#math/analysis # Theorem $\mathbb{R}$ is `complete`, or `Cauchy complete`: every Cauchy sequence in $\mathbb{R}$ converges to a limit in $\mathbb{R}$. ## Proof **Part 1:** First, we show that $(a_{n})$ is bounded. Choose $\epsilon=1$ in the [[1B2 - Cauchy Condition]]. This implies there exists an $N\in \mathbb{N}$ such that for all $n,k\geq{}N$ we have $\lvert a_{n}-a_{k} \rvert<1$. Now, choose a $K$ large enough such that $ -K \leq{}a_{1},\ldots,a_{N}\leq{}K. $ Note this is doable since $a_{1},\ldots,a_{N}$ is finite. Since we chose $\epsilon=1$, we know that for the rest of the sequence, $a_{N+1},a_{N+2},\ldots$ , that $\lvert a_{n}-a_{k} \rvert<1$. Thus, choose $M=K+1$. For $n\geq{}N$, take $k=N$, then apply the triangle inequality on $a_{n}=(a_{n}-a_{N})+a_{N}$ to get $ \lvert a_n\rvert\leq{}\lvert a_n-a_N\rvert+\lvert a_N\rvert<1+K=M. $ Together with the bound on $a_1,\ldots,a_N$, this gives $ -M<{}a_{n}<{}M. \quad\mathbf{(1)}$ So, $(a_{n})$ is a bounded sequence. **Part 2:** Now, we find the $b$ that we wish to show $(a_{n})$ converges to. Define a set $X$ with $ X=\{ x\in \mathbb{R}: \exists \text{ infinitely many }n\text{ such that }a_{n}\geq{}x\}. $ Thus $X$ is the set of all real thresholds that the sequence $(a_{n})$ exceeds infinitely often.^[While ordering isn't mentioned, if you take the set of those infinitely many $n\in \mathbb{N}$, you could order it. Hence specifying the ordering is inconsequential.] Note by $(1)$, we have $-M\in X$ and $M\not\in X$. Thus $X$ is both a non-empty subset of $\mathbb{R}$ and is bounded above by $M$. By the [[1B01 - Least Upper Bound]] property, we have $b=\text{l.u.b. } X$ with $-M\leq{}b\leq{}M$. **Part 3:** We will show that $(a_{n})$ does converge to $b$ as $n\to \infty$. To do this we must prove [[1B1 - Sequence Convergence]]. I.e. given $\epsilon>0$ there is an $N$ for which all $n\geq{}N$ we have $\lvert a_{n}-b \rvert<\epsilon$. Since $(a_{n})$ is Cauchy and $\frac{\epsilon}{2}$ is positive, there exists an $N$ such that if $n,k\geq{}N$ we have $ \lvert a_{n}-a_{k} \rvert <\frac{\epsilon}{2}. $ Since $b-\frac{\epsilon}{2}<b$, we know it is not an upper bound for $X$. Thus, there is an $x\in X$ with $b-\frac{\epsilon}{2}<{}x$. And for infinitely many $n$ we have $a_{n}\geq{}x.$ Additionally, since $b$ is an upper bound for $X$ and $b+\frac{\epsilon}{2}>b$, it does not belong to $X$. By definition of $X$, we then know that there are finitely many $n$ for which $a_{n}\geq{}b+\frac{\epsilon}{2}$. Thus, after discarding those finitely many indices, there are infinitely many $n$ for which $ b-\frac{\epsilon}{2}<{}x\leq{}a_{n}<{}b+\frac{\epsilon}{2}. \quad\quad\mathbf{(2)}$ Since we have infinitely many of these $n$ there are infinitely many that are $\geq{}N$. **Part 4:** Close it out utilizing the [[Triangle Inequality]]. Pick one, say $a_{n_{0}}$, with $n_{0}\geq{}N$ and $b-\frac{\epsilon}{2}<{}a_{n_{0}}<{}b+\frac{\epsilon}{2}$. Then for all $n\geq{}N$, we have $ \lvert a_{n}-b \rvert \leq{}\lvert a_{n}-a_{n_{0}} \rvert +\lvert a_{n_{0}}-b \rvert < \frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\quad\quad\mathbf{(3)} $ Thus $(a_{n})$ converges.