Real Analysis: Sequences
Bounded Sequences
Bounded Sequences
A sequence $\{a_n\}$ in a metric space $X$ is bounded if there exists a closed neighborhood $\overline{N}_r(x)$ of some radius $r$ centered at some point $x \in X$ such that $a_n \in \overline{N}_r(x)$ for all $n \in \mathbb{N}$. In other words, a sequence is bounded if the distance between any two of its elements is finite.
Boundedness can be seen as the second best alternative to convergence. While a bounded sequence may or may not converge to a single value, at least it stays within a given range of values. However, we need not settle for second best forever. As we'll see in the next sections on monotonic sequences and subsequences, sometimes showing that a sequence is bounded is a key step along the way towards demonstrating some of its convergence properties.
The fact that boundedness is a requirement for convergence gives us another tool for detecting divergence  if a sequence is not bounded, then it surely does not converge.
Bounded Sequences in $\mathbb{R}$
We can refine our definition of boundedness to incorporate the ordering properties of the real numbers (or indeed any ordered set). A real sequence $\{a_n\}$ is bounded above if there is some $b$ such that $a_n < b$ for all $n \in \mathbb{N}$. In this case, $b$ is called an upper bound of $\{a_n\}$. Likewise, $\{a_n\}$ is bounded below if there is a $b$ such that $a_n > b$ for all $n \in \mathbb{N}$, in which case $b$ is instead called a lower bound of $\{a_n\}$. A real sequence is bounded if it is both bounded above and bounded below.
Order Limit Theorem
The order limit theorem states two important and intuitive properties of the convergence of real sequences related to bounds. Let $\{a_n\}$ and $\{b_n\}$ be sequences in $\mathbb{R}$ that converge to $p$ and $q$, respectively, and let $c \in \mathbb{R}$. The order limit theorem states the following:

If $a_n \leq c$ for all $n \in \mathbb{N}$, then $p \leq c$.

If $a_n \leq b_n$ for all $n \in \mathbb{N}$, then $p \leq q$.
Bolzano Weierstrass Theorem
The Bolzano Weierstrass Theorem states that every bounded sequence in $\mathbb{R}$ has at least one convergent subsequence. This is in fact the simplest version of the theorem. The result generalizes to sequences $\mathbb{R}^n$ and finally to the equivalence of sequentially compact subsets of $\mathbb{R}^n$ with closed and bounded subsets of $\mathbb{R}^n$.
Problems
Show that if $\{a_n\}$ converges, then $\{a_n\}$ is bounded.
Assume $\{a_n\}$ converges to a value $p$. Then there is some $M \in \mathbb{N}$ such that $d(a_n, p) < 1$ for all $n > M$. The sequence $a_{M+1}, a_{M+2}, \ldots$ is therefore bounded by the closed neighborhood $\overline{N}_{1}(p)$. Likewise, the subset $\{a_1, \ldots, a_{M}\}$ is finite and therefore bounded by a closed neighborhood $\overline{N}_{r}(a_1)$ where $r = \text{Diam}(a_1, \ldots, a_M)$. Since the union of two bounded sets is bounded, it follows that $\{a_1, \ldots, a_M\} \cup \{a_{M+1}, a_{M+2}, \ldots\}$ = $\{a_n\}$ is bounded.
Give an example of a sequence that is bounded but not convergent.
The sequence $\{(1)^n\} = 1, 1, 1, 1, \ldots$ is bounded above by $1$ and bounded below by $1$, but does not converge.
Show that the definition of boundedness in terms of upper and lower bounds for real sequences is equivalent to the definition in terms of neighborhoods. Namely, show that a real sequence $\{a_n\}$ is both bounded above and bounded below if and only if there there is some closed neighborhood $\overline{N}_{r}(p)$ such that $a_n \in \overline{N}_r(p)$ for all $n \in \mathbb{N}.$
Assume $\{a_n\}$ is bounded. Then it is bounded above by $u$ and bounded below by $v$. It follows that $d(v, a_n) = a_n  v \leq u  v$ for all $n \in \mathbb{N}$, and therefore $a_n \in \overline{N}_{uv}(v)$.
