Sunday, September 13

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• Suppose \(\left\{{a_n}\right\}\) is not bounded above.
• Then any \(k\in {\mathbb{N}}\) is not an upper bound for \(\left\{{a_n}\right\}\).
• So choose a subsequence \(a_{n_k} > k\), then by order-limit laws, \begin{align*} a_{n_k} > k \implies \liminf_{k\to\infty} a_{n_k} > \liminf_{k\to\infty} k = \infty .\end{align*}

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• Suppose \(\left\{{a_n}\right\}\) is bounded by \(M\), so \(a_n < M < \infty\) for all \(n\in {\mathbb{N}}\).

• Then if \(\left\{{a_{n_k}}\right\}\) is a subsequence, we have \(a_{n_k} \in \left\{{a_n}\right\}\), so \(a_{n_k} < M\) for all \(k\in {\mathbb{N}}\).

• But then \begin{align*} a_{n_k} < M \implies \limsup_{k\to\infty} a_{n_k} \leq M ,\end{align*}

• Now note that if \(\lim_{k\to\infty} a_{n_k}\) exists, \begin{align*} \lim_{k\to\infty} a_{n_k} < \limsup_{k\to\infty} a_{n_k} \leq M < \infty ,\end{align*} so every subsequence is bounded and thus can not converge to \(\infty\).

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• Suppose \(x_n \leq M\) for all \(n\), we will show that every subsequential limit is also bounded by \(M\).

• Let \begin{align*}S \coloneqq\left\{{x\in {\mathbf{R}}{~\mathrel{\Big\vert}~}x \text{ is a subsequential limit of } \left\{{x_n}\right\}}\right\}\end{align*} be the set of subsequential limits.

  • Note that \(\inf S \coloneqq\liminf_{n\to\infty} x_n\) by definition (i).

• Let \(\left\{{x_{n_k}}\right\}\in S\) be an arbitrary convergent subsequence (since we are only concerned about subsequences with well-defined limits).

• Then for every \(k\) we have \(x_{n_k} \in \left\{{x_n}\right\}\), so \begin{align*} {\left\lvert {x_{n_k}} \right\rvert} \leq M .\end{align*}

• By order limit laws, \begin{align*} {\left\lvert {x_{n_k}} \right\rvert} \leq M \implies \lim_{k\to\infty} {\left\lvert {x_{n_k}} \right\rvert} \leq M ,\end{align*}

• Since the map \(x\mapsto {\left\lvert {x} \right\rvert}\) is continuous, using the sequential definition of continuity we can pass the limit through the absolute value to obtain \begin{align*} {\left\lvert { \lim_{k\to\infty} x_{n_k}} \right\rvert} \leq M .\end{align*}

• Since the subsequence was arbitrary, we find that \(M\) is an upper bound for \(S\) and so \(\sup S \leq M\).

• But \begin{align*} \inf S \leq \sup S \leq M \implies \inf S \leq M .\end{align*}

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• Suppose \({\left\lvert {x_n} \right\rvert} \leq M\) for every \(n\), we will directly show that \({\left\lvert {\lim_{n\to\infty}\inf_{k\geq n} x_n} \right\rvert} \leq M\).\

• By order-limit laws, for every fixed \(n\) we have \begin{align*} {\left\lvert {x_{n}} \right\rvert} \leq M \iff -M \leq x_{n} \leq M \implies -M \leq \inf_{k>n} {x_{k}} \leq M ,\end{align*} where we’ve used the fact that \(x_n \geq -M\) for all \(n\) implies that \(\inf_{k\geq n} x_k \geq -M\).

• Again applying order-limit laws, \begin{align*} -M \leq \inf_{k\geq n} {x_{k}} \leq M \implies -M \leq \lim_{n\to\infty} \inf_{k\geq n} {x_{k}} \leq M \iff {\left\lvert {\lim_{n\to\infty} \inf_{k\geq n} x_{n_k} } \right\rvert} \leq M .\end{align*}

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Note that here we define \(S\) to be the set of all subsequential limits of \(\left\{{x_n}\right\}\) and \begin{align*}\liminf_n x_n \coloneqq\inf S.\end{align*}

• Suppose toward a contradiction that \(\beta < \liminf_{n} x_n\) but there does not exist any \(N\) such that \(n\geq N \implies x_n > \beta\).

• Then for all \(N\) there exists an \(n> N\) with \(x_n \leq \beta\), so the set \begin{align*}B \coloneqq\left\{{n \in {\mathbb{N}}{~\mathrel{\Big\vert}~}x_n \leq \beta}\right\}\end{align*} is countably infinite.

• Then by Bolzano-Weierstrass, since \(B\) is bounded it contains a convergent subsequence \(x_{n_k}\) which satisfies \begin{align*} x_{n_k} \leq \beta \quad \forall k \implies L\coloneqq\lim_{k\to\infty} x_{n_k} \leq \beta \end{align*} where we’ve used order-limit laws.

• We now have \(L\in S\), a subsequential limit satisfying \(L\leq \beta\) and since \(\inf S\) is a lower bound for \(S\), \begin{align*} \inf S \leq L \leq \beta .\end{align*} which contradicts \(\beta < \liminf_n x_n\).

