Problem 1. a) Show that any closed interval $I=[a, b]$ is the countable infinite intersection of open intervals.
b) Show that any open interval $I=(a, b)$ is the countable infinite union of closed intervals.
Proof . (a) $[a, b]=\bigcap_{n=1}^{\infty}(a-1 / n, b+1 / n)$.
(b) Let $N$ be large enough so that $b-a>\frac{2}{N}$ (so that we do not get empty set in the union, but it is also fine if you include many emtpy sets). Then $(a, b)=\bigcup_{n=N}^{\infty}[a+1 / n, b-1 / n]$.
Problem 2. Define an open subset of $\mathbb{R}, U \subset \mathbb{R},$ to be set that has the property that for every $x \in U$, there is an open interval $I$ so that $x \in I \subset U$. The empty set is considered open as it vacuously satisfies this condition.
a) Show that if $\left\{U_{\lambda}\right\}_{\lambda \in A}$ is any collection of open subsets, then $U=\bigcup_{\lambda \in A} U_{\lambda}$ is open.
b) Show that if $U_{1}, \ldots, U_{n}$ is a finite collection of open subsets, then $U=U_{1} \cap \ldots \cap U_{n}$ is open.
c) Show by example, that there is a countable infinite collection of open subsets $\left\{U_{n}\right\}_{n \in \mathbb{N}}$ so that $V=$ $\bigcap_{i=1}^{\infty} U_{i}$ is not open.
Proof . (a) Let $x \in U,$ then by definition there is some $\lambda$ such that $x \in U_{\lambda} .$ Since $U_{\lambda}$ is open we have an open interval $I_{x}$ such that $x \in I_{x} \subset U_{\lambda} \subset U,$ so $U$ is open.
(b) Let $x \in U,$ then $x \in U_{i}$ for each $i=1, \ldots, n .$ By definition there are open intervals $I_{1}, \ldots I_{n}$ such that $x \in I_{i} \subset U_{i} .$ Without loss of generality, and after possibly shrinking $I_{i},$ we can assume $I_{i}=\left(x-a_{i}, x+a_{i}\right)$ for $a_{i} \in \mathbb{R}$. Now choose a real number $a<\min \left\{a_{1}, \ldots, a_{n}\right\},$ it follows that $(x-a, x+a) \subset I_{i} \subset U_{i}$ for each $i=1, \ldots, n,$ and hence $(x-a, x+a) \subset \bigcap_{i=1}^{n} U_{i}=U,$ so $U$ is
open.
(c) The example is the same as Problem 1 (a). We will show that the closed interval (as the name suggests) $[a, b]$ is not open. In fact, any open interval around $b$ will not be fully contained in $[a, b]$. Indeed, any interval around $b$ contains an interval of the form $(b-\varepsilon, b+\varepsilon)$ for some $\varepsilon>0,$ and it is clear that $b+\varepsilon / 2 \notin[a, b]$.
Problem 3. Let $U$ be an open subset as defined above. Show that if $U$ is non-empty, then $U$ contains some rational number.
Proof . Since $\mathbb{Q}$ is dense in $\mathbb{R}$, given any $x \in U$ and an interval $(x-a, x+a) \subset U, \mathbb{Q} \cap(x-a, x+a)$ is not empty, so there is at least some rational number in it.
Problem 4.

Determine if the following series converge and if they do find their limit. Please justify your answer rigorously.
a) $\left\{\frac{n}{n+1}\right\}_{n=1}^{\infty}$.
b) $\left\{\frac{(-2)^{n}}{n^{2}}\right\}_{n=1}^{\infty}$

