Real Analysis: Sequences

Problems

1. Show that $\left\{\frac{1}{n}\right\}$ is a Cauchy sequence.

   Pick $\varepsilon > 0.$ By the Archimedean property, there exists an $N \in \mathbb{N}$ such that $N \varepsilon > 1.$ For all $n > N$ it then follows that $\varepsilon > \frac{1}{N} > \frac{1}{n}.$ Next, note that $0 < \frac{1}{m} < \frac{1}{n}$ for all $m > n.$ Therefore $\frac{1}{n} - \frac{1}{m} < \varepsilon.$ Therefore $\left\{\frac{1}{n}\right\}$ is a Cauchy sequence.

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2. Show that every convergent sequence in $\mathbb{R}$ with the Euclidean metric is a Cauchy sequence.

   Assume $\{a_n\}$ is a convergent sequence in $\mathbb{R}$ under the Euclidean metric. Then there is some $L \in \mathbb{R}$ such that for every $\dfrac{\varepsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that $|a_n - L| < \dfrac{\varepsilon}{2}$ for all $n > N.$ It follows that $a_n \in \left(L - \dfrac{\varepsilon}{2}, L + \dfrac{\varepsilon}{2}\right)$ whenever $n > N.$ Thus for any $n, m > N,$ we see that $|a_n - a_m| < \varepsilon.$ Therefore $\{a_n\}$ is a Cauchy sequence.

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3. Show that every convergent sequence in any metric space $(X, d)$ is Cauchy.

   Let $\{a_n\}$ be a Cauchy sequence in $(X, d).$ Then there is some $L \in X$ such that for every $\frac{\varepsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that $d(a_n, L) < \frac{\varepsilon}{2}$ for every $n > N.$ By the triangle inequality, $d(a_n, a_m) \leq d(a_n, L) + d(L, a_m) < \varepsilon$ for all $n, m > N.$ It follows that $\{a_n\}$ is a Cauchy sequence.

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4. Show that every Cauchy sequence is bounded.

   Let $\{a_n\}$ be a Cauchy sequence in a metric space $(X, d).$ Then there exists an $N \in \mathbb{N}$ such that $d(a_n, a_m) < 1$ for all $n, m \in \mathbb{N}$ where $n > N$ and $m > N.$ It follows that $\{a_n\}$ is bounded for $n > N.$ Likewise, since the set $\{a_1, \ldots, a_N\}$ is finite, it is bounded. Since the union of two bounded sets is bounded, it follows that $\{a_n\}$ is bounded.

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5. Cauchy Criterion for $\mathbb{R}$: Show that every Cauchy sequence in $\mathbb{R}$ converges.

   Let $\{a_n\}$ be a Cauchy sequence in $\mathbb{R}.$ It follows that $\{a_n\}$ is bounded. By the Bolzano-Weierstrass theorem, $\{a_n\}$ has a convergent subsequence, call it $\{a_{n_k}\}.$ Let $\lim\limits_{n_k \rightarrow \infty} a_{n_k} = L.$ We show that $\lim\limits_{n \rightarrow \infty} a_n = L.$

   Since $\{a_n\}$ is Cauchy, for any $\frac{\varepsilon}{2} > 0,$ there exists an $N \in \mathbb{N}$ such that $d(a_n, a_m) < \frac{\varepsilon}{2}$ for all $n \geq N$ and $m \geq N.$ Since $\{a_{n_k}\}$ converges to $L,$ it follows that there exists an $M \in \mathbb{N}$ such that $d(a_{n_k}, L) < \frac{\varepsilon}{2}$ whenever $n_k > M.$ Take $n$ and $n_k$ to be greater than $\max\{N, M\}.$ By the triangle inequality,

   $d(a_n, L) < d(a_n, a_{n_k}) + d(a_{n_k}, L)\\d(a_n, L) < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2}\\d(a_n, L) < \varepsilon$

   Therefore $\lim\limits_{n \rightarrow \infty} a_n = L.$

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6. Show that a Cauchy sequence in a metric space $(X, d)$ converges if and only if it has a subsequence that converges in $X.$

