Real Analysis: Sequences

Problems

1. Let $\{a_n\} = \left\{ \begin{array}{ll} 1 & x \text{ is even}\\\frac{3}{x} & x \text{ is odd}\end{array} \right.$

   1. Graph the first several terms of $\{a_n\}.$

   2. Determine $\limsup\limits_{n \rightarrow \infty}\{a_n\}$ and $\liminf\limits_{n \rightarrow \infty}\{a_n\}.$ Are they equal?

   1. 

   2. Note that for all $n > 3,$ $a_n \leq 1$ by definition, and so $\limsup\limits_{n \rightarrow \infty}\{a_n\} = 1.$
      To determine the limit inferior, we partition the original sequence into its even and odd-numbered subsequences. The odd-numbered subsequence is always $1$ by definition, and so its infimum is $1.$ The even-numbered subsequence is $\frac{3}{n}.$ We can see that $\inf\left\{\frac{3}{n}\right\} = 0.$ From here we note that $\inf\{0, 1\} = 0,$ and so $\liminf\{a_n\} = 0.$

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2. Consider two sequences $\{a_n\}$ and $\{b_n\}$ where $a_n \leq b_n.$ Show the following:

   1. $\liminf\limits_{n \rightarrow \infty}\{a_n\} \leq \liminf\limits_{n \rightarrow \infty} \{b_n\}.$

   2. $\limsup\limits_{n \rightarrow \infty}\{a_n\} \leq \limsup\limits_{n \rightarrow \infty} \{b_n\}.$

   Consider $N \in \mathbb{N}.$ For all $n > N,$ it follows by construction that $a_k \leq b_k.$ Therefore $\inf\limits_{k>N}{a_k} \leq \inf\limits_{k>N}{b_k}$ and $\sup\limits_{k>N}{a_k} \leq \sup\limits_{k>N}{b_k}.$ By the algebraic limit theorem, it follows that $\liminf\limits_{n \rightarrow \infty}\{a_n\} \leq \liminf\limits_{n \rightarrow \infty}\{b_n\}$ and $\limsup\limits_{n \rightarrow \infty}\{a_n\} \leq \limsup\limits_{n \rightarrow \infty}\{b_n\}.$

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3. Show that if $\{a_n\}$ is monotonically increasing, then $\limsup\limits_{n \rightarrow \infty} a_n = \sup a_n.$

   Assume $\{a_n\}$ is monotonically increasing. Then $a_n \leq a_{n+1}$ for all $n \in \mathbb{N}.$ It follows that $\sup\limits_{n > m} a_n = \sup\limits_{n > m+1} a_n ,$ as $a_m < \sup\limits_{n > m} a_n.$ Therefore $\limsup\limits_{n \rightarrow \infty} a_n = \sup a_n.$

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4. Show that if $\limsup\limits_{n \rightarrow \infty} a_n = L,$ then for every $\varepsilon > 0$ there exists an $N \in\mathbb{N}$ such that $a_n < L + \varepsilon$ whenever $n > N.$

   Select $\varepsilon > 0.$ Then there is some $N \in \mathbb{N}$ such that $\sup\limits_{m\geq n}\{a_m\} - L < \varepsilon$ whenever $n \geq N.$ By definition of infimum, we see that $a_n \leq \sup\limits_{m\geq n} \{a_m\}$ for all $n \geq N.$ It follows that $a_n - L < \varepsilon,$ and thus that $a_n < L + \varepsilon.$

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5. Double Bounded Limit Theorem: Show that $\{a_n\}$ converges to $p$ if and only if the limit supremum and limit infimum of $\{a_n\}$ both exist and $\limsup\limits_{n \rightarrow \infty} a_n = \liminf\limits_{n \rightarrow \infty} a_n.$

