Let $\epsilon \in \R_{>0}$ be given. Since $\sequence {a_n}$ is Cauchy, a natural number $N$ exists such that: :$\size {a_n - a_m} < \epsilon$ for every $m, n \ge N$ We aim to show that $\sequence {a_n}$ converges. That is, that numbers $a$ in $\R$ and $N'$ in $\N$ exist such that: :$\size {a_n - a} < \epsilon$ for every $n > N'$ Let $\sequence {\epsilon_i}_{i \mathop \in \N}$ be a sequence of strictly positive real numbers that satisfies: :$\epsilon_0 = \epsilon$ :$\epsilon_{i + 1} < \epsilon_i$ :$\displaystyle \lim_{i \mathop \to \infty} \epsilon_i = 0$ Since $\sequence {a_n}$ is Cauchy, for each $\epsilon_i$ a natural number $N_i$ exists such that: :$\size {a_n - a_m} < \epsilon_i$ for every $m, n \ge N_i$ Let us study the sequence $\sequence {N_i}_{i \mathop \in \N}$. First, we consider $N_0$. We can choose $N_0 = N$ because $(1)$ $\size {a_n - a_m} < \epsilon$ for every $m, n \ge N$ and $(2)$ $\epsilon = \epsilon_0$. Next, we consider the relation between $N_{i + 1}$ and $N_i$. We have, for $i \in \N$: :$\size {a_n - a_m} < \epsilon_i$ for every $m, n \ge N_i$ and: :$\size {a_n - a_m} < \epsilon_{i + 1}$ for every $m, n \ge N_{i + 1}$ There are two cases for $\size {a_n - a_m}$ when $m, n \ge N_i$. Either: :$\size {a_n - a_m} < \epsilon_{i + 1}$ for every $m, n \ge N_i$ or: :$\size {a_{n'} - a_{m'} } \ge \epsilon_{i + 1}$ for some $m', n' \ge N_i$ In the first case, we can choose $N_{i + 1} = N_i$ since $\size {a_n - a_m} < \epsilon_{i + 1}$ for every $m, n \ge N_i$. In the second case, suppose that $m', n' \ge N_{i + 1}$. This cannot be true since we would then have $\size {a_{n'} - a_{m'} } < \epsilon_{i + 1}$ since $\size {a_n - a_m} < \epsilon_{i + 1}$ for every $m, n \ge N_{i + 1}$. Therefore, at least one of $m', n'$, call it $k'$, must be less than $N_{i + 1}$. This means that: {{begin-eqn}} {{eqn | l = N_i       | o = \le       | r = k' < N_{i + 1} }} {{eqn | ll= \leadsto       | l = N_i       | o = <       | r = N_{i + 1} }} {{end-eqn}} Thus, we have established that there exists a sequence $\sequence {N_i}_{i \mathop \in \N}$ that satisfies: :$N_0 = N$ :$N_{i + 1} \ge N_i$ Since $\sequence {a_n}$ is Cauchy, we have for every $i \in \N$: {{begin-eqn}} {{eqn | l = \size {a_n - a_m}       | o = <       | r = \epsilon_i       | c = for every $m, n \ge N_i$ }} {{eqn | ll= \leadsto       | l = \size {a_n - a_{N_i} }       | o = <       | r = \epsilon_i       | c = for every $n \ge N_i$ by choosing $m = N_i$ }} {{eqn | ll= \leadsto       | l = \size {a_n - a_{N_i} }       | o = <       | r = \epsilon_i       | c = for every $n > N_i$ }} {{eqn | ll= \leadsto       | l = a_{N_i} − \epsilon_i       | o = <       | r = a_n < a_{N_i} + \epsilon_i       | c = for every $n > N_i$ by Negative of Absolute Value: Corollary 1 }} {{end-eqn}} Define a real sequence $\sequence {U_i}_{i \mathop \in \N}$ by: :$U_0 = a_{N_0} + \epsilon_0$ :$U_{i + 1} = \min \set {U_i, a_{N_{i + 1} } + \epsilon_{i + 1} }$ We observe that $U_{i + 1}$ is the minimum of two numbers, one of which is $U_i$. Therefore $\sequence {U_i}$ is decreasing. === $U_i$ is an upper bound for $\sequence {a_n}_{n \mathop > N_i}$ === We prove this by using the Principle of Mathematical Induction. $a_{N_0} + \epsilon_0$ is an upper bound for $\sequence {a_n}_{n \mathop > N_0}$ because $a_n < a_{N_0} + \epsilon_0$ whenever $n > N_0$. Therefore, $U_0$ is an upper bound for $\sequence {a_n}_{n \mathop > N_0}$ since $U_0 = a_{N_0} + \epsilon_0$. This concludes the first induction step. We need to prove that $U_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop > N_{i + 1} }$ if $U_i$ is an upper bound for $\sequence {a_n}_{n \mathop > N_i}$. $U_{i + 1}$ equals either $U_i$ or $a_{N_{i + 1} } + \epsilon_{i + 1}$. Assume that $U_{i + 1} = U_i$. $U_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop > N_i}$ since $U_i$ is an upper bound for $\sequence {a_n}_{n \mathop > N_i}$ by presupposition. We have that $\sequence {a_n}_{n \mathop > N_{i + 1} }$ is a subset of $\sequence {a_n}_{n \mathop > N_i}$ because $N_{i + 1} \ge N_i$. Therefore, $U_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop > N_{i + 1} }$ because $U_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop > N_i}$. Assume that $U_{i + 1} = a_{N_{i + 1} } + \epsilon_{i + 1}$. $U_{i + 1}$ is an upper bound for $\sequence {a_n}_{n \mathop > N_{i + 1} }$ because $a_n < a_{N_{i + 1} } + \epsilon_{i + 1}$ whenever $n > N_{i + 1}$. This concludes the proof that $U_i$ is an upper bound for $\sequence {a_n}_{n \mathop > N_i}$ for every $i \in \N$. {{qed|lemma}} === $\sequence {U_i}$ converges === By Real Cauchy Sequence is Bounded, $\sequence {a_n}$ is bounded. Therefore, $\sequence {a_n}$ has a lower bound $b$. For every $i \in \N$, $\sequence {a_n}_{n \mathop > N_i}$ is a subsequence of $\sequence {a_n}$. Therefore, $b$ is also a lower bound for $\sequence {a_n}_{n \mathop > N_i}$. $U_i$ is an upper bound for $\sequence {a_n}_{n \mathop > N_i}$. Therefore, $b$ is less than or equal to $U_i$ for every $i$. So, $\sequence {U_i}$ is bounded below. Since $\sequence {U_i}$ is decreasing, its first element is an upper bound for $\sequence {U_i}$. Since $\sequence {U_i}$ is bounded below and above, it is bounded. By Monotone Convergence Theorem, a bounded, monotonic sequence converges, and therefore $\sequence {U_i}$ converges. {{qed|lemma}} Now, define a real sequence $\sequence {L_i}_{i \mathop \in \N}$ by: :$L_0 = a_{N_0} - \epsilon_0$ :$L_{i + 1} = \max \set {L_i, a_{N_{i + 1} } - \epsilon_{i + 1} }$ An analysis of $\sequence {L_i}$, not given here because it is similar to the one above of $\sequence {U_i}$, produces the following results: :$\sequence {L_i}$ is increasing :$L_i$ is a lower bound for $\sequence {a_n}_{n \mathop > N_i}$ for every $i \in \N$ :$\sequence {L_i}$ converges === The limits of $\sequence {U_i} $ and $\sequence {L_i}$ as $i \to \infty$ are equal === Since $U_i$ and $L_i$ are, respectively, upper and lower bounds for $\sequence {a_n}_{n \mathop > N_i}$ for every $i \in \N$, we have for all $i \in \N$: {{begin-eqn}} {{eqn | l = 0       | o = \le       | r = U_{i + 1} - L_{i + 1} }} {{eqn | r = \min \set {U_i, a_{N_{i + 1} } + \epsilon_{i + 1} } - \max \set {L_i, a_{N_{i + 1} } - \epsilon_{i + 1} }       | c = Definitions of $U_{i + 1}$ and $L_{i + 1}$ }} {{eqn | o = \le       | r = a_{N_{i + 1} } + \epsilon_{i + 1} - \max \set {L_i, a_{N_{i + 1} } - \epsilon_{i + 1} }       | c = because $\min \set {U_i, a_{N_{i + 1} } + \epsilon_{i + 1} } \le a_{N_{i + 1} } + \epsilon_{i + 1}$ }} {{eqn | o = \le       | r = a_{N_{i + 1} } + \epsilon_{i + 1} - \paren {a_{N_{i + 1} } - \epsilon_{i + 1} }       | c = because $\max \size {L_i, a_{N_{i + 1} } - \epsilon_{i + 1} } \ge a_{N_{i + 1} } - \epsilon_{i + 1}$ }} {{eqn | r = 2 \epsilon_{i + 1} }} {{end-eqn}} which shows that $U_i - L_i \to 0$ as $i \to \infty$ since $\displaystyle \lim_{i \mathop \to \infty} \epsilon_i = 0$. We have: {{begin-eqn}} {{eqn | l = \lim_{i \mathop \to \infty} U_i - \lim_{i \mathop \to \infty} L_i       | r = \lim_{i \mathop \to \infty} (U_i - L_i)       | c = Combined Sum Rule for Real Sequences }} {{eqn | ll= \leadstoandfrom       | l = \lim_{i \mathop \to \infty} U_i - \lim_{i \mathop \to \infty} L_i       | r = 0       | c = as $\displaystyle \lim_{i \mathop \to \infty} (U_i - L_i) = 0$ }} {{eqn | ll= \leadstoandfrom       | l = \lim_{i \mathop \to \infty} U_i       | r = \lim_{i \mathop \to \infty} L_i }} {{end-eqn}} {{qed|lemma}} Let $\displaystyle a := \lim_{i \mathop \to \infty} U_i = \lim_{i \mathop \to \infty} L_i$. We have for every $i \in \N$: {{begin-eqn}} {{eqn | l = L_i       | o = \le       | r = a \le U_i       | c = as $\sequence {L_i}$ is increasing and $\sequence {U_i}$ is decreasing }} {{eqn | ll= \leadstoandfrom       | l = 0       | o = \le       | r = a - L_i \le U_i - L_i }} {{end-eqn}} Also, we have for every $n > N_i$ for every $i \in \N$: {{begin-eqn}} {{eqn | l = L_i       | o = \le       | r = a_n \le U_i       | c = as $L_i$ is a lower bound for $\sequence {a_n}_{n > N_i}$ and $U_i$ is an upper bound for $\sequence {a_n}_{n \mathop > N_i}$ }} {{eqn | ll= \leadstoandfrom       | l = 0       | o = \le       | r = a_n - L_i \le U_i - L_i }} {{end-eqn}} A natural number $j$ exists such that, for every $i \ge j$: {{begin-eqn}} {{eqn | l = \size {U_i - L_i}       | o = <       | r = \dfrac \epsilon 2       | c = as $\displaystyle \lim_{k \mathop \to \infty} (U_k - L_k) = 0$ }} {{end-eqn}} Putting all this together, we find for every $n > N_j$: {{begin-eqn}} {{eqn | l = \size {a_n - a}       | r = \size {a_n - L_j + L_j - a} }} {{eqn | o = \le       | r = \size {a_n - L_j} + \size {L_j - a}       | c = Triangle Inequality for Real Numbers }} {{eqn | o = \le       | r = \size {U_j - L_j} + \size {L_j - a}       | c = as $0 \le a_n - L_j \le U_j - L_j$ }} {{eqn | o = \le       | r = \size {U_j - L_j} + \size {U_j - L_j}       | c = as $0 \le a - L_j \le U_j - L_j$ }} {{eqn | r = 2 \size {U_j - L_j} }} {{eqn | o = <       | r = \epsilon       | c = as $\size {U_j - L_j} < \frac \epsilon 2$ }} {{end-eqn}} which finishes the proof that $\sequence {a_n}$ is convergent. {{qed}}