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\def\proof{\intro{Proof}}
\def\endproof{\leavevmode\hskip\parfillskip\vrule height 6pt width 6pt
depth 0pt{\parfillskip0pt\medskip}}
\let\bu\bullet
{\bigbf\centerline{Holden Rohrer}\bigskip\centerline{MATH 4317\qquad
HOMEWORK \#4, Part II}}
\question{Give an example of each of the following, or argue that such a
request is impossible.}
\item{a.} A sequence that has a subsequence that is bounded but contains
no subsequence that converges.
\intro{Disproof}
This is impossible.
Let $(a_n)$ be a sequence bounded by $M.$
We will show by contradiction that for any $\epsilon > 0,$ there must be
a susbequence $(a_{\phi(n)})$ such that $|a_n-l|<\epsilon$ for some $l.$
First, we break the interval $[-M,M] \supseteq \{a_n\}$ into the
partition $$[-M,-M+\epsilon)\cup
[-M+\epsilon,-M+2\epsilon)\cup\cdots\cup [M-\epsilon,M] =
S_1\cup S_2\cup\cdots\cup S_n.$$
For contradiction, we assume that there is no subsequence meeting the
Cauchy criterion for $\epsilon.$
This means $(a_n)\cap S_i$ is finite for all $S_i$ because otherwise
these would be subsequences satisfying the Cauchy Criterion for
$\epsilon.$
$$S_1\cup\ldots\cup S_n = [-M,M],$$
so $[-M,M]\cap (a_n) = (a_n)$ has a length less than or equal to the sum
of lengths across $S_i\cap (a_n),$ which makes it finite, but the
sequence is infinite, giving a contradiction.
\endproof
\vfil\eject
\item{b.} A sequence that does not contain $0$ or $1$ as a term but
contains subsequences converging to each of these values.
\proof
Let $$a_n = 1/2 + (-1)^n{n\over 2(n+1)}.$$
\endproof
\vfil\eject
\item{c.} A sequence that contains subsequences converging to every
point in the infinite set $\{1, 1/2, 1/3, 1/4, 1/5,\ldots\},$ and no
subsequences converging to points outside of this set.
\intro{Disproof}
Such a sequence would necessarily contain a subsequence converging to
$0.$
Let $\epsilon > 0.$
We can easily choose some element in the set $0<s = 1/n < \epsilon/2.$
Then, because we have a subsequence converging to this point, for some
$N,$ we get $$|a_N - s| < \epsilon/2 \Longrightarrow s-\epsilon/2 < a_N
< s+\epsilon/2.$$
We already know $0 < s < \epsilon/2,$ so $|s-\epsilon/2| < \epsilon/2 <
|s+\epsilon/2| < \epsilon,$ and $|a_N| <
\max\{|s-\epsilon/2|,|s+\epsilon/2|\} < \epsilon,$ showing that we must
have a subsequence converging to $0.$
\endproof
\vfil\eject
\item{d.} A sequence that contains subsequences converging to every
point in the infinite set below, and no subsequences converging to
points outside of this set.
$$\{2^{-m} + 2^{-n} : 1\leq m < n\} = \bigcup_{m=1}^\infty \{2^{-m} +
2^{-m-1}, 2^{-m}+2^{-m-2},\ldots\}.$$
\intro{Disproof}
For a very similar reason as (c), we cannot have such a sequence
containing such subsequences without containing their limit point $0$
(and possibly others).
We will dispense with the long $\epsilon$ proof and simply note that we
must have some $m$ such that $2^{-m+1}<\epsilon/4,$ and
$2^{-m}+2^{-{m+1}} = 3(2^{-m+1})<3\epsilon/4,$ and choosing a point
$a_N$ from
our sequence within $\epsilon/4$ because of the subsequence already
established, we again get an $a_N<\epsilon,$ showing it converges to
$0,$ which is outside the set.
\bye
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