   Assume $\{a_n\}$ converges to $p$. Then for every $\varepsilon > 0$, there exists an $N \in \mathbb{N}$ such that $d(a_n, p) < \varepsilon$ for every $n > N.$ Therefore $p - \varepsilon < a_n < p + \varepsilon.$ Therefore $\sup \{a_{n>N}\} \leq p + \varepsilon$ and $\inf\{a_{n > N}\} \geq p - \varepsilon$. It follows that $|\sup \{a_{n>N}\} - p| \leq \varepsilon$ and $|\inf\{a_{n > N}\} - p| \leq \varepsilon$. Therefore $\limsup\limits_{n \rightarrow \infty} \{a_n\} = p = \liminf\limits_{n \rightarrow \infty} \{a_n\}.$

   Conversely, assume $\limsup\limits_{n \rightarrow \infty} \{a_n\} = p = \liminf\limits_{n \rightarrow \infty} \{a_n\}.$ Then for any $\varepsilon > 0$, there exist some $N, M \in \mathbb{N}$ such that $|\sup\{a_n\} - p| < \varepsilon$ and $|\inf\{a_m\} - p| < \varepsilon$ for all $n > N$ and $m > M$. Pick $K = \text{max}(N, M)$ and such that $|\sup\{a_k\} - p| < \varepsilon$ and $|\inf\{a_k\} - p| < \varepsilon$ whenever $k > K.$ It follows that $p - \varepsilon < a_k < \sup\{a_k\} < p + \varepsilon$ and $p - \varepsilon < \inf\{a_k\} < a_k < p + \varepsilon.$ This simplifies to $p - \varepsilon < a_k < p + \varepsilon,$ and in turn $|a_k - p| < \varepsilon,$ which by definition means that $\lim\limits_{k \rightarrow \infty} \{a_k\} = p.$

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6. Let $\{a_n\}$ be a real sequence, and let $A$ be the set of all subsequential limits of $\{a_n\},$ including possibly $\pm\infty.$ Show that $A$ is nonempty.

   Proof by contradiction. Assume $A$ is empty. Then $\infty \notin A$ and $-\infty \notin A.$ It follows that $\{a_n\}$ has no subsequence that tends to either positive or negative infinity. Therefore $\{a_n\}$ is bounded. It follows from the Bolzano-Weierstrass theorem that $\{a_n\}$ has at least one convergent subsequence whose limit is $L.$ But this is a contradiction, because it implies $L \in A,$ contradicting the emptiness of $A$. Therefore $A$ was not empty after all.

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7. Show that the set of subsequential limits for a sequence is closed.

   Let $S = \{ s : s = \lim\limits_{k \rightarrow \infty} a_{n_k} \text{ for some subsequence } \{a_{n_k}\} \}.$ We show that if $p$ is a limit point of $S,$ then $p \in S.$

   First, note that if $S$ has no limit points, then $S$ vacuously contains all of its limit points and is therefore closed.

   Alternatively, let $p$ be a limit point of $S.$ We inductively create the sequence $\{a_{n_k}\}$ such that $\lim\limits_{k \rightarrow \infty} a_{n_k} = p.$

   First, choose some $n_0 \in \mathbb{N}$ such that $a_{n_0} \neq p.$ Let $\delta = d(p, a_{n_0}).$

   Next, assume $n_0, \ldots, n_k$ exist such that $d(p, a_{n_k}) \leq \left(\dfrac{1}{2}\right)^k \delta.$ Because $p$ is a limit point of $S,$ there exists a point $s_{k+1} \in S$ such that $d(p, s_{k+1}) < \left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)^{k+1}\delta.$ Because $s_{k+1}$ is a subsequential limit of $\{a_n\},$ it follows that there is some point $a_{n_{k+1}}$ such that $n_{k+1} > n_k$ and $d(a_{n_{k+1}}, s_{k+1}) < \left(\dfrac{1}{2}\right)\left(\dfrac{1}{2}\right)^{k+1}\delta.$ By the triangle inequality, $d(p, a_{n_{k+1}}) < \left(\dfrac{1}{2}\right)^{k+1}\delta.$ It follows by induction that $\lim\limits_{k \rightarrow \infty} a_{n_k} = p.$ Therefore $p \in S,$ and so $S$ is closed.