   Let $\{a_n\}$ be a Cauchy sequence in a metric space $(X, d).$ If $\{a_n\}$ converges, then it has a convergent subsequence, namely itself. Conversely, assume $\{a_n\}$ has a convergent subsequence, $\{a_{n_k}\}$ whose limit is $L \in X.$ Then for any $\frac{\varepsilon}{2} > 0$ there exists an $N \in \mathbb{N}$ such that $d(a_{n_k}, L) < \frac{\varepsilon}{2}$ whenever $n_k > N.$ Also, for any $\frac{\varepsilon}{2} > 0$ there exists an $M \in \mathbb{N}$ such that $d(a_n, a_{n_k}) < \frac{\varepsilon}{2}$ whenever $n > M.$ Take $n$ and $n_k$ to be greater than $\max\{N, M\}.$ It follows from the triangle inequality that

   $d(a_n, L) < d(a_n, a_{n_k}) + d(a_{n_k}, L)\\d(a_n, L) < \dfrac{\varepsilon}{2} + \dfrac{\varepsilon}{2}\\d(a_n, L) < \varepsilon$

   Therefore $\{a_n\}$ converges.

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7. Cauchy Criterion for $\mathbb{R}^n$: Show that every Cauchy sequence in $\mathbb{R}^n$ converges under the Euclidean metric.

   Let $\{\mathbf{a}_n\}$ be a Cauchy sequence in $\mathbb{R}^k$ under the Euclidean metric. Then for every $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $d(\mathbf{a}_n, \mathbf{a}_m) < \varepsilon$ for all $n \geq N$ and $m \geq N.$ It follows that $d(a_{n, i}, a_{m, i}) < \varepsilon$ for every $1 \leq i \leq k.$ Therefore the $i$th components of $\{\mathbf{a}_n\}$ form Cauchy sequences $\{a_{n,i}\}$ in $\mathbb{R}.$ By the Cauchy criterion on $\mathbb{R},$ each converges to a value $L_i.$ It follows by the convergence criterion for $\mathbb{R}^n$ that $\{\mathbf{a}_n\}$ converges to $\mathbf{L} = \{L_1, \ldots, L_k\}.$

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8. Equivalent definition, Part 1: Show that if a metric space is Cauchy complete, then it has the nested subset property.

   Let $(X, d)$ be a Cauchy complete metric space, and let $\{S_n\}$ be a nested sequence of subsets.

   Part 1: Construct Cauchy sequences out of points of each $S_n.$

   From each $S_n,$ select two points $a_n, b_n \in S_n.$ It follows that $d(a_n, b_n) \leq \text{Diam}(S_n).$ Since $S_{n+1} \subseteq S_n,$ it also follows that $\text{Diam}(S_{n+1}) \leq \text{Diam}(S_n),$ and so $d(a_n, a_m) \leq \text{Diam}(S_n)$ for all $m > n.$ Since $\{S_n\}$ is a nested sequence of subsets, for every $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $\text{Diam}(S_n) < \varepsilon$ for all $n > N.$ Thus $d(a_n, a_m) < \varepsilon$ and $d(b_n, b_m) < \varepsilon$ for $n, m \geq N.$ It follows that $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences. By completeness on $(X, d),$ $\{a_n\}$ and $\{b_n\}$ converge. Let $\lim\limits_{n \rightarrow \infty} a_n = L_a$ and $\lim\limits_{n \rightarrow \infty} b_n = L_b.$

   Part 2: Show that $L_a = L_b.$

   Proof by contradiction. Assume $L_a \neq L_b.$ Set $\varepsilon = d(L_a, L_b) > 0.$ Then there exists an $N_1 \in \mathbb{N}$ such that $d(a_n, b_n) < \frac{\varepsilon}{4}$ whenever $n \geq N_1.$ Likewise, there exists an $N_2 \in \mathbb{N}$ such that $d(a_n, L_a) < \frac{\varepsilon}{4}$ whenever $n > N_2,$ as well as an $N_3 \in \mathbb{N}$ such that $d(b_n, L_b) < \frac{\varepsilon}{4}$ whenever $n > N_3.$ Set $N = \max\{N_1, N_2, N_3\}.$ It follows by the triangle inequality that $d(L_a, L_b) \leq d(L_a, a_n) + d(a_n, b_n) + d(b_n, L_b) = \frac{3\varepsilon}{4}.$ But this is a contradiction. Therefore $L_a = L_b$ after all.

