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.

The Bolzano-Weierstrass theorem is a specific case of the equivalence of compactness and sequential compactness in metric spaces, and it is very important that you understand this concept. — Math1000, Mar 03 '17 at 22:14
@Crostul because this is an inquiry-based learning course, the text does not prove the theorem. — Ephraim, Mar 04 '17 at 16:06
The set of rationals has the property: (A)"every Cauchy sequence is bounded" but does not have the property (B)"every Cauchy sequence converges". So you will need more than just (A) to prove (B). — GEdgar, Apr 12 '17 at 23:49
@GEdgar isn't that why we would use the completeness axiom for the real numbers to show that there is a least upper bound? — Ephraim, Apr 13 '17 at 03:02

score 8 · Accepted Answer · answered Mar 03 '17 at 22:16

8

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.

answered Mar 03 '17 at 22:16

user281392

685

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I largely doubt the OP knows what "limit superior" means. Anyway, good job. – Crostul Mar 03 '17 at 22:17
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In that case, I'd recommend the wikipedia article https://en.wikipedia.org/wiki/Limit_superior_and_limit_inferior which seems to be well-written. – user281392 Mar 03 '17 at 22:18
@user281392 thank you for the proof and for the article. The logic is sound, and I'll look into what a "limit superior" is (using your article). – Ephraim Mar 04 '17 at 16:04
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@Ephraim One could probably avoid using the words "superior (or inferior) limit". An infinite bounded set in $\mathbb{R}$ has a limit point. If one of the $x_n$ appears infinitely often then you can show that it is the limit; otherwise ${x_n}$ is infinite and choosing $x$ to a limit point of the set ${x_n}$ works. Using $\limsup$ just makes the proof shorter. – user281392 Mar 04 '17 at 16:14
Sorry to comment on an old answer, but I love this proof, but I wanted to be clear, does the second bound $|x_n-x|<\epsilon/2$ come from the fact the sequence is Cauchy so the limit superior is a term of the sequence hence for indices past a fixed point it is arbitrarily close to this term in the sequence? I just wasn’t sure how to justify this. – Rio Alvarado Oct 18 '19 at 22:33

score 6 · Answer 2 · answered Mar 03 '17 at 22:24

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

score 3 · Answer 3 · answered Mar 03 '17 at 23:42

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.

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

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

3 Answers3