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Copy file name to clipboardExpand all lines: conversion2025/mathpix_to_llm_to_in2lambda_to_JSON.ipynb
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" Your task is to extract all individual questions and their worked solutions from the provided markdown content.\n",
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"\n",
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" 1. **Content Extraction:**\n",
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" - Identify a suitable `name` for the set of questions.\n",
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" - You may choose a suitable `name` for the set of questions.\n",
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" - Identify the `year` if mentioned; otherwise, use \"0\".\n",
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" - For each question, carefully extract the full question text into `question_content` and the corresponding full solution/answer text into `solution_content`. They may not be in the same section.\n",
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" - Using the question numbers to identify the full questions, for each question:\n",
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" - Carefully extract the full question text into `question_content`.\n",
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" - Carefully extract the corresponding full solution/answer text into `solution_content`. They may not be in the same section.\n",
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" - If no solution is found, leave `solution_content` as an empty string `\"\"`.\n",
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" - Preserve all image tags like ``, making sure they are placed with their respective \"question_content\" and \"solution_content\". Do not duplicate it.\n",
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" - For Each Question extract all image references (e.g., `filename.jpg`) found within the `question_content` and `solution_content` and place them in the `images` list.\n",
"name": "Real Analysis - Sequences and Convergence",
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"year": "0",
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"questions": [
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{
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"title": "Equivalences of properties of sequences",
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"content": "Consider the following properties of a sequence of real numbers $(a_n)_{n\\ge0}$\u2026:\n\n1. $a_n\\to a$, or\n\n2. \"$a_n$ eventually equals $a$\" -- i.e.\n$\\exists N\\in\\mathbb N_{>0}$ such that\n$\\forall n\\ge N,\\ a_n = a$, or\n\n3. \"$(a_n)$ is bounded\" -- i.e. $\\exists\\,R\\in\\mathbb R$ such that\n$|a_n|<R\\\\\\forall n\\in \\mathbb N_{>0}$.\n\nFor each statement (a-e) below, which of (i-iii) is it equivalent\nto? Proof?",
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"parts": [
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"1. $\\exists\\, N\\in\\mathbb N_{>0}$ such that\n$\\forall n\\ge N,\\\\forall\\epsilon>0,\\ |a_n-a|<\\epsilon$.",
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"2. $\\forall\\epsilon>0$ there are only finitely many\n$n\\in\\mathbb N_{>0}$ for which $|a_n-a|\\ge\\epsilon$.",
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"3. $\\forall N\\in\\mathbb N_{>0},\\\\exists\\,\\epsilon>0$ such that\n$n\\ge N\\\\Rightarrow\\ |a_n-a|<\\epsilon$.",
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"4. $\\exists\\,\\epsilon>0$ such that\n$\\forall N\\in\\mathbb N_{>0},\\ |a_n-a|<\\epsilon\\\\forall n\\ge N$.",
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"5. $\\forall R >0\\\\exists\\, N\\in\\mathbb N_{>0}$ such that\n$n\\ge N\\\\Rightarrow\\ a_n\\in(a-\\frac1R,a+\\frac1R)$."
