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\begin{document}

\hbox to \hsize{\textbf{Math 331\hfill Homework 2}}
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\medskip

\centerline{\textit{This homework is due by the end of the day on Friday, September 11}}
\centerline{\textit{Problems are on Sections 1.3 and 1.4, not including Bolzano-Weirstrass.}}

\bigskip

\def\F{{\mathbb F}}
\def\O{{\mathcal O}}

\begin{problem}
[From Problem 1.3.7 in the textbook]\ \ 
Suppose that $(\F,+,\cdot)$ is a field, and $S\subseteq \F$. We say that $S$ is
a subfield of $\F$ if it is a field under the same addition and multiplication as $\F$.
To show that $S$ is a subfield of $\F$, it is enough to show that $0\in S$, $1\in S$,
and $S$ is closed under addition, multiplication, taking additive inverses, and
taking multiplicative inverses.  The other field axioms are automatically true for
multiplication and addition in $S$, since they are already true in $\F$.

Let $\Q[\sqrt 2] = \{r+s\sqrt 2\,|\, r,s\in\Q\}$.  Show that $\Q[\sqrt 2]$ is
a subfield of $\R$.    (Hints: Note that $r$ and $s$ can be zero in $r+s\sqrt 2$.
To show that $S$ is closed under taking multiplicative inverses of non-zero elements,
you might need to show that if $r+s\sqrt 2\ne 0$, then $r^2-2s^2\ne 0$.)
\end{problem}
\begin{answer}
\end{answer}

\begin{problem}
[Problem 1.3.11 from the textbook]\ \ Let $(F,+,\cdot)$ be an ordered field.  Use the
definition of $x<y$ and the order axioms to prove the
transitive property of $<$.  That is, show that for any $a,b,c\in\F$, if $a<b$ and $b<c$,
then $a<c$. [Note: Since $\F$ is not necessarily $\R$, you can't use common facts that you
know about $\R$.  You can only use the actual definition and axioms.]
\end{problem}
\begin{answer}
% Erase this line and put your answer here!
\end{answer}

\begin{problem}
[Problem 1.3.12 from the textbook.  We proved that $1>0$ in class; this problem gives
an alternative proof]\ \ Suppose $(\F,+,\cdot)$ is an ordered field, and $a\in\F$.
Prove: If $a\ne0$ then $a^2>0$.  Conclude that $1>0$. [You can use Theorem 1.3.6,
which says that $(-a)(-b)=ab$ in any field.]
\end{problem}
\begin{answer}
% Erase this line and put your answer here!
\end{answer}

\begin{problem}
\textbf{(a)} Let $\O_1,\O_2\dots,\O_k$ be some finite number of open subsets of
$\R$.  Prove that their intersection, $\bigcap_{i=1}^k\O_i$, is open.  (Hint: Use the characterization
of open that involves $\eps>0$.  Start by taking arbitrary $x\in\bigcap_{i=1}^k\O_i$.)
\textbf{(b)} Show that the intersection of an infinite number of open sets is not necessarily open
by finding $\bigcap_{n=1}^\infty\big(-1-\frac{1}{n},1+\frac{1}{n}\big)$. (Justify your answer!)
\end{problem}
\begin{answer}
% Erase this line and put your answer here!
\end{answer}


\begin{problem}
Consider the \textbf{unbounded} closed interval $[0,\infty)$.  Find an open cover of this
interval that has no finite subcover.  (Prove your answer!)
\end{problem}
\begin{answer}
% Erase this line and put your answer here!
\end{answer}


\begin{problem}
[Problem 1.4.3 from the textbook]\ \ Suppose that $\{\O_\alpha\,|\,\alpha\in A\}$ is an open cover
of the interval $[0,1)$.  Suppose furthermore that $1\in\bigcup_{\alpha\in A}\O_\alpha$.
Prove that there is finite subcover of $[0,1)$ from $\{\O_\alpha\,|\,\alpha\in A\}$.
[This question tests your understanding of the proof of the Heine-Borel Theorem.]
\end{problem}
\begin{answer}
% Erase this line and put your answer here!
\end{answer}

\begin{problem}
Let $f(x)$ be a real-valued function that is defined on an interval $I$.  We say that $f$ is
bounded above on $I$ if there is a number $M$ such that $f(x)<M$ for all $x\in I$.

Suppose that $f(x)$ is defined on the bounded, closed interval $[a,b]$.  Suppose that for
every $x\in[a,b]$, there is an $\eps>0$ such that $f$ is bounded above on the
inteval $(x-\eps,x+\eps)$.  Use the Heine-Borel theorem to prove that
$f$ is bounded above on $[a,b]$.  (Hint: Compare this an example about functions 
that was done in class.)
\end{problem}
\begin{answer}
% Erase this line and put your answer here!
\end{answer}





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