[problem:The Harmonic Series Is Not Cauchy | 1] Prove directly from the definition that the sequence of partial sums $S_n = \sum_{k=1}^n 1/k$ is not a Cauchy sequence. [/problem]

[solution] We must find $\varepsilon_0 > 0$ such that for every $N \in \mathbb{N}$, there exist $m > n \geq N$ with $|S_m - S_n| \geq \varepsilon_0$. Take $\varepsilon_0 = 1/2$. For any $N$, set $n = N$ and $m = 2N$. Then: \begin{align*} |S_{2N} - S_N| = \sum_{k=N+1}^{2N} \frac{1}{k} \geq \sum_{k=N+1}^{2N} \frac{1}{2N} = N \cdot \frac{1}{2N} = \frac{1}{2} = \varepsilon_0. \end{align*} Since for every $N$ we found $m = 2N > n = N \geq N$ with $|S_m - S_n| \geq 1/2$, the sequence $(S_n)$ is not Cauchy. By the [General Principle of Convergence](/theorems/172), $\sum 1/k$ diverges. Note the contrast with the sequence of *terms*: $a_n = 1/n \to 0$, so consecutive partial sums get close ($|S_{n+1} - S_n| = 1/(n+1) \to 0$). The Cauchy condition demands more — all pairs $(n, m)$, not just consecutive ones — and this stronger demand detects the divergence that the term test misses. $\checkmark$ [/solution]

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[problem:Absolute Convergence Characterises Completeness | 2] Let $(V, \|\cdot\|)$ be a normed vector space. Prove that the following are equivalent: (a) $V$ is complete (i.e., $V$ is a [Banach space](/page/Banach%20Space)). (b) Every absolutely convergent series in $V$ converges: if $\sum_{n=1}^\infty \|v_n\| < \infty$, then $\sum_{n=1}^\infty v_n$ converges in $V$. [/problem]

[solution] **(a) $\Rightarrow$ (b).** Assume $V$ is complete and $\sum \|v_n\| < \infty$. The partial sums $S_N = \sum_{n=1}^N v_n$ form a Cauchy sequence: for $m > n$, \begin{align*} \|S_m - S_n\| = \left\|\sum_{k=n+1}^m v_k\right\| \leq \sum_{k=n+1}^m \|v_k\|. \end{align*} The right side is the tail of the convergent series $\sum \|v_n\|$, so it tends to $0$ as $n \to \infty$ (uniformly in $m > n$). Therefore $(S_N)$ is Cauchy, and completeness gives convergence. **(b) $\Rightarrow$ (a).** Assume every absolutely convergent series converges. Let $(x_n)$ be a Cauchy sequence in $V$. We construct an absolutely convergent series whose partial sums form a subsequence of $(x_n)$. **Step 1: Extract a fast subsequence.** By the Cauchy property, for each $k \geq 1$, choose $n_k$ such that $\|x_n - x_m\| < 2^{-k}$ for all $n, m \geq n_k$. We may assume $n_1 < n_2 < n_3 < \cdots$ (by choosing inductively). **Step 2: Telescope.** Define $v_1 = x_{n_1}$ and $v_k = x_{n_k} - x_{n_{k-1}}$ for $k \geq 2$. Then $\sum_{k=1}^K v_k = x_{n_K}$ (telescoping). The series is absolutely convergent: \begin{align*} \sum_{k=2}^\infty \|v_k\| = \sum_{k=2}^\infty \|x_{n_k} - x_{n_{k-1}}\| < \sum_{k=2}^\infty 2^{-(k-1)} = 1 < \infty. \end{align*} **Step 3: Conclude.** By hypothesis (b), $\sum v_k$ converges to some $x \in V$. Since $\sum_{k=1}^K v_k = x_{n_K}$, the subsequence $x_{n_K} \to x$. A Cauchy sequence with a convergent subsequence converges to the same limit (triangle inequality: $\|x_n - x\| \leq \|x_n - x_{n_K}\| + \|x_{n_K} - x\|$, both terms small for large $n, K$). Therefore $x_n \to x$, proving completeness. $\checkmark$ [/solution]

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[problem:Completion Of The Rationals | 2] Carry out the Cauchy sequence construction of $\mathbb{R}$ from $\mathbb{Q}$ in detail: define the equivalence relation, verify the metric is well-defined, embed $\mathbb{Q}$ isometrically, and prove the resulting space is complete. [/problem]