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],
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"images": [],
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"parts_solutions": [
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"(a) $\\iff$ (ii) because \"$\\forall\\epsilon>0,\\ |a_n-a|<\\epsilon$\"\nis the same statement as \"$a_n = a$.\"\n\n(Proof: if $a_n\\ne a$ then set $\\epsilon:= |a_n-a| > 0$ so that\n$|a_n-a| <\\epsilon$ is not true.)",
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"(b) $\\iff$ (i). Suppose (b) is true. Fix any $\\epsilon > 0$ and\nlet $n_1,\\ldots, n_r$ be the finite number of $n_i$ with\n$|a_{n_i}- a|\\ge\\epsilon$.\n\nSet $N :=\\max\\{n_1,\\ldots, n_r\\} + 1$. Then $\\forall n\\ge N$ we have\n$|a_n -a| < \\epsilon$, so $a_n\\to a$.\n\nSuppose (i) is true. Fix any $\\epsilon > 0$, then\n$\\exists\\, N\\in\\mathbb N_{>0}$ such that\n$|a_n -a| < \\epsilon\\\\forall n\\ge N$. In particular if\n$|a_n- a|\\ge\\epsilon$ then $n<N$ so there are only finitely many\nsuch $n\\in\\mathbb N_{>0}$.}",
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"(c) $\\iff$ (iii). Suppose (c) is true and take $N = 1$. Then\n$\\exists\\,\\epsilon > 0$ such that\n$|a_n -a| < \\epsilon\\\\forall n\\ge 1$. So, by the triangle\ninequality, $|a_n| < |a| + \\epsilon$. Putting $R := |a| + \\epsilon$\ngives (iii).\n\nSuppose (iii) is true, i.e. $\\exists\\, R\\in\\mathbb R$ such that\n$|a_n|<R\\\\\\forall n\\in N$. By the triangle inequality, $|a_n- a| <\nR + |a|\\\\,\\forall n\\ge N$. Putting $\\epsilon := R + |a|$ proves\n(c).",
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"(d) \\iff (iii). Suppose (d) is true and take $N = 1$. Then\n$|a_n -a| < \\epsilon\\\\forall n\\ge 1$. So, by the triangle\ninequality, $|a_n| < |a| +\\epsilon$. Putting $R := |a| + \\epsilon$\ngives (iii).\n\nSuppose (iii) is true, i.e. $\\exists\\, R\\in\\mathbb R$ such that\n$|a_n|<R\\\\\\forall n\\in N$. By the triangle inequality,\n$|a_n - a| <\nR + |a|\\\\forall n\\ge N$. Putting $\\epsilon:= R + |a|$ proves (d).",
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"(e) \\iff (i): just replace \\epsilon by \\frac1R in the definition\nof convergence."
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]
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},
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{
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"title": "Convergence in the Complex Plane",
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"content": "Given a sequence $(a_n)_{n\\ge1}$ of *complex* numbers, define what\na_n\\to a means. For $x,y\\in\\mathbb R$ and $z:=x+iy\\in\\mathbb C$\nshow $\\max(|x|,|y|)\\\\le\\ |z|\\\\le\\\\sqrt2\\max(|x|,|y|),$ and\n$$a_n\\to a+ib\\in\\mathbb C\\quad\\iff\\quad \\mathrm{Re}(a_n)\\to a \\quad\\mathrm{and\\quad Im}(a_n)\\to b.$$",
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"parts": [
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"Given a sequence $(a_n)_{n\\ge1}$ of *complex* numbers, define what\na_n\\to a means. For $x,y\\in\\mathbb R$ and $z:=x+iy\\in\\mathbb C$\nshow $\\max(|x|,|y|)\\\\le\\ |z|\\\\le\\\\sqrt2\\max(|x|,|y|),$ and\n$$a_n\\to a+ib\\in\\mathbb C\\quad\\iff\\quad \\mathrm{Re}(a_n)\\to a \\quad\\mathrm{and\\quad Im}(a_n)\\to b.$$"
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],
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"images": [],
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"parts_solutions": [
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"The inequalties\n$$\\max(x^2,y^2)\\\\le\\ x^2+y^2\\\\le\\\\max(x^2,y^2)+\\max(x^2,y^2)$$\ngive $$\\max(|x|,|y|)^2\\\\le\\ |z|^2\\\\le\\ 2\\max(|x|,|y|)^2.$$ Suppose\na_n\\to a+ib and fix any $\\epsilon>0$. Then\n$\\exists\\, N\\in\\mathbb N_{>0}$ such that\n$$n\\ge N\\\\Rightarrow\\ |a_n-(a+ib)|<\\epsilon\\\\Rightarrow\\\\max(|\\mathrm{Re}(a_n)-a|,|\\mathrm{Im}(a_n)-b|)<\\epsilon,$$\nusing the first stated inequality. Therefore\n$|\\mathrm{Re}(a_n)-a|<\\epsilon$ and $|\\mathrm{Im}(a_n)-b|<\\epsilon$\nas required.\n\nConversely, suppose\n$\\mathrm{Re}(a_n)\\to a \\quad\\mathrm{and\\quad Im}(a_n)\\to b$ and fix\nany $\\epsilon>0$. Then $\\exists\\, N\\in\\mathbb N_{>0}$ such that\n$n\\ge N\\\\Rightarrow\\ |\\mathrm{Re}(a_n)-a|<\\epsilon/\\sqrt 2$ and\n$|\\mathrm{Im}(a_n)-b|<\\epsilon/\\sqrt 2$. Thus\n$$|a_n-(a+ib)|\\ <\\\\sqrt 2\\max(|\\mathrm{Re}(a_n)-a|,|\\mathrm{Im}(a_n)-b|)\\ <\\\\sqrt 2\\epsilon/\\sqrt 2\\ =\\\\epsilon,$$\nusing the second stated inequality."
