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\lhead{\large \textbf{Math 461}}
\chead{\Large \textbf{Homework 3}}
\rhead{\large \textbf{DUE 9/22/2023}}
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{\Large\underline{\textbf{Exercises:}}} (\S $9$, \S $10$, and \S $11$)



\begin{enumerate}

\item Let $f\colon A\to B$ be a function.
	\begin{enumerate}
	\item Use the axiom of choice to show that if $f$ is surjective, then there exists $g\colon B\to A$ with $f\circ g(b)=b$ for all $b\in B$.
	
	\item Without using the axiom of choice show that if $f$ is injective, then there exists $h\colon B\to A$ with $h\circ f(a)=a$ for all $a\in A$.
	
	\end{enumerate}
	
\item Show that the well-ordering theorem implies the axiom of choice.

\item Let $S_\Omega$ be the minimal uncountable well-ordered set from \S10.
	\begin{enumerate}
	\item Show that $S_\Omega$ has no largest element.
	
	\item Show that for every $x\in S_\Omega$, the subset $\{y\in S_\Omega\mid x<y\}$ is uncountable.
	
	\item Consider the subset
		\[
			X:=\{x\in S_\Omega\mid ~{\rm the ~open~ interval}~(a,x)\neq\emptyset \text{ for all } a<x\}.
		\]
	Show that $X$ is uncountable. [\textbf{Hint:} proceed by contradiction and use the fact that for any $y\in S_\Omega$ there exists $z\in S_\Omega$ with the open interval $(y,z)=\emptyset$.]
	\end{enumerate}


\item In this exercise you will use Zorn's lemma to prove the following fact from linear algebra: every vector space $V$ has a basis. For a subset $A\subset V$, recall: the \textbf{span} of $A$ is the set of all finite linear combinations of vectors in $A$; $A$ is said to be \textbf{independent} if the only way to write the zero vector as a linear combination of elements in $A$ is via the trivial linear combination with all zero scalar coefficients; and $A$ is said to be a \textbf{basis} for $V$ if it is independent and its span is all of $V$.
	\begin{enumerate}
	\item Suppose $A\subset V$ is independent. Show that if $v$ is not in the span of $A$, then $A\cup\{v\}$ is independent.
	
	\item Show that the collection of independent subsets of $V$, ordered by inclusion, has a maximal element.
	
	\item Show that $V$ has a basis.
	\end{enumerate}
	
	
\end{enumerate}




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