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Lemma: Every Cauchy sequence is bounded. Proof: Let $(a_{n})$ be Cauchy. We choose $ 0<\epsilon_{0}$. So $ \forall \; n>m\geq N_{0}$ we have that $\vert a_{n}-a_{m} \vert < \epsilon_{0}$. Therefore $(a_{n})$ is bounded for all $ m \geq N_{0} $ by $ \epsilon_{0} $. Since $ \mathbb{N}_{N_{0}}$ is finite, it is bounded.
Theorem. Every Cauchy sequence in $\R$ is bounded. Proof 1. Let $\sequence {a_n}$ be a Cauchy sequence in $\R$. Then there exists $N \in \N$ such that: $\size {a_m - a_n} < 1$ for all $m, n \ge N$. In particular, by the Triangle Inequality, for all $m \ge N$: So $\sequence {a_n}$ is bounded, as claimed. $\blacksquare$ Proof 2.
I am supposed to prove, that every Cauchy sequence is bounded. From the definition of Cauchy sequence: $\left ( \forall \varepsilon > 0 \right )\left ( \exists n_{0} \in\mathbb{N} \right ) \left ( \forall m,n \in\mathbb{N}:m,n\geq n_{0} \right )d\left ( x_{n},x_{m} \right )< \varepsilon $ .
Every Cauchy sequence is bounded. Once you get far enough in a Cauchy sequence, you might suspect that its terms will start piling up around a certain point because they get closer and closer to each other.
A sequence $\{x_n\}$ in a metric space is said to be bounded if it is contained in some open ball $B(a,r)$. Prove that every Cauchy sequence in a metric space is bounded. Proof . Put $\{x_n\}\in(X,d)$ s.t. $\forall\epsilon>0\ \exists k\in\mathbb{N}\ s.t. \ d(x_n,x_m)<\epsilon$ whenever $n,m\geq k$. Now for $x_n\geq k$ we have $x_n\in{B(x_n ...
9.5 Cauchy =⇒ Convergent [R] Theorem. Every real Cauchy sequence is convergent. Proof. Let the sequence be (a n). By the above, (a n) is bounded. By Bolzano-Weierstrass (a n) has a convergent subsequence (a n k) → l, say. So let ε > 0. Then ∃N 1 such that r > N 1 =⇒ |a nr −l| < ε/2 ∃N 2 such that m,n > N 2 =⇒ |a m −a n| < ε/2 ...
Define the sequence (sn) by s1 = 1 and sn+1 = 1 + sn for n ≥ 1. Prove (sn) converges and find its limit. Idea: To show (sn) converges, show that (sn) is bounded by . and that (sn) is. Then, to find the limit, use the fact that if (sn) is a convergent sequence, then lim sn =. n→∞.