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]
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},
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{
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"title": "Squeeze Theorem for Sequences",
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"content": "Suppose that $a_n\\le b_n\\le c_n\\\\forall n$ and that $a_n\\to a$ and\n$c_n\\to a$. Prove that $b_n\\to a$.",
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"parts": [
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"Suppose that $a_n\\le b_n\\le c_n\\\\forall n$ and that $a_n\\to a$ and\n$c_n\\to a$. Prove that $b_n\\to a$.",
"content": "Suppose that $a_n\\to0$ and $(b_n)$ is bounded. Prove that\n$a_nb_n\\to0$.",
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"parts": [
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"Suppose that $a_n\\to0$ and $(b_n)$ is bounded. Prove that\n$a_nb_n\\to0$."
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],
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"images": [],
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"parts_solutions": [
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"$\\exists\\, B>0$ such that $|b_n|\\le B\\\\forall n$.\n\nGiven $\\epsilon>0,\\\\exists\\, N\\in\\mathbb N_{>0}$ such that\n$n\\ge N\\\\Rightarrow\\ |a_n|<\\epsilon/B$.\n\nTherefore $|a_nb_n|=|a_n||b_n|\\le(\\epsilon/B)B=\\epsilon$, as\nrequired."
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]
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},
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{
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"title": "Boundedness of the sequence and bound on $b_n$",
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"content": "$\\!\\!$ Suppose that $(a_n)$ and $(b_n)$ are sequences of real\nnumbers such that $a_n\\to a$ and $b_n\\to b\\ne0$. Prove that the set\n$\\{a_n\\colon n\\in\\mathbb N_{>0}\\}$ is bounded and that\n$$\\exists\\, N\\in\\mathbb N_{>0} \\quad\\mathrm{such\\ that}\\quad n\\ge N\\\\Rightarrow\\ |b_n|>|b|/2.$$",
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"parts": [
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"Prove that the set $\\{a_n\\colon n\\in\\mathbb N_{>0}\\}$ is bounded.",
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"Prove that $$\\exists\\, N\\in\\mathbb N_{>0} \\quad\\mathrm{such\\ that}\\quad n\\ge N\\\\Rightarrow\\ |b_n|>|b|/2.$$"
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],
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"images": [],
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"parts_solutions": [
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"",
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"Set $\\epsilon=|b|/2>0$. Then $\\exists\\, N\\in\\mathbb N_{>0}$ such that $$n\\ge N\\\\Rightarrow\\ |b_n-b|<\\epsilon\\\\Rightarrow\\ |b|<|b_n|+\\epsilon\\\\Rightarrow\\ |b_n|>|b|-\\epsilon=|b|/2.$$"
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]
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},
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{
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"title": "Limit of quotient of sequences",
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"content": "Therefore $(a_n/b_n)_{n\\ge N}$ is a sequence of real numbers; prove\nit tends to $a/b$.",
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"parts": [
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"Therefore $(a_n/b_n)_{n\\ge N}$ is a sequence of real numbers; prove\nit tends to $a/b$."