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Note that here we define \begin{align*} \liminf_n x_n \coloneqq\lim_{n\to\infty} S_n {\quad \operatorname{where } \quad} S_n \coloneqq\inf \left\{{x_k {~\mathrel{\Big\vert}~}k\geq n}\right\} .\end{align*}

• Write \(L\coloneqq\lim_{n\to\infty} S_n\) and suppose \(\beta < L\).

• Then we have \begin{align*} \forall{\varepsilon}>0,\, \exists N{\quad \operatorname{such that} \quad} n\neq N \implies {\left\lvert {S_n - L} \right\rvert} < {\varepsilon} .\end{align*}

• Since \(\beta < L \iff L-\beta > 0\), we can set \({\varepsilon}\coloneqq L-\beta\) to produce an \(N\) such that \begin{align*} n\geq N \implies {\left\lvert {L-S_n} \right\rvert} < L-\beta \iff \beta - L < S_n - L < L-\beta .\end{align*}

• Just taking the first part of this composite inequality we have \begin{align*} n\geq N \implies \beta - L < S_n - L \iff \beta < S_n \coloneqq\inf_{k\geq n} x_k \leq x_n ,\end{align*} supplying the \(N\) for which \(n\geq N \implies \beta < x_n\) as desired.

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• Suppose toward a contradiction that \(\beta < \liminf_n x_n\) but there is no \(N\) such that \(n\geq N \implies x_n> \beta\).

• Then for all \(N\) there exists an \(n\) with \(x_n \leq \beta\), so if we form the set \begin{align*} B_n \coloneqq\left\{{k\in {\mathbb{N}}{~\mathrel{\Big\vert}~}k\geq n \text{ and } x_k \leq \beta}\right\} ,\end{align*} then \(B_n\) is countably infinite for every \(n\)

• But then \(B_n\subseteq \left\{{k\in {\mathbb{N}}{~\mathrel{\Big\vert}~}k\geq n}\right\}\) for every \(n\) implies that \begin{align*} \inf_{k\geq n} x_k \leq \inf_{k\in B_n} x_k \leq \beta \qquad \forall n ,\end{align*} since an infimum over a larger set can only get smaller.

• Applying order-limit laws, we then have \begin{align*} \inf_{k\geq n} \leq \beta \,\,\forall n \implies \lim_{n\to \infty}\inf_{k\geq n} x_n \leq \beta ,\end{align*} but this contradicts \(\liminf_{n} x_n > \beta\).

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• Suppose \(\left\{{x_n}\right\}\) is bounded and \(\limsup {\left\lvert {x_n} \right\rvert} = 0\).
• Then using the supremum definition, \(\lim_{n\to\infty} \sup_{k\geq n} {\left\lvert {x_k} \right\rvert} = 0\).
• Note that \begin{align*} \lim_{n\to\infty} x_n = 0 \iff \forall {\varepsilon}\quad \exists N \text{ such that } n\geq N \implies {\left\lvert {x_n} \right\rvert} < {\varepsilon} .\end{align*}
• So let \({\varepsilon}>0\) be arbitrary.
• By the definition of the limit appearing in the \(\limsup\), there exists an \(N_0\) such that \begin{align*}n\geq N_0 \implies \sup_{k\geq n} {\left\lvert {x_k} \right\rvert} < {\varepsilon}.\end{align*}
• But then taking \(N = N_0\) in the first equation yields the result, since \begin{align*} n\geq N_0 \implies {\left\lvert {x_n} \right\rvert} \leq \sup_{k\geq n} {\left\lvert {x_k} \right\rvert} < {\varepsilon} .\end{align*}

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• Note that \(-1 \leq \sin(x) \leq 1\) and \(\sin(x) = \pm 1 \iff x = 2k\pi \pm {\pi \over 2}\).
• Since \(\pi\) is irrational, \(\sin(n)\) will never be of this form, so \(-1 < \sin(n) < 1\).
• Taking floors, we have \(-1 \leq {\left\lfloor \sin(n) \right\rfloor} \leq 0\), which in fact means that \(\sin(n) \in \left\{{-1, 0}\right\}\) can only take on one of two values.
• The set of subsequential limits is then just \(\left\{{-1, 0}\right\}\).
• Claim: \(\limsup {\left\lfloor \sin(n) \right\rfloor} = 0\).
  • It suffices to show that \({\left\lfloor \sin(n) \right\rfloor} = 0\) infinitely often
  • But note that there is an integer in any interval of the form \([2k\pi, 2k\pi + \pi]\) for \(k\in {\mathbb{N}}\), since it is of length \(\pi > 1\).
  • In these intervals, \(0 < \sin(n) < 1\), and so \({\left\lfloor \sin(n) \right\rfloor} = 0\), and there infinitely many such intervals.
  • So form a subsequence \(x_{n_k} = {\left\lfloor \sin(n_k) \right\rfloor}\) by choosing \(n_k\) to be any integer in
• Claim: \(\liminf {\left\lfloor \sin(n) \right\rfloor} = -1\).
  • By the exact same argument applied to intervals of the form \([3k\pi,3k\pi + \pi]\) where \(-1 < \sin(n) < 0\), we find that \({\left\lfloor \sin(n) \right\rfloor} = -1\) infinitely often.