$\quad \blacksquare{}$ ```tikz \begin{document} \begin{tikzpicture}[scale=1.05] % ================= CASE 1 ================= \draw (-0.2,3) -- (10.2,3); % fixed reference points \foreach \x in {0,3,5,7,10} \fill (\x,3) circle (2pt); \node[below] at (0,3) {$-M$}; \node[below] at (3,3) {$b-\frac{\epsilon}{2}$}; \node[below] at (5,3) {$b$}; \node[below] at (7,3) {$b+\frac{\epsilon}{2}$}; \node[below] at (10,3) {$M$}; % possible a_n and a_n0 \fill (4.2,3) circle (2.5pt); \fill (5.9,3) circle (2.5pt); \node[above] at (4.2,3) {$a_n$}; \node[above] at (5.9,3) {$a_{n_0}$}; \node[left] at (-0.5,3) {Case 1*:}; % ================= CASE 2 ================= \draw (-0.2,1.5) -- (10.2,1.5); \foreach \x in {0,3,5,7,10} \fill (\x,1.5) circle (2pt); \node[below] at (0,1.5) {$-M$}; \node[below] at (3,1.5) {$b-\frac{\epsilon}{2}$}; \node[below] at (5,1.5) {$b$}; \node[below] at (7,1.5) {$b+\frac{\epsilon}{2}$}; \node[below] at (10,1.5) {$M$}; % reversed relative positions \fill (4.1,1.5) circle (2.5pt); \fill (6.1,1.5) circle (2.5pt); \node[above] at (4.1,1.5) {$a_{n_0}$}; \node[above] at (6.1,1.5) {$a_n$}; \node[left] at (-0.5,1.5) {Case 2*:}; % ================= CASE 3 ================= \draw (-0.2,0) -- (10.2,0); \foreach \x in {0,3,5,7,10} \fill (\x,0) circle (2pt); \node[below] at (0,0) {$-M$}; \node[below] at (3,0) {$b-\frac{\epsilon}{2}$}; \node[below] at (5,0) {$b$}; \node[below] at (7,0) {$b+\frac{\epsilon}{2}$}; \node[below] at (10,0) {$M$}; % both on same side of b \fill (3.8,0) circle (2.5pt); \fill (4.5,0) circle (2.5pt); \node[above] at (3.8,0) {$a_n$}; \node[above] at (4.5,0) {$a_{n_0}$}; \node[left] at (-0.5,0) {Case 3:}; \end{tikzpicture} \end{document} ``` Note that in case 3 we have $\lvert a_{n}-b \rvert=\lvert a_{n}-a_{n_{0}} \rvert+\lvert a_{n_{0}}-b \rvert$. This does not weaken $\mathbf{(3)}$, since the right-hand side is still less than $\epsilon$. Thus, the inequality expressed in $\mathbf{(3)}$ describes all three cases and proves convergence in each. For case 1, note that $a_{n_{0}}$ is above $b$. This is possible, but $\mathbf{(2)}$ guarantees only that there are infinitely many terms within $\frac{\epsilon}{2}$ of $b$, not that infinitely many of them are above $b$. # Cauchy Convergence Criterion A sequence $(a_{n})$ in $\mathbb{R}$ converges if and only if it is a Cauchy sequence. $ \begin{aligned} &\exists b\in\mathbb{R}\;\forall\epsilon>0\;\exists N\in\mathbb{N}\; \forall n\geq N,\quad \lvert a_n-b\rvert<\epsilon \\ &\iff \\ &\forall\epsilon>0\;\exists N\in\mathbb{N}\; \forall n,k\geq N,\quad \lvert a_n-a_k\rvert<\epsilon. \end{aligned} $