I saw in an article that for every real number, there exists a Cauchy sequence of rational numbers that converges to that real number. This was stated without proof, so I'm guessing it is a well-known theorem in analysis, but I have never seen this proof. So could someone give a proof (preferably with a source) for this fact? It can use other theorems from analysis, as long as they aren't too obscure.
[Math] Proof that all real numbers have a rational Cauchy sequence
cauchy-sequencesproof-writingrational numbersreal numbersreal-analysis
Related Solutions
The proof that $\mathbb{R}$ does indeed have the Least Upper Bound property really depends upon how you're defining the real numbers; for example, if $\mathbb{R}$ is constructed using Dedekind cuts, then the proof is rather straight-forward and easy. If you're constructing $\mathbb{R}$ using equivalence classes of Cauchy sequences, then it's involved (at least the proofs I've seen/done).
Thus, I'll use this answer to address the concerns of why the Least Upper Bound property needs to be mentioned at all, and why it doesn't trivially hold true for everything.
Firstly, the Least Upper Bound property is essentially the reason calculus can be done; as we shall see, there are "gaps" in the rational numbers. The ability to take limits, which is central to everything done in Real Analysis, is closely related to the Least Upper Bound property. To show why this is the case, I'll quote some equivalences:
- Least Upper Bound Property
- Bolzano-Weierstrass Theorem (all cauchy sequences are convergent) and the Archimedean property
- Monotone Convergence Theorem and the Archimedean property
- Nested Intervals Theorem
All of the above are equivalent, and all are central to Real Analysis.
As far as the Least Upper Bound property not holding more generally in, for example, $\mathbb{Q}$, consider the following set: $$\{ q\in \mathbb{Q} \mid q>0 \wedge q^2<2 \}$$ It is clearly bounded above; 2 is an upper bound, for example. Does this set have a least upper bound? In $\mathbb{R}$ it certainly would, and would be $\sqrt{2}$; since $\mathbb{Q}$ is dense in $\mathbb{R}$, if there were a least upper bound in $\mathbb{Q}$ for this set, then it would also be a least upper bound of the set in $\mathbb{R}$. But we know that $\sqrt{2}$ is irrational, so this cannot be the case. Thus, $\mathbb{Q}$ cannot have the Least Upper Bound property. This is why the Real numbers are necessary.
The main difference is that $\Bbb R$ is complete, while $\Bbb Q$ is not, which means that every Cauchy sequence of real numbers is convergent, while Cauchy sequence of rationals does not need to converge in $\Bbb Q$.
Thus, when you have your problem stated for real Cauchy sequences, it is almost trivial: sequence $(a_n)$ is Cauchy, therefore convergent, and since its subsequence has limit $L$, $(a_n)$ converges to $L$ as well.
On the other hand, if you switch back to rationals, you do not know that $(a_n)$ is convergent a priori. However, here comes the most important part: $\Bbb R$ is metric space completion of $\Bbb Q$ - meaning that every Cauchy sequence in $\Bbb Q$ will converge to some real number and any real number is a limit of Cauchy sequence in $\Bbb Q$.
Of course, the whole story about $\Bbb Q$ and $\Bbb R$ can be completely bypassed (but you specifically asked for the difference) and one can show that Cauchy sequence is convergent if and only if it has a convergent subsequence.
Best Answer
Let $\alpha$ be any real number.
Define $$a_n = \frac{\lfloor n \alpha \rfloor}{n}$$ where $\lfloor , \rfloor $ denotes the floor function.
Then $a_n \in \mathbb Q$. Moreover we have $$\frac{n \alpha -1}{n} \leq a_n \leq \alpha$$ and hence $a_n \to \alpha$.