[solution] **Step 1: Define the space.** Let $\mathcal{C}$ be the set of all Cauchy sequences of rationals. Define the equivalence relation $(a_n) \sim (b_n) \iff |a_n - b_n| \to 0$. Set $\hat{\mathbb{Q}} = \mathcal{C} / {\sim}$. **Step 2: The metric is well-defined.** For $[(a_n)], [(b_n)] \in \hat{\mathbb{Q}}$, define $d([(a_n)], [(b_n)]) = \lim_{n \to \infty} |a_n - b_n|$. The limit exists: the sequence $c_n = |a_n - b_n|$ is Cauchy in $\mathbb{R}$ (which we may assume exists, or work within $\mathbb{Q}$ using the fact that $|c_n - c_m| \leq |a_n - a_m| + |b_n - b_m| \to 0$, and bounded monotone sequences converge). Well-definedness on equivalence classes: if $(a_n) \sim (a_n')$ and $(b_n) \sim (b_n')$, then $||a_n - b_n| - |a_n' - b_n'|| \leq |a_n - a_n'| + |b_n - b_n'| \to 0$, so the limits agree. **Step 3: Isometric embedding.** Define $\iota: \mathbb{Q} \to \hat{\mathbb{Q}}$ by $\iota(q) = [(q, q, q, \ldots)]$. Then $d(\iota(p), \iota(q)) = \lim |p - q| = |p - q|$, so $\iota$ is an isometry. **Step 4: Density.** For any $[(a_n)] \in \hat{\mathbb{Q}}$ and $\varepsilon > 0$, choose $N$ with $|a_n - a_m| < \varepsilon/2$ for $n, m \geq N$. Then $d([(a_n)], \iota(a_N)) = \lim_{n \to \infty} |a_n - a_N| \leq \varepsilon/2 < \varepsilon$. So $\iota(\mathbb{Q})$ is dense. **Step 5: Completeness.** Let $(x^{(k)})_{k=1}^\infty$ be a Cauchy sequence in $\hat{\mathbb{Q}}$, where each $x^{(k)} = [(a_n^{(k)})_n]$. By density, for each $k$, choose $q_k \in \mathbb{Q}$ with $d(x^{(k)}, \iota(q_k)) < 1/k$. The sequence $(q_k)$ is Cauchy in $\mathbb{Q}$: $|q_k - q_l| \leq d(\iota(q_k), x^{(k)}) + d(x^{(k)}, x^{(l)}) + d(x^{(l)}, \iota(q_l)) < 1/k + d(x^{(k)}, x^{(l)}) + 1/l \to 0$. Define $x = [(q_k)] \in \hat{\mathbb{Q}}$. Then $d(x^{(k)}, x) \leq d(x^{(k)}, \iota(q_k)) + d(\iota(q_k), x) < 1/k + d(\iota(q_k), x)$. Since $d(\iota(q_k), x) = \lim_{j \to \infty} |q_k - q_j| \to 0$ as $k \to \infty$ (because $(q_j)$ is Cauchy), we get $d(x^{(k)}, x) \to 0$. $\checkmark$ [/solution]

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[problem:Contraction On An Incomplete Space | 2] Give an example of a contraction mapping $T: X \to X$ on a metric space $(X, d)$ that has no fixed point. Identify precisely where the proof of the Banach Fixed-Point Theorem fails. [/problem]

[solution] **The example.** Let $X = (0, 1)$ with the usual metric, and define $T(x) = x/2$. Then $|T(x) - T(y)| = |x - y|/2$, so $T$ is a contraction with $\lambda = 1/2$. The equation $T(x) = x$ gives $x/2 = x$, so $x = 0$. But $0 \notin (0, 1)$, so $T$ has no fixed point in $X$. **Where the proof fails.** The Banach fixed-point theorem constructs the sequence $x_0, T(x_0), T^2(x_0), \ldots$ Starting from any $x_0 \in (0, 1)$: $T^n(x_0) = x_0 / 2^n \to 0$. The sequence is Cauchy (it converges in $\mathbb{R}$, hence is Cauchy in $\mathbb{R}$, hence Cauchy in $(0, 1)$). But the limit $0$ is not in $X = (0, 1)$. The proof of Banach's theorem invokes completeness at exactly this point — to guarantee that the Cauchy sequence has a limit *in the space*. The incomplete space $(0, 1)$ has a "gap" at $0$ where the fixed point should be. Replacing $X$ with $[0, 1]$ (which is complete) restores the theorem: $T(0) = 0$ is the unique fixed point. $\checkmark$ [/solution]