\documentclass{article}
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\title{Math 101 \\ Selected Solutions for Problem Set 3}
\author{Stefan Karpinski}

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\begin{document}

\maketitle



\Problem{W 5.3.11}


Assuming that $0 \not\in \N$, then
\begin{align*}
	\Union_{{n \in \N}} A_{n} = (0,1).
\end{align*}
To prove this note first that $A_{n} \subset (0,1)$ for all $n$,
so that we have $\Union_{{n \in \N}} A_{n} \subseteq (0,1)$.  For
any $x \in (0,1)$, we know by the Archimidean property of the
real numbers that $2^{-n} < x$ for some $n$ in \N.  Pick such $n$ to
be minimal.  Then $x \in A_{n}$.  So $\Union_{{n \in \N}} A_{n}
\supseteq (0,1)$ also and we know the sets are equal as
claimed.  Finally, we see that
\begin{align*}
	[0,1] - \Union_{{n \in \N}} A_{n} = \{0,1\}.
\end{align*}
Many people tried to use limits in this problem and various
arguments about how $\lim_{{n \rightarrow \infty}} 2^{-n} = 0$, but it
never quite gets there.  Since we haven't even considered
the notion of a limit in the class, we certainly can't use
limits to show anything at this point.  It is good that this
problem brings limits to mind as they are at least relevant
in spirit, but this problem is simply a matter of
demonstrating the equality of two sets---which pretty much
never has anything to do with limits.


\Problem{W 8.1.1}


It is standard to require the existence of a multiplicative
identity in the definition of a ring.  With this
requirement, we can simply observe that $-1 x = -x$.  While
this may seems obvious, it is not and requires proof:
\begin{align*}
	x + (-1) x = 1 x + (-1) x = (1 - 1) x = 0 x = 0.
\end{align*}
Then both parts of this problem become a simple matter of
using this fact and and the distributive law.

However, Wolf does not require the existence of a
multiplicative identity in his definition of a ring.  We can
still do the problem, mimicking the trick we just used with
$-1$.  To show that $(-x) + (-y)$ is the additive inverse of $x +
y$, we need to check that
\begin{align*}
	(x + y) + ((-x) + (-y))
	& = x + y + (-x) + (-y) \\
	& = (x + (-x)) + (y + (-y)) \\
	& = 0 + 0 \\
	& = 0.
\end{align*}
The proof for part $(b)$ is similar.


\Problem{W 8.1.3}


Here I give only the answers.  Remember that for your
problem sets, unless explicitly stated otherwise, you need
to justify and explain your answers.  Don't be dismayed if,
as in this problem, this seems to mean that you have to
repeatedly verify lots of annoying axioms.  If you think for
a bit there's usually a way to avoid such tedious mindless
work by one or two incisive observations---that's what
mathematics is all about.  In this case, if you just
``happened'' to chose four problems where the example
structure was \textit{not} a ring then all you needed to do was to
point out one ring axiom that failed.  If it is obvious that
a particular axiom is satisfied then its enough to say so
(but do say it, don't just not say nothing), but make sure
that you do address those axioms that are not obviously


\Part{(a)}


A ring without identity.


\Part{(b)}


Not a ring:  distributivity fails.


\Part{(c)}


Not a ring:  additive associativity fails.


\Part{(d)}


This is a field, which is ``essentially the same as''
($i.e$. isomorphic to) the reals with the normal operations.


\Part{(e)}


This is a ring with additive identity $(0,0)$ and
multiplicative identity $(1,1)$.  Sometimes you will
therefore see strange things like
\begin{align*}
	0 = (0,0) \\
	1 = (1,1)
\end{align*}
which really means
\begin{align*}
	0_{\R\times\R} = (0_{\R} , 0_{\R}) \\
	1_{\R\times\R} = (1_{\R} , 1_{\R})
\end{align*}
where $0_{R}$ just means the additive identity in the ring $R$ and
similarly, $1_{R}$ just means the multiplicative identity in the
ring $R$.  If the context is understood, we simply write $0$ and
$1$.

However, this is not a field since non-zero elements such as
$(0,1)$ and $(1,0)$ have no multiplicative inverses.  In
general, these very same two elements show that the cross
product ($a.k.a$. direct product) of two fields is never a
field, since every field contains $0$ and $1$, and these two
elements are always non-zero and never have (multiplicative)
inverses.


\Part{(f)}


This is a ring:  all the axioms go work because they work
\textit{pointwise} on the functions.  It's not a field since only
functions that are never zero have multiplicative inverses.


\Part{(g)}


Not a ring:  distributivity fails.


\Part{(h)}


This is a ring with identity, but not a field.


\Problem{W 8.1.5}



\Part{(a)}


Again here, when the problem says determine, you have to
justify what you say and explain.  All you need to remark
for most of the axioms ($i.e$. axioms $V3 - V7$ and $V12 - V17$)
is that they hold because they do in the entirety of \R.
Then it remains to show that axioms $V1, V2, V9$ (pretty
clear) and $V11$ still hold, while axioms $V8$ and $V10$ fail.


\Part{(b)}


There are many right answers.  The set of non-negative
reals is one.  \Q -$ \{0\}$ is another.


\Problem{W 8.1.10 and 8.1.11}


Recall that ``the smallest subset of $X$ satisfying the
predicate $P'$' is a subset $A \subseteq X$ satisfying $P$ such
that for every $B \subseteq X$ satisfying $P, A \subseteq B$.
By this definition, as we saw in class, if such a smallest
element exists then it must be the intersection of all $B
\subseteq X$ satisfying $P$.  But the most important part is
that if the intersection of all $B$ satisfying $P$ does not also
satisfy $P$ there is no smallest element satisfying $P$ (since
if there were it would have to be that intersection).

In the first problem $(8.1.10)$ our predicate is ``is a closed
interval containing $3, 7$ and $15$.''  The intersection at hand
is 		
\begin{align*}
	A = \Inter_{{a \leq 3, 15 \leq b}} [a,b].
\end{align*}
Since $[3,15] \subseteq [a,b]$ in every case, $[3,15] \subseteq
A$ also.  And since $[3,15]$ is infact one of the intersected
closed intervals, $A \subseteq [3,15]$ also, proving that $A =
[3,15]$, which is in fact a closed interval containing $3, 7$
and $15$, so it is our desired smallest such set.

In the second problem $(8.1.11)$ our predicate is ``is an open
interval containing $3, 7$ and $15$.''  The intersection at hand
is
\begin{align*}
	B = \Inter_{{a < 3, 15 < b}} (a,b).
\end{align*}
Again since $[3,15] \subseteq [a,b]$ in every case, $[3,15]
\subseteq B$ also.  This time $[3,15]$ is not one of the
intersected sets so we can't make the same argument as
before.  Instead we note that for $x \not\in [3,15]$ we know that
for $0 < \epsilon < \min \{|3 - x|,|15 - x|\}$
\begin{align*}
	x \not\in (3 - \epsilon, 15 + \epsilon),
\end{align*}
so $x \not\in B$.  Thus, $B = [3,15]$.  But this is not an open
interval and thus does not satisfy our predicate.  We
conclude that there is no smallest open interval containing
$3, 7$ and $15$.

\end{document}
