What I have been given is that all Cauchy Sequences are bounded (proven by the definition of Cauchy sequences), definition of Cauchy sequences, definition of convergent sequences.

Note that all Cauchy Sequences are in real set. Also, I don't yet know about spaces and all, so please don't use those terminologies and all. Also, please don't use subsequences and all, if you want, please specify their meanings and definitions.

Can I prove that Cauchy Sequences converge?

My try (Not at all successful):

Given that, for every $\epsilon>0$, there exists a natural number $N$, such that $$|a_m-a_n|<\epsilon \space\space\forall\space \space n,m\geq N$$

So $|a_n-L|\leq|a_n-a_m|+|a_n-L|\leq\epsilon+|a_m-L|$ where L is any real number.

I think that it can be done after that, but I just cannot get anything ahead.

  • $\begingroup$ Do you know the meaning of $limsup $and $liminf$ of a bounded sequence? $\endgroup$ – dmtri Aug 15 '18 at 9:36
  • $\begingroup$ No. If you use their definitions, I can follow. $\endgroup$ – Aditya Agarwal Aug 15 '18 at 9:36
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    $\begingroup$ The problem with your argument is that $L$ is not defined. $\endgroup$ – uniquesolution Aug 15 '18 at 9:37
  • $\begingroup$ You can find a lengthy but absolutely elementary proof here: proofwiki.org/wiki/… $\endgroup$ – user526015 Aug 15 '18 at 9:37
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    $\begingroup$ @AdityaAgarwal which real number? Any real number? Does it hold for $L$, $2L$, $3L$ $\endgroup$ – uniquesolution Aug 15 '18 at 9:43

The reason why Cauchy sequences converge has less to do with elementary definitions of the limit, than with the topological property of completeness, whose very definition is precisely this: every Cauchy sequence converges. Equipped only with the information that a certain sequence is Cauchy, you cannot prove that it converges without reference to the completeness of the underyling space containing it, because if you could, you could copy the argument to a non-complete metric space, where some Cauchy sequences do not converge.

The fact that every bounded sequence has a convergent subsequence also uses completeness, indirectly.

As your "argument" shows, the main difficulty lies in producing the candidate for the limit. It is simply not part of the information carried by just being a Cauchy sequence. This is because the only candidate for the limit may simply not exist in the underlying space - as in the case of rational numbers with the usual metric. This is the essence of the definition of completeness: filling in the "holes" created by those "non existent" limits of Cauchy sequences. This is in fact how completion of a metric space is constructed.

The proof in wikiproof referred to in the comments is not correct.


In the construction of the reals from the rationals, there is a theorem called the general principle of convergence which states that in the reals that all Cauchy sequences converge.

(General Principle of Convergence) All Cauchy sequences converge.

This can be deduced from your other properties such as L.U.B(Least Upper Bound Property) or Bounded Monotonic Sequence Property.

(Least Upper Bound Property) If every non-empty susbset with an upper bound has a least upper bound.


(Monotonic Sequence Property) Every monotonically increasing/decreasing sequence with an upper bound/lower bound converges.

In fact, it can be shown that the following three are equivalent.

  1. (General Principle of Convergence) plus Archimedean Principle(which states the naturals are unbounded)

  2. (Least Upper Bound Property)

  3. (Monotonic Sequence Property)

Usually, one starts with the Least Upper Bound Property or the Monotonic Sequence Property and deduce the other equivalences.

  • $\begingroup$ How? Please elaborate? $\endgroup$ – Aditya Agarwal Aug 15 '18 at 9:38
  • $\begingroup$ If you don't mind could you just briefly tell me how your lecturer tried to construct the real numbers? (They are all equivalent but it depends on what you take as your starting premise.) $\endgroup$ – daruma Aug 15 '18 at 9:40
  • $\begingroup$ We haven't yet constructed the real numbers. :| $\endgroup$ – Aditya Agarwal Aug 15 '18 at 9:41
  • $\begingroup$ You explicitly said don't use subsequences but the proof does require subsequences. Firstly, you find that the sequence is eventually bounded. You show that there is a convergent subsequence by Bolzano Weierstrass. Then, you show that this limit is acutally the limit of the original sequence. (I will edit it later if you want a fully written out but as I said, it may require things you haven't talked about yet.) $\endgroup$ – daruma Aug 15 '18 at 9:48

Every bounded sequence of real numbers have limit numbers. That is numbers that the terms of sequence can arbitrarily come close to them. This set of limits numbers has minimum and maximum. The sequence has then a limit if these 2 coincidice.

  • $\begingroup$ You may want to clarify "limit numbers". (Clearly, not all bounded sequence have limits such as $0,1,0,1,...$ but yes, they do have limsup and liminf if that is what you mean) $\endgroup$ – daruma Aug 15 '18 at 9:49
  • $\begingroup$ Can you find a number so as infinite many terms of the sequence are close to it as you like? Then this is a limit number of the sequence. $\endgroup$ – dmtri Aug 15 '18 at 9:55

It depends on what you are using as your primary definition of the real numbers - all definitions are equivalent, but proofs like this one are part of demonstrating that this is true. Let's assume you know that every non-empty set of reals which is bounded above/below has a least upper/greatest lower bound.

Let $U$ be the set of real numbers which are greater than an infinite number of elements of the sequence, and $D$ be the set of real numbers which are less than an infinite number of elements of the sequence. Show that $U$ is non-empty and bounded below, hence has a greatest lower bound $L_u$. Similarly $D$ has a least upper bound $L_d$. Now show that $L_u=L_d=L$ the limit you are looking for.


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