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8. Let $A$ be the set of subsequential limits of $\{a_n\}$ in the extended real numbers. Show that $\sup(A) \in A.$

   There are three cases to consider:

   • If $\sup(A) = \infty,$ then $A$ is not bounded above. Therefore $\{a_n\}$ is not bounded above, so it follows that there is some subsequence $\{a_{n_k}\}$ such that $\lim\limits_{n \rightarrow \infty} a_{n_k} = \infty.$

   • If instead $\sup(A) = c$ for some $c \in \mathbb{R},$ then $A$ is nonempty and it contains at least one subsequential limit of $\{a_n\},$ namely the one whose limit equals $c.$ Since the set of all subsequential limits of a sequence form a closed set, and since closed sets in turn contain their boundary points, which in the case of real subsets includes their infimum and supremum, it follows that $c \in A.$

   • Finally, if $\sup(A) = -\infty,$ then there is no other subsequential limit, as $A = \{-\infty\}.$

   In all cases, $\sup(A) \in A.$

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9. Equivalent definition: Show that if $S$ is the set of all subsequential limits of $\{a_n\},$ including possibly $\infty$ and $-\infty,$ then $\limsup\limits_{n \rightarrow \infty} a_n = \sup(S).$

   Let $L = \limsup\limits_{n \rightarrow \infty} a_n.$ There are three cases to consider: $L = \infty,$ $L  \in \mathbb{R},$ and $L = -\infty.$

   • Assume $L = \infty.$ Then $\{a_n\}$ is unbounded above. It follows that for every $2^k \in \mathbb{R},$ there is some $a_{n_k} > 2^k.$ Selecting $n_{k+1} > n_k$ produces a subsequence $\{a_{n_k}\}$ that diverges to $\infty,$ and so $L \in S.$ Because $L = \infty$ it follows that $L = \sup(S).$

   • Assume $L \in \mathbb{R}.$ Because $L = \limsup\limits_{n \rightarrow \infty} a_n,$ for every $2^{-k} > 0$ there exists an $n_k \in \mathbb{N}$ such that $|a_{n_k} - L| < 2^{-k}.$ Choosing $n_{k+1}$ such that $n_{k+1} > n_k$ forms a subsequence $\{a_{n_k}\}$ such that $\lim\limits_{k \rightarrow \infty} a_{n_k} = L.$ Since $L$ is a subsequential limit of $\{a_n\},$ it follows that $L \in S.$
     To show that $L = \sup(S),$ proceed by contradiction. Assume $L \neq \sup(S).$ Then there is some subsequence $\{a_{n_j}\}$ such that $\lim\limits_{j \rightarrow \infty} a_{n_j} = P > L.$ This implies that for any $2^{-p}$ there exists some $M_p \in \mathbb{N}$ such that $|a_{n_j} - P| < 2^{-p}$ whenever $n_j \geq M_p.$ Select $j_0$ such that $2^{-j_0} < (P - L).$ Then $|a_{n_j} - P| < (P - L)$ whenever $j \geq M_{j_0}.$ Therefore $L + 2^{-j_0} < a_{n_j}.$ But this implies that $\lim\limits_{j \rightarrow \infty} a_{n_j} \geq L + 2^{-j_0},$ and thus that $\limsup\limits_{n \rightarrow \infty} a_n \geq L + 2^{j_0},$ which is a contradiction. Therefore $L = \sup(S)$ after all.

   • Assume $L = -\infty.$ Then $\lim\limits_{n \rightarrow \infty} a_n = -\infty,$ and so $S = \{L\}.$ Therefore $L = \sup(S).$

   In all three cases, we have shown that $L = \sup(S).$

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