   Let $L = L_a = L_b.$ 

   Part 3: Show that $L \in \bigcap\limits_{n=0}^{\infty} S_n.$

   Proof by contradiction. Assume $L \notin \bigcap\limits_{n=0}^{\infty} S_n.$ Then there is some $N_1 \in \mathbb{N}$ such that $L \notin S_{N_1}.$ Therefore $L \in S_{N_1}^c.$ Since $S_{N_1}$ is closed, it follows that $S_{N_1}^c$ is open. Therefore there exists some $r > 0$ such that $B_r(L) \subseteq S_{N_1}^c.$ Pick $\varepsilon = r.$ Then there exists some $N_2 \in \mathbb{N}$ such that $d(a_n, L) < \varepsilon$ whenever $n > N_2.$ Set $N = \max\{N_1, N_2\}.$ When $n > N,$ we see that $d(a_n, L) < r,$ and so $a_n \in B_r(L).$ But this implies that $a_n \in S_{N_1}^c,$ a contradiction since $a_n \in S_n \subseteq S_{N_1}.$ Therefore $L \in \bigcap\limits_{n=0}^{\infty} S_n$ after all.

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9. Equivalent definition, Part 2: Show that if a metric space has the nested subset property, it is Cauchy complete.

   Let $(X, d)$ be a metric space with the nested subset property. Let $\{a_n\}$ be a Cauchy sequence.

   Part 1: Construct a nested sequence of closed subsets.

   Since $\{a_n\}$ is Cauchy, it is bounded. It follows that for every $n \in \mathbb{N},$ there exists an $r_n \in \mathbb{R}$ such that $r_n = \text{Diam}(\{a_m : m \geq n\}).$ Since subsequent terms in a sequence form a subset of their union with the preceding term, it follows that $r_{n+1} \leq r_n.$ Therefore $\overline{B}_{r_m}(a_m) \subseteq \overline{B}_{r_n}(a_n)$ for all $m \geq n.$ Since $\{a_n\}$ is Cauchy, for every $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $d(a_n, a_m) < \varepsilon$ for all $n, m \geq N.$ Therefore $r_n \leq \varepsilon,$ and so $\{\overline{B}_{r_n}(a_n)\}$ forms a nested sequence of subsets. By the nested subset property on $(X, d),$ the intersection $\bigcap\limits_{n=0}^{\infty} \overline{B}_{r_n}(a_n)$ is nonempty and contains a single value, call it $L.$

   Part 2: Show that the $\lim\limits_{n \rightarrow \infty} a_n = L.$

   Since $\{a_n\}$ is Cauchy, for any $\varepsilon > 0$ there exists an $N \in \mathbb{N}$ such that $d(a_n, a_m) < \varepsilon$ whenever $n, m \geq N.$ By construction, $\text{Diam}(\overline{B}_{r_n}(a_n)) < \varepsilon$ as well. Since $L \in \overline{B}_{r_n}(a_n),$ it follows that $d(a_n, L) < \varepsilon.$ Therefore $\lim\limits_{n \rightarrow \infty} a_n = L.$

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10. Equivalent definition: Show that if the least upper bound property on $\mathbb{R}$ is swapped for the Cauchy criterion, then the least upper bound property is implied.

    Define $\mathbb{R}^*$ to have all the properties of $\mathbb{R}$ except that the least upper bound property has been exchanged for the Cauchy criterion. We show that $\mathbb{R}^*$ has the least upper bound property.

    Part 1: We construct a sequence $\{a_n\}$ whose elements are not upper bounds of $A$ and a sequence $\{b_n\}$ whose elements are upper bounds of $A.$

    Let $A$ be a nonempty subset of $\mathbb{R}^*$ that is bounded above by $b_1,$ and let $a_1 \in \mathbb{R}^*$ such that $a_1$ is not an upper bound of $A.$ Set $\delta_1 = b_1 - a_1,$ the distance between $a_1$ and $b_1.$ Consider $c_1 = \frac{a_1 + b_1}{2}.$ If $c_1$ is an upper bound for $A,$ then set $a_2 = a_1$ and $b_2 = c_1.$ Otherwise, set $a_2 = c_1$ and $b_2 = b_1.$ In either event, set $\delta_2 = b_2 - a_2 = \frac{1}{2}\delta_1.$ Repeat this process of halving to form the two sequences $\{a_n\}$ and $\{b_n\}.$ Note that $\delta_n = 2^{-n+1}\delta_1.$

    Part 2: We show that $\{a_n\}$ and $\{b_n\}$ are Cauchy sequences.