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],
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"images": [],
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"parts_solutions": [
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"$$\\left|\\frac{a_n}{b_n}-\\frac ab\\right|\\ =\\ \\left|\\frac{a_nb-ab_n}{bb_n}\\right|\\ =\\ \\left|\\frac{(a_n-a)b+a(b-b_n)}{bb_n}\\right|\\ \\le\\ \\left|\\frac{(a_n-a)}{b_n}\\right|+\n\\left|\\frac{a(b-b_n)}{bb_n}\\right|.$$ From above we can find\n$N_1\\in\\mathbb N_{>0}$ such that\n$n\\ge N_1\\ \\Rightarrow\\ |b_n|\\ge|b|/2$, which in turn implies that\n$$\\left|\\frac{a_n}{b_n}-\\frac ab\\right|\\ \\le\\ \\frac{|a_n-a|}{|b|/2}+\n|a|\\frac{|b-b_n|}{|b|.|b|/2}\\ =\\ \\frac{2}{|b|}|a_n-a|+\n\\frac{2|a|}{b^2}|b-b_n|.$$ Now fix any $\\epsilon>0$. There exists\n$N_2\\in\\mathbb N_{>0}$ such that\n$n\\ge N_2\\ \\Rightarrow\\ |a_n-a|<|b|\\epsilon/4$. And there exists\n$N_3\\in\\mathbb N_{>0}$ such that\n$n\\ge N_3\\ \\Rightarrow\\ |b_n-b|<|b|^2\\epsilon/4(1+|a|)$.\\n\nTherefore if we set $N:=\\max\\{N_1,N_2,N_3\\}$ then\n$$n\\ge N\\ \\ \\Rightarrow\\ \\ \\left|\\frac{a_n}{b_n}-\\frac ab\\right|\\ <\\ \\frac{2|b|\\epsilon/4}{|b|}+\\n\\frac{2|a|}{b^2}\\frac{b^2\\epsilon}{4(1+|a|)}\\ <\\ \\epsilon/2+\\epsilon/2\\ =\\ \\epsilon.$$"
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]
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},
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{
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"title": "Sorta-Cauchy sequences and divergence",
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"content": "We call a sequence *sorta-Cauchy* if it satisfies the condition\n$$\\forall\\epsilon>0\\\\exists\\, N\\in\\mathbb N_{>0}\\\\ n\\ge N\\\\Rightarrow\\ |a_n-a_{n+1}|<\\epsilon.$$",
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"parts": [
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"Give an example of a sorta-Cauchy sequence which diverges to\n$+\\infty$. Conclude that sorta-Cauchy is not as strong as Cauchy."
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],
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"images": [],
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"parts_solutions": [
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"Any $a_n$ that increases so slowly to infinity that $a_{n+1}-a_n$ converges to zero. Eg $a_n=\\sqrt n$ or $a_n=\\log n$ or $a_n=\\sum_{i=1}^n\\frac 1i$."
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]
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},
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{
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"title": "Cauchy sequence in \\mathbb Q not converging in \\mathbb Q",
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"content": "Give an example of a Cauchy sequence in $\\mathbb Q$ which does not\nconverge in $\\mathbb Q$.",
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"parts": [
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"We know that in $\\mathbb R$, a sequence is Cauchy if and only if it\nis convergent. Show that it is impossible to prove this using only\nthe arithmetic and order axioms of $\\mathbb R$ (i.e. all the axioms\nexcept the completeness axioms -- the one about the existence of\nleast upper bounds)."
