Real Analysis: Sequences

Problems

1. 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.

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2. 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.

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3. 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 $a_n \in \overline{N}_{u-v}(v)$.

   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 $p + r$.

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4. 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.

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5. 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.

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6. 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 $\lim\limits_{n \rightarrow \infty} a_n \geq c.$

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7. 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 $\lim\limits_{n \rightarrow \infty} a_n \leq \lim\limits_{n \rightarrow \infty} b_n$.

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8. 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 $\lim\limits_{n \rightarrow \infty} a_n \geq \lim\limits_{n \rightarrow \infty} b_n$.

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9. 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_n||b_n - q| + |q||a_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 $\frac{\varepsilon}{2|a_n|} > 0$ there exists an $N \in \mathbb{N}$ such that $|b_n - q| < \frac{\varepsilon}{2|a_n|}$ for all $n > N$, and likewise that for any $\frac{\varepsilon}{2|q|} > 0$ there exists an $M \in \mathbb{N}$ such that $|a_m - p| < \frac{\varepsilon}{2|q|}$ 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 B|b_n - q| + |q||a_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}{2|q|}\right) \\ |a_kb_k - pq| < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2} \\ |a_kb_k - pq| < \varepsilon $

   Therefore $\lim\limits_{n \rightarrow \infty} a_kb_k = \left(\lim\limits_{n \rightarrow \infty}a_n\right)\left(\lim\limits_{n \rightarrow \infty} a_kb_k \right).$

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10. 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 $\lim\limits_{n \rightarrow \infty} b_n = p$.

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11. 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 $n_k > n_{k-1}$. 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 $x$.

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