Is it possible to prove that all Cauchy sequences of real numbers converge without using the Bolzano-Weierstrass theorem?

Question

Question: Prove that a sequence of real numbers is convergent if and only if it is a Cauchy sequence.

I'm currently learning real analysis through an inquiry based course, and I'm trying to prove the above statement in the backwards direction. I've already proved that every Cauchy sequence is bounded (using similar logic to this proof), so now I'm trying to see how I can use that information in my proof.

Most proofs that I have seen across the internet use the "Bolzano-Weierstrass theorem," which is something that is not in the text (and it seems like a pretty involved proof), so I'm trying to see if there's another way to complete this proof.

We are allowed to assume that a monotone sequence is convergent iff it is bounded, but the text doesn't say much about monotone sequences, so I'm not sure if that information is helpful or not.

Thanks for any help you can give me in understanding this concept. I'm happy to elaborate where I can.

Answer 1

Let $(x_n)$ be convergent, with limit $x$. It is also a Cauchy sequence by the estimate $|x_p-x_q|\leq|x_p-x|+|x_q-x|$.

For the other direction, suppose $(x_n)$ is Cauchy. Since it's bounded, we can define $x\in\mathbb{R}$ to be the limit superior of the $x_n$. Let's check that the $x_n$ converge to $x$:

Fix $\varepsilon>0$. Let $N\in\mathbb{N}$ be such that $|x_p-x_q|<\varepsilon/2$ for all $p,q>N$. Let $n>N$ be such that $|x_n-x|<\varepsilon/2$. Then, for all $k>n$, we have $|x_k-x|\leq |x_k-x_n|+|x_n-x|<\varepsilon/2+\varepsilon/2=\varepsilon$. Convergence follows.

Answer 2

You can prove it by nesting intervals.

Suppose $(x_n)$ is a Cauchy sequence. Then it is bounded (easy lemma), say it is contained in $[a_0,b_0]$ and set $d=b_0-a_0$.

Since the sequence is Cauchy, there exists $k_1$ so that, for $m,n>k_1$, $|x_m-x_n|<d/4$; in particular, there exists a subinterval $[a_1,b_1]\subseteq[a_0,b_0]$ with $b_1-a_1=d/2$ so that $x_m\in[a_1,b_1]$ for all $m>k_1$.

We can now start a recursive procedure: if we have found a subinterval $[a_r,b_r]$ with $b_r-a_r=d/2^r$ and $k_r$ so that $x_m\in[a_r,b_r]$ for $m>k_r$, then we can find, with the same method as above, $[a_{r+1},b_{r+1}]\subseteq[a_r,b_r]$ and $k_{r+1}$ so that $b_{r+1}-a_{r+1}=d/2^{r+1}$ and, for $m>k_{r+1}$, $x_m\in[a_{r+1},b_{r+1}]$.

Note that the sequence $(a_r)$ is increasing and the sequence $(b_r)$ is decreasing; they converge to the same limit $l$, as $\lim_{n\to\infty}(b_r-a_r)=\lim_{r\to\infty}d/2^r=0$.

Prove that $\lim_{n\to\infty}x_n=l$.

Answer 3

Spivak's Calculus contains a low-tech proof that every real sequence contains a monotone subsequence:

Definition: If $(x_{n})$ is a real sequence, define a peak point to be an index $N$ such that $x_{m} \leq x_{N}$ for all $m > N$.

In words, if $N$ is a peak point of $(x_{n})$, then no subsequent value exceeds $x_{N}$.

Theorem: If $(x_{n})$ is a real sequence, there exists a monotone subsequence $(x_{n_{k}})$.

Proof (sketch): If $(x_{n})$ has infinitely many peak points, it's easy to construct an increasing sequence $(n_{k})$ of peak points. The corresponding subsequence is non-increasing by the definition of a peak point.

If $(x_{n})$ has only finitely many peak points, let $n_{1}$ be an index larger than every peak point. Since $n_{1}$ is not a peak point, there exists an index $n_{2} > n_{1}$ such that $x_{n_{2}} > x_{n_{1}}$. Continue inductively, constructing a strictly increasing subsequence.

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The Bolzano-Weierstrass theorem follows easily, since it's readily proved that a bounded, monotone sequence of reals is convergent.