Conversely, assume there is a neighborhood $\overline{N}_r(p)$ such that $a_n \in \overline{N}_r(p)$ for all $n \in \mathbb{N}$. Then $p  r \leq a_n \leq p + r$, and thus $\{a_n\}$ is bounded below by $p  r$ and above by $p + r$.
Show that boundedness is dependent on the underlying metric by showing that $\{n^2\}$ is bounded under the discrete metric but not under the Euclidean metric.
The closed neighborhood $\overline{N}_2(0)$ contains all real numbers of distance $2$ or less from $0$. Under the discrete metric, every nonzero real number is distance $1$ from $0$, and $0$ is of course distance $0$ from itself, so it follows that $\overline{N}_2(0)$ contains $\mathbb{R}$ (and is in fact equal to it). Therefore $\{n^2\}$ is bounded by $\overline{N}_2(0)$.
Alternatively, we proceed by contradiction and assume that under the Euclidean metric, $\{n^2\}$ is bounded by some closed neighborhood $\overline{N}_r(x)$ given some radius $r > 0$ at some point $x \in \mathbb{R}$. By the Archimedean principle, there is a natural number $n$ greater than $r+1$. Since $x^2 > x$ for all $x > 1$, it follows that $n^2 > r$. If $0 \in \overline{N}_r(x)$, then $n \notin \overline{N}_r(x)$, and likewise if $n \in \overline{N}_r(x)$, then $0 \notin \overline{N}_r(x)$. Therefore $\{a_n\}$ is not bounded by $\overline{N}_r(x)$ after all, and thus $\{a_n\}$ is unbounded.
Order Limit Theorem 1: Show that if $a_n \leq c$ for all $n \in \mathbb{N}$, then $\lim\limits_{n \rightarrow \infty} a_n \leq c$.
Assume $\lim\limits_{n \rightarrow \infty} a_n = p$.
Proof by contradiction. Assume $p > c$. Then there exists an $N \in \mathbb{N}$ such that $a_n  p < \frac{p  c}{2}$ for all $n > N$.
First, assume $a_n  p > 0$. Then $a_n > p$.
$ a_n  p < \dfrac{p  c}{2} \\ a_n  p < \dfrac{p  c}{2} \\ a_n < \dfrac{3p  c}{2} \\ a_n < \dfrac{3a_n  c}{2} \\ a_n < \dfrac{3a_n}{2}  \dfrac{c}{2} \\ \dfrac{a_n}{2} <  \dfrac{c}{2} \\ a_n > c $
But $a_n \leq c$, which is a contradiction.
Alternatively, assume $a_n  p < 0$. Then $a_n < p$.
$ a_n  p < \dfrac{p  c}{2} \\ p  a_n < \dfrac{p  c}{2} \\ p < \dfrac{p + 2a_n  c}{2} \\ \dfrac{p}{2} < \dfrac{2a_n  c}{2} \\ \dfrac{a_n}{2} < \dfrac{2a_n  c}{2} \\ \dfrac{a_n}{2} < \dfrac{c}{2} \\ c < a_n $
But $a_n \leq c$, which is a contradiction. Thus our assumption that $p > c$ is wrong, so it must be that $p \leq c$ after all.
Show that if $a_n \geq c$ for all $n \in \mathbb{N}$, then $\lim\limits_{n \rightarrow \infty} \geq c$.
If $a_n \geq c$, then $a_n \leq c.$ By the order limit theorem, $\lim\limits_{n \rightarrow \infty} a_n \leq c.$ By the algebraic limit theorem, $\lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} a_n.$ Therefore $\lim\limits_{n \rightarrow \infty}a_n \leq c$, and so $\lim\limits_{n \rightarrow \infty} a_n \geq c.$
Order Limit Theorem 2: Show that if $a_n \leq b_n$ for all $n \in \mathbb{N}$, then $\lim\limits_{n \rightarrow \infty} a_n \leq \lim\limits_{n \rightarrow \infty} b_n$.
