\documentclass{article}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{hyperref}
\usepackage{enumerate}
\pagenumbering{gobble}
\usepackage{tikz}
\usepackage{mathrsfs}
\begin{document}
\vspace*{-1.4in}
\flushleft \underline{Homework 9 -- MATH 2510 Spring 2019}\\
\vspace{0.1in}
Recall: an ordinal is a transitive well-ordered set. The first few ordinals are: $0=\emptyset, 1=\{\emptyset\}, 2=\{0,1\}, 3=\{0,1,2\}, 4=\{0,1,2,3\}, \ldots$. The first infinite ordinal is $\omega = \{0,1,2,3,\ldots\}$. We think of ordinals as being ``ordered by $\epsilon$", i.e. if $\alpha \in \beta$, then we think of $\alpha < \beta$. A function $f \colon A \rightarrow B$ is called one-to-one provided that whenever $f(x)=f(y)$, it follows that $x=y$. For sets $A$ and $B$, we say that $\mathrm{card}(A) \preceq \mathrm{card}(B)$ provided there is a one-to-one function $f \colon A \rightarrow B$. A cardinal number is an ordinal number with the property that if $\beta < \alpha$ (ordinals) then it follows that $\mathrm{card}(\beta) \preceq \mathrm{card}(\alpha)$ (cardinals). Let $\alpha$ and $\beta$ be cardinal numbers, we define cardinal addition by
$$\alpha \oplus \beta = \mathrm{card} \left( (\alpha \times \{0\}) \bigcup \left(\beta \times \{1\} \right)\right) .$$
Cardinal multiplication is defined by
$$\alpha \otimes \beta = \mathrm{card}(\alpha \times \beta).$$
If $A$ and $B$ are sets, then ${}^BA$ denotes the set of functions whose domain is $B$ and whose codomain is $A$.
$${}^BA=\left\{f \colon B \rightarrow A \Bigg| f \text{ is a function}\right\}.$$
Cardinal exponentiation is defined by
$$\alpha^{\beta} = |{}^{\beta}\alpha|.$$
Consider an ordinal $\alpha$ -- we say that a subset $\beta \subset \alpha$ is cofinal with $\alpha$ if it obeys the following property:
$$\forall a \in \alpha \exists b \in \beta ( a \in b).$$
The cofinality of a set $\alpha$ is defined to be the minimum cardinality of all sets which are cofinal with $\alpha$, i.e.
$$\mathrm{cf}(\alpha)=\mathrm{min} \{ |\beta| \colon \beta \text{ is cofinal with } \alpha\}.$$
\begin{enumerate}[1.]
\item Show that the ordinals $\mathrm{card}(\omega + 3)=\mathrm{card}(\omega + \omega)$ by demonstrating appropriate one-to-one functions. \\
\textit{hint}: recall that we can draw $\omega +3$ in the following way: \\
\begin{tikzpicture}
\draw (0,-1) -- (0,1);
\draw (0.25,-0.8) -- (0.25,0.8);
\draw (0.45,-0.6) -- (0.45,0.6);
\draw (0.6,-0.4) -- (0.6,0.4);
\draw (0.7,-0.2) -- (0.7,0.2);
\draw (0.75,-0.15) -- (0.75,0.15);
\draw (0.8,-0.1) -- (0.8,0.1);
\draw (1,-1) -- (1,1);
\draw (1.25,-0.8) -- (1.25,0.8);
\draw (1.45,-0.6) -- (1.45,0.6);
\end{tikzpicture} \\
and we can draw $\omega+\omega$ in the following way: \\
\begin{tikzpicture}
\draw (0,-1) -- (0,1);
\draw (0.25,-0.8) -- (0.25,0.8);
\draw (0.45,-0.6) -- (0.45,0.6);
\draw (0.6,-0.4) -- (0.6,0.4);
\draw (0.7,-0.2) -- (0.7,0.2);
\draw (0.75,-0.15) -- (0.75,0.15);
\draw (0.8,-0.1) -- (0.8,0.1);
\draw (1,-1) -- (1,1);
\draw (1.25,-0.8) -- (1.25,0.8);
\draw (1.45,-0.6) -- (1.45,0.6);
\draw (1.6,-0.4) -- (1.6,0.4);
\draw (1.7,-0.2) -- (1.7,0.2);
\draw (1.75,-0.15) -- (1.75,0.15);
\draw (1.8,-0.1) -- (1.8,0.1);
\end{tikzpicture}
\item Consider the ordinal number $\alpha=\{0,1,2,3\}$. \\
\begin{enumerate}
\item List all subsets of $\alpha$ (i.e. find the power set $\mathscr{P}(\alpha)$).
\item Cross out all subsets of $\alpha$ that are \textbf{not cofinal} with $\alpha$.
\item Write down the cardinalities of all cofinal subsets of $\alpha$.
\item Write $\mathrm{cf}(\alpha)$.
\item (Bonus): What is $\mathrm{cf}(\omega+1)$? (\textit{hint}: recall that $\omega+1=\{0,1,2,\ldots,\omega\}$)
\end{enumerate}
\item Calculate the requested cardinal arithmetic operation. Carefully do it by the definition (i.e. show your sets, etc).
\begin{enumerate}
\item $1 \oplus 2$
\item $3 \otimes 3$
\item $4^2$
\end{enumerate}
\iffalse
\item Consider a new theory called ``differential algebra theory" which has a vocabulary $\mathcal{V}=\{\partial, \cdot, +\}$ where $\partial$ is a unary function, and $+$ and $\cdot$ are binary functions. \\
\textbf{Axiom 1}: $(\forall x)(\forall y)(\partial(x \cdot y)=(\partial x)\cdot y + y \cdot (\partial x))$ \\
\textbf{Axiom 2}: $(\forall x)(\forall y)(\partial(x+y)=\partial(x)+\partial(y))$ \\
Prove (informally is ok)
$$\partial((f \cdot g)\cdot h) =((\partial f)\cdot g+f \cdot(\partial g))\cdot h + (f \cdot g)\cdot (\partial h).$$
Make clear where you use which axiom.
\fi
\end{enumerate}
\end{document}