Proof . (a) We claim that the limit is $1 .$ Indeed, given $\varepsilon>0,$ let $N>0$ be such that $\frac{1}{N}<\varepsilon,$ then for all $n>N$ we have
$$\left|\frac{n}{n+1}-1\right|=\frac{1}{n+1}<\frac{1}{N}<\varepsilon$$
So the limit is 1 by definition.
(b) We claim that the limit does not exist. Suppose for a contradiction that $L \in \mathbb{R}$ is the limit of the sequence. We must exhibit an $\varepsilon$ such that there is no $N \in \mathbb{N}$ such that
$$\left|\frac{(-2)^{n}}{n^{2}}-L\right|<\varepsilon$$
for all $n>N$. We will show this using $\varepsilon=1$. Let $n$ be even and by triangle in equality we have
$$\left|\frac{(-2)^{n}}{n^{2}}-L\right| \geq \frac{2^{n}}{n^{2}}-L$$
Now since $2^{n}$ grows much faster than $n^{2}$, given any $N \in \mathbb{N},$ we can choose an even number $n>N$ large so that
$$\frac{2^{n}}{n^{2}}>L+2$$
and it follows
$$\left|\frac{(-2)^{n}}{n^{2}}-L\right|>L+2-L>2>1$$
a contradiction, so the limit must not be equal to $L$ for any $L \in \mathbb{R} .$ Hence the limit must not exist.
Problem 5.

Let $\left\{x_{n}\right\}_{n=1}^{\infty}$ be a convergent monotone sequence. If $\lim _{n \rightarrow \infty} x_{n}=x_{k}$ for a fixed $k$, then $x_{n}=x_{k}$ for all $n \geq k$

Proof .

Without loss of generality we may suppose $\left\{x_{n}\right\}$ is nondecreasing (otherwise consider $\left\{-x_{n}\right\}$ ). Suppose for a contradiction that there is some $N>k$ such that $x_{N}>x_{k}$ (since the sequence is nondecreasing $\left.x_{N} \geq x_{k}\right) .$ Let $\varepsilon=x_{N}-x_{k},$ it follows from monotonicity that
$$\left|x_{n}-x_{k}\right|=x_{n}-x_{k} \geq x_{N}-x_{k}>\varepsilon$$
for every $n \geq N,$ so $x_{k}$ clearly cannot be the limit of the sequence, a contradiction.

Problem 6.

a) Let $\left\{I_{n}\right\}_{n \in \mathbb{N}}$ be a collection of closed bounded intervals with $I_{n+1} \subset I_{n}$ for all $n \geq 1$. Show that $\bigcap_{n=1}^{\infty} I_{n}$ is non-empty.
b) Let $\left\{J_{n}\right\}_{n \in \mathbb{N}}$ be a collection of open bounded intervals with $J_{n+1} \subset J_{n}$ for all $n \geq 1$. Show by example that it is possible for $\bigcap_{n=1}^{\infty} J_{n}$ to be empty.

Proof . (a) Let $I_{n}=\left[a_{n}, b_{n}\right] .$ The nested condition ensures that $\left\{a_{n}\right\}$ is a monotone nondecreasing sequence. Note also that $\left\{a_{n}\right\}$ is bounded above by $b_{1},$ so by the monotone convergence theorem we have $a_{n} \rightarrow a$ for some $a \in \mathbb{R} .$ We claim that $a \in \bigcap_{n=1}^{\infty} I_{n} .$ Note that it suffices to show that $a \leq b_{n}$ for every $n$. Suppose for a contradiction that $a>b_{n 0}$ for some $n_{0} .$ Let $\varepsilon=a-b_{n 0} .$ Since $a_{n} \rightarrow a,$ there is $N \in \mathbb{N}$ such that $n>N$ implies $a-a_{n}<\varepsilon=a-b_{n 0} .$ On the other hand, since the intervals are nested we have $a-b_{n 0}<a-b_{n}$ for every $n>n_{0} .$ Thus choosing $n>\max \left\{N, n_{0}\right\}$ we have
$$a-a_{n}<\varepsilon=a-b_{n_{0}}<a-b_{n} \Longrightarrow b_{n}<a_{n}$$
a contradiction, since the intervals are assumed to be nonempty. This proves $a \leq b_{n}$ for every $n$ and so $a \in I_{n}$ for every $n,$ which in turn implies $a \in \bigcap_{n=1}^{\infty} I_{n}$.
(b) Consider $J_{n}=(0,1 / n),$ then $\bigcap_{n=1}^{\infty} J_{n}=\emptyset$.