By the first part of the order limit theorem, since $b_n  a_n \geq 0$, it follows that $\lim\limits_{n \rightarrow \infty} b_n  a_n \geq 0$. By the algebraic limit theorem, $\lim\limits_{n \rightarrow \infty} b_n  a_n = \lim\limits_{n \rightarrow \infty} b_n  \lim\limits_{n \rightarrow \infty} a_n$. It follows that $\lim\limits_{n \rightarrow \infty} b_n  \lim\limits_{n \rightarrow \infty} a_n \geq 0$. Therefore $\lim\limits_{n \rightarrow \infty} a_n \leq \lim\limits_{n \rightarrow \infty} b_n$.
Show that if $a_n \geq b_n$ for all $n \in \mathbb{N}$, then $\lim\limits_{n \rightarrow \infty} a_n \geq \lim\limits_{n \rightarrow \infty} b_n$.
If $a_n \geq b_n$, then $a_n \leq b_n$. By the order limit theorem, $\lim\limits_{n \rightarrow \infty} a_n \leq \lim\limits_{n \rightarrow \infty} b_n$. By the algebraic limit theorem, $\lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} a_n$. Therefore $\lim\limits_{n \rightarrow \infty} a_n \leq \lim\limits_{n \rightarrow \infty} b_n$, so $\lim\limits_{n \rightarrow \infty} a_n \geq \lim\limits_{n \rightarrow \infty} b_n$.
Alternate proof: Given a sequence $\{a_n\}$ that converges to $p$ and a sequence $\{b_n\}$ that converges to $q$, use the fact that convergent sequences are bounded and the identity $a_nb_n  pq = a_nb_n  a_nq + a_nq  pq$ to prove the Algebraic Limit Theorem for products, namely that $\lim\limits_{n \rightarrow \infty} a_nb_n = \left(\lim\limits_{n \rightarrow \infty} a_n\right)\left(\lim\limits_{n \rightarrow \infty}b_n\right).$
Start with the given identity and use absolute value inequalities to tease out the terms $a_n  p$ and $b_n  q$:
$ a_nb_n  pq = a_nb_n  a_nq + a_nq  pq \\ a_nb_n  pq \leq a_nb_n  a_nq + a_nq  pq \\ a_nb_n  pq \leq a_nb_n  q + qa_n  p $
Now, since $\{a_n\}$ and $\{b_n\}$ both converge, we would like to replace the $a_n  p$ and $b_n  q$ terms with terms involving $\varepsilon$ such that the coefficients cancel and the righthand side simplifies to $\varepsilon$. We might say that for any $\frac{\varepsilon}{2a_n} > 0$ there exists an $N \in \mathbb{N}$ such that $b_n  q < \frac{\varepsilon}{2a_n}$ for all $n > N$, and likewise that for any $\frac{\varepsilon}{2q} > 0$ there exists an $M \in \mathbb{N}$ such that $a_m  p < \frac{\varepsilon}{2q}$ for all $m > M.$ We would substitute these terms in and everything would cancel out to $\varepsilon$ as desired.
However, there is a probelm with this maneuver. Unlike the $q$ term, the $a_n$ term is not a constant  it's a sequence term, thus we cannot use it directly in generating generating our $\varepsilon$ term. However, because $\{a_n\}$ is a convergent sequence, we know that it is bounded. Let $B$ be an upper bound for $\{a_n\}$. Now we substitue $B$ into the inequality:
$ a_nb_n  pq \leq Bb_n  q + qa_n  p $
Now we're in business. For any $\frac{\varepsilon}{2B}$ there exists an $N \in \mathbb{N}$ such that $b_n  q < \frac{\varepsilon}{2B}$ for all $n > N$. We can keep our inequality for $b_m  q$ involving $M$ as before. Pick $K = \max\{N, M\}.$ Then for all $k > K$ we have:
$ a_kb_k  pq < B\left(\dfrac{\varepsilon}{2B}\right) + q\left(\dfrac{\varepsilon}{2q}\right) \\ a_kb_k  pq < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2} \\ a_kb_k  pq < \varepsilon $
Therefore $\lim\limits_{n \rightarrow \infty} a_kb_k = \left(\lim\limits_{n \rightarrow \infty}a_n\right)\left(\lim\limits_{n \rightarrow \infty} a_kb_k \right).$
Squeeze Theorem: Let $\{a_n\}$, $\{b_n\}$, and $\{c_n\}$ be sequences in $\mathbb{R}$ with the property that $a_n \leq b_n \leq c_n$ for all $n \in \mathbb{N}$. Show that if $\lim\limits_{n \rightarrow \infty} a_n = \lim\limits_{n \rightarrow \infty} c_n = p$, then $\lim\limits_{n \rightarrow \infty} b_n = p$.
