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\centerline{Mathematics 242 -- Principles of Analysis}
\centerline{Selected Solutions, Problem Set 2}
\centerline{September 22, 2000}
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\noindent
\S 1.5/16.  We want to show that if 
$$A = \{x\in {\bf R} : 0 < x < 1\ {\rm and}\ x = \sqrt{2}r\ {\rm for}\ r \in {\bf Q}\}$$
then ${\rm lub}(A) = 1$.  To do this we must show that $1$ is an upper
bound for $A$, and that no $\beta < 1$ is an upper bound for $A$.  The
fact that $1$ is an upper bound for $A$ is ``built into'' the definition of
$A$, so to speak -- all $x \in A$ have $x < 1$.  Hence we need to show that
given any $\beta < 1$, there exists some $x = \sqrt{2}r$ (an element of $A$)
satisfying $\beta < x < 1$.  Consider the interval
$({\beta\over \sqrt{2}},{1\over \sqrt{2}})$. By Theorem 1.5.6 (or Problem 15),
there exists a rational number $r$ in this interval, so 
$${\beta\over \sqrt{2}} < r < {1\over \sqrt{2}}.$$
Multiplying through by $\sqrt{2} > 0$, the inequalities are preserved, so 
$$\beta < r \sqrt{2} < 1.$$
Hence $\beta$ is not an  upper bound for $A$, since the number
$r \sqrt{2} \in A$ but it is larger than $\beta$.  Therefore, 
${\rm lub}(A) = 1$.
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\noindent
\S 2.1/2a.  By inspection, $\lim_{n\to\infty}{16\over n^3} = 0$.  To 
prove this using the $\epsilon-n_0$ definition, we proceed as follows.
Let $\epsilon > 0$ be any positive number.  If 
$n_0 > {\root 3\of {16\over\epsilon}}$, and $n \ge n_0$, then 
$n > {\root 3\of {16\over\epsilon}}$, so $n^3 > {16\over\epsilon}$, so
${16\over n^3} < \epsilon$.  Since $\left|{16\over\epsilon} - 0\right| < \epsilon$
for all $n \ge n_0$, this shows that the definition of convergence is satisfied 
with limiting value $a = 0$.
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\noindent
\S 2.1/16 c.  We must show that if $\lim_{n\to\infty} y_n = a$, 
$\lim_{n\to\infty} z_n = a$, and $x_{2k} = y_k$, $x_{2k-1} = z_k$, then
$lim_{n\to \infty} x_n = a$ also.  To prove this, let $\epsilon > 0$
be any positive number.  Since $\lim_{n\to\infty} y_n = a$, there is
some $k_0 \in {\bf N}$ such that for all $k \ge k_0$, $|y_k - a| < \epsilon$.
If $n = 2k$ is an even integer, this shows that 
$|x_n - a| = |x_{2k} - a| = |y_k - a| < \epsilon$ for all $n = 2k \ge 2k_0$.
Similarly, since $\lim_{n\to\infty} z_n = a$, there is
some $\ell_0 \in {\bf N}$ such that for all $\ell \ge \ell_0$, $|z_\ell - a| < \epsilon$.
If $n = 2\ell-1$ is an even integer, this shows that 
$|x_n - a| = |x_{2\ell-1} - a| = |z_\ell - a| < \epsilon$ for all 
$n = 2\ell - 1 \ge 2\ell_0 - 1$.  If $n \ge {\rm max}(2k_0,2\ell_0-1)$, 
this shows that $|x_n - a| < \epsilon$ in all cases (i.e. whether $n$
is even or odd).  Therefore $\lim_{n\to\infty} x_n = a$, since the 
definition is satisfied.
\bye