    Note that each $b_n$ is an upper bound for $A$ and therefore also for $\{a_n\},$ and note that no $a_n$ is an upper bound for $A,$ and thus a low bound for $\{b_n\}.$

    Pick $\varepsilon > 0.$ By the Archimedean property on $\mathbb{R}^*,$ there exists an $N \in \mathbb{N}$ such that $2^{-N}\delta_1 < \varepsilon.$ It follows that for any $n, m > N$ where $n < m$ that $|a_n - a_m| < |a_n - b_n| = 2^{-N}\delta_1 < \varepsilon.$ Therefore $\{a_n\}$ is a Cauchy sequence. By the Cauchy criterion on $\mathbb{R}^*,$ it follows that $\{a_n\}$ converges to a point $L_1.$

    Likewise, pick $\varepsilon > 0.$ By the Archimedean property on $\mathbb{R}^*,$ there exists an $N \in \mathbb{N}$ such that $2^{-N}\delta_1 < \varepsilon.$ It follows that for any $n, m > N$ where $n < m$ that $|b_n - b_m| < |a_n - b_n| = 2^{-N}\delta_1 < \varepsilon.$ Therefore $\{b_n\}$ is a Cauchy sequence. By the Cauchy criterion on $\mathbb{R}^*,$ it follows that $\{b_n\}$ converges to a point $L_2.$

    Part 3: We show that $\lim\limits_{n \rightarrow \infty} a_n = L = \lim\limits_{n \rightarrow \infty} b_n.$

    Proof by contradiction. Set $\varepsilon = L_2 - L_1.$ Since $\{a_n\}$ and $\{b_n\}$ converge, there exists an $N_1 \in \mathbb{N}$ such that $|a_n - L_1| < \frac{\varepsilon}{4}$ and $|b_n - L_2| < \frac{\varepsilon}{4}$ whenever $n \geq N_1.$ Likewise, by the Archimedean property on $\mathbb{R}^*,$ there exists an $N_2 \in \mathbb{N}$ such that $2^{-N+1}\delta_1 < \frac{\varepsilon}{4}.$ Set $N = \max\{N_1, N_2\}.$ Then $|a_n - b_n| < \frac{\varepsilon}{4}$ and $|a_n - L_1| < \frac{\varepsilon}{4}$ and $|b_n - L_2| < \frac{\varepsilon}{4}$ whenever $n > N.$ By the triangle inequality, this means that $|L_2 - L_1| \leq |L_1 - a_n| + |a_n - b_n| + |b_n - L_2| = \frac{3\varepsilon}{4}.$ But this is a contradiction. Therefore the limits must have been identical after all.

    Set $L = L_1 = L_2.$

    Part 4: We show that $L$ is an upper bound of $A.$

    We use trichotomy to show that $L$ cannot be otherwise. Assume there is a $c \in A$ such that $L < c.$ Set $\varepsilon = c - L.$ Then there exists an $N \in \mathbb{N}$ such that $|b_n - L| < \varepsilon$ whenever $n \geq N.$ This implies that $L < b_N < c$ But this is a contradiction, since $b_N$ is an upper bound of $A.$ Therefore $L \geq c$ after all.

    Part 5: We show that $L = \sup(A).$

    Next we show that $L$ is the least upper bound of $A.$ Proof by contradiction. Assume there is some $x < L$ such that $x$ is an upper bound of $A.$ Let $\varepsilon = L - x.$ Then there exists an $N \in \mathbb{N}$ such that $|a_n - L| < \varepsilon$ whenever $n \geq N.$ It follows that $x < a_N \leq L.$ But this is a contradiction, since $a_N$ is not an upper bound of $A.$ Therefore $L$ is the least upper bound of $A$.

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