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],
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"images": [],
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"parts_solutions": [
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"Suppose that we could prove that a sequence is Cauchy if and only if\nit is convergent with only the arithmetic and order axioms. We know\nin $\\mathbb Q$ that the order and arithmetic axioms hold, this would\nsuggest that any Cauchy sequence in $\\mathbb Q$ must converge in\n$\\mathbb Q$. This is a contradiction since we have just shown that\nthere are Cauchy sequences in $\\mathbb Q$ that do not converge in\n$\\mathbb Q$ in the first part of this question."
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]
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},
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{
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"title": "Bounded sequence and supremum of sets",
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"content": "Let $(a_{n})_{n \\in \\mathbb N_{>0}}$ be a bounded sequence.",
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"parts": [
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"For each $n \\in \\mathbb N_{>0}$, define the set\n$S_{n} = \\{ a_{j}: j \\ge n\\}$. Prove that, for every\n$n \\in \\mathbb N_{>0}$, there exists some $b_{n} \\in \\mathbb{R}$\nsuch that $b_{n} = \\sup(S_{n})$.",
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""
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],
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"images": [],
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"parts_solutions": [
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"Let $M$ be an upper bound for $(a_{n})_{n \\in \\mathbb N_{>0}}$, then the set $S_{n}$ is non-empty with $M$ as an upper bound, so by the completeness axiom $S_{n}$ has a supremum.",
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""
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]
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},
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{
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"title": "Existence of the Limit Supremum of a Sequence",
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"content": "Let $B = \\{ b_{n}: n \\in \\mathbb N_{>0}\\}$ where $b_{n}$ is\ndefined as above. Prove that there exists some\n$l \\in \\mathbb{R}$ such that $l = \\inf(B)$.",
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"parts": [
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"(Remark: $l$ is\ncalled the limit supremum of the sequence\n$(a_{n})_{n \\in \\mathbb N_{>0}}$, and the usual notation is\n$l = \\limsup_{n \\rightarrow \\infty} a_{n}$)."
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],
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"images": [],
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"parts_solutions": [
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"The set $B$ is clearly non-empty. Let $\\tilde{M}$ be a lower\nbound for $(a_{n})_{n \\in \\mathbb N_{>0}}$. Then\n$b_{n} = \\sup(S_{n}) \\ge \\tilde{M}$. This means $\\tilde{M}$ is a\nlower bound for $B$ and so by the completeness axiom, $B$ has an\ninfimum."
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]
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},
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{
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"title": "Limit superior of sequences",
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"content": "For each of the sequences below, find the value of\n$\\limsup_{n \\rightarrow \\infty} a_{n}$ and give justification\nfor your answer.",
"title": "Limit Superior and Infimum of a Sequence",
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"content": "$a_{n} = \\frac{(-1)^n}{n}$",
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"parts": [
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"$a_{n} = \\frac{(-1)^n}{n}$"
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],
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"images": [],
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"parts_solutions": [
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"$\\limsup_{n \\rightarrow \\infty} a_{n} = 0$. Note that, for\nevery $j \\in \\mathbb N_{>0}$, one has\n$b_{j} = \\sup(S_{j}) = \\frac{1}{j}$ for $j$ even and\n$\\frac{1}{j+1}$ for $j$ odd. Therefore we argue that\n$\\inf(B) = 0$. Clearly $0$ is a lower bound. To show that it\nis an infimum, we show that for any $\\epsilon > 0$, there\nexists $b \\in B$ such that $b - \\epsilon < 0$. Note that, by\nthe Archimedean Axiom, we can find an $j \\in \\mathbb N_{>0}$\nsuch that $j \\ge \\frac{1}{\\epsilon}$, so that\n$\\frac{1}{j} < \\epsilon$. It then follows that\n$b_{j} - \\epsilon < 0$, which finishes the proof since\n$b_{j} \\in B$."
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