|+|\mathcal{I}| \cdot |S| = 
   \aleph_0+(\aleph_0 \cdot |\mathbb{F}|) \cdot \aleph_0 = \aleph_0 \cdot |\mathbb{F}|$$

\vspace{-0.05in}

\THM The dimension of the space of rational functions with coefficients in a field $\mathbb{F}$ thought of as a vector space over $\mathbb{F}$ is $\dim(\mathbb{F}(t)) = \aleph_0 \cdot |\mathbb{F}|$ (i.e., $\aleph_0$ for finite fields and $|\mathbb{F}|$ otherwise).

\vspace{0.05in}

In particular, $\beta=\{1\} \cup \left\{ t^k,\frac1{(t-r)^k},\frac1{(t^2+bt+c)^k},\frac{t}{(t^2+bt+c)^k} \,\middle|\, k \in \mathbb{Z}_{>0} \text{ and } r,b,c \in \mathbb{R} \text{ where } b^2<4c\right\}$ is a basis for $\mathbb{R}(t)$ and $\dim(\mathbb{R}(t))=|\mathbb{R}|=\mathfrak{c}$. 
And $\beta = \{1\} \cup \left\{ t^k,\frac1{(t-z)^k} \,\middle|\, k \in \mathbb{Z}_{>0} \text{ and } z \in \mathbb{C} \right\}$ is a basis for $\mathbb{C}(t)$ and $\dim(\mathbb{C}(t))=|\mathbb{C}|=\mathfrak{c}$. 

\begin{thebibliography}{my-bib}
\bibitem[BC]{parfrac} T. Bradley and W. Cook, ``Two Proofs of the Existence and Uniqueness of the Partial Fraction Decomposition'', International Mathematical Forum. 7 (2012), no. 31, 1517--1535.\\
 Avaialable online: \href{http://www.m-hikari.com/imf/imf-2012/29-32-2012/cookIMF29-32-2012.pdf}{http://www.m-hikari.com/imf/imf-2012/29-32-2012/cookIMF29-32-2012.pdf}
\end{thebibliography}

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