Since $a_n \leq b_n$, by the order limit theorem, $\lim\limits_{n \rightarrow \infty} a_n = p \leq \lim\limits_{n \rightarrow \infty} b_n$. Likewise, since $b_n \leq c_n$, the order limit theorem again tells us that $\lim\limits_{n \rightarrow \infty} b_n \leq p = \lim\limits_{n \rightarrow \infty} c_n$. Since $p \leq \lim\limits_{n \rightarrow \infty} b_n \leq p$, it follows that $\lim\limits_{n \rightarrow \infty} b_n = p$.
Bolzano Weierstrass Theorem: Show that every bounded sequence has a convergent subsequence.
Hint: Use the nested interval theorem.
We first construct a subsequence $\{a_{n_k}\}$, then show that it converges.
Let $\{a_n\}$ be a bounded sequence. Then $\{a_n\}$ is bounded by the interval $I_0 = [p,q]$ where $p = \inf\{a_n\}$ and $q = \sup\{a_n\}$. Select the point $s_0 = \frac{\inf(I_0) + \sup(I_0)}{2}$ in the middle of $I_0$, and consider the intervals $I_{1,l} = [p, s_0]$ and $I_{1,u} = [s_0, q]$. It follows that $\{a_n\}$ contains an infinite number of points in at least one of these intervals (since if they were both finite, $\{a_n\}$ would be finite). Select one such interval and call it $I_1$, and select any element in the sequence $a_{n_1}$ such that $a_{n_1} \in I_1$.
In general, select the point $s_{n} = \frac{\inf(I_n) + \sup(I_n)}{2}$ in the middle of $I_n$ and consider the intervals $I_{n+1,l} = [\inf(I_n), s_{n}]$ and $I_{n+1,u} = [s_{n}, \sup(I_N)]$. Again it must be that an infinite number of terms in $\{a_n\}$ lie in at least one of these intervals, as otherwise $I_n$ would contain a finite number of terms, which it does not. Select one such interval and call it $I_{n+1}$, and select any element of the sequence $a_{n_k}$ such that $a_{n_k} \in I_{n+1}$ and $n_k > n_{k1}$. It follows that $I_1 \supset I_2 \supset I_3 \ldots,$ and $\{a_{n_k}\}$ is a subsequence of $\{a_n\}$.
To show that $\{a_{n_k}\}$ converges, we show that $\{I_n\}$ form a nested sequence of intervals. It follows by the construction of $\{I_n\}$ that $I_{n+1} \subset I_n$ for all $n \in \mathbb{N}$. Likewise, $\mu(I_{n+1}) = \frac{1}{2}\mu(I_{n})$, and thus $\mu(I_n) = \left(\frac{1}{2}\right)^n\mu(I_0)$. For any $\varepsilon > 0$, pick $N$ such that $\left(\frac{1}{2}\right)^NI_0 < \varepsilon$. It follows that $\mu(I_N) < \varepsilon$. Therefore $\{I_n\}$ forms a nested sequence of intervals.
By the nested interval theorem, it follows that there is a single value $x \in I_n$ for all $n \in \mathbb{N}$. Let $h = p  q$. It follows that $a_{n_k}  x < \mu(I_k)$ for all $k \geq n$. Thus for any $\varepsilon > 0$, select an $N \in \mathbb{N}$ such that $\left(\frac{1}{2}\right)^Nh < \varepsilon$. It follows that $a_{n_k}  x < \varepsilon$. Therefore the subsequence $\{a_{n_k}\}$ converges to $x$.