Considering the following proof and its converse: If a sequence converges, then it is Cauchy. That is, if $\lim_{n\to \infty}a_{n} = L$, then given $m>N$, we have that $|a_{m}-a_{n}| < \epsilon$

I tried using the $\epsilon-N$ definition for $\lim_{n\to \infty}a_{n} = L$ and that if something is less than $\epsilon$, then it must be less than $\left.\frac\epsilon2\right.\rbrace$

I am also stuck proving the converse: If a sequence is Cauchy, then it converges.

Hints, please?

  • $\begingroup$ For the converse, you can find an upper bound for the sequence and then use the least upper bound property of $\mathbb R$. $\endgroup$ – Nishant Apr 8 '14 at 5:01
  • $\begingroup$ Proving the converse is significantly more difficult. You can bound a Cauchy sequence fairly easily though, and then applying the Bolzano-Weierstrass Theorem allows you to reverse engineer a point of convergence. $\endgroup$ – CunningTF Apr 8 '14 at 10:28
  • $\begingroup$ The converse direction is, by definition, the claim that $\Bbb R$ is a topologically complete space. This is a non-trivial property of any metric space, and for the real numbers equivalent to any of the completeness axioms, usually order completeness. The proposition that order completeness implies topological completeness is the Bolzano-Weierstraß theorem. $\endgroup$ – LutzL Apr 8 '14 at 13:50
  • $\begingroup$ I asked my professor about the converse and he said it is sufficient to show that every Cauchy sequence is bounded. Is this because a bounded sequence must converge eventually? $\endgroup$ – user3175426 Apr 8 '14 at 15:08

Suppose $(x_n)$ converges, say $x_n \to L $. Let $\epsilon > 0 $ be given. By definition, we can find $N \in \mathbb{N}$ such that if $n \geq N$, then

$$ | x_n - L | < \frac{ \epsilon}{2} $$

Similarly, take $M \in \mathbb{N}$ such that if $m \geq M$, then

$$ |x_m - L | < \frac{ \epsilon }{2} $$

Let $K = \max\{ N,M \} $. Then, for all $n,m \geq K$, we have (using triangle inequality)

$$ |x_n - x_m | \leq |x_n - L | + |L - x_m| < \frac{ \epsilon}{2} + \frac{ \epsilon}{2} = \epsilon $$

Hence, by definition, $(x_n)$ is a cauchy sequence.

  • 2
    $\begingroup$ Why would you look for $M$ ? It has the same role as $N$... $\endgroup$ – Gabriel Romon Apr 8 '14 at 5:09

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