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RTP

Let $C$ be a set of Cauchy sequences. $\forall x \in {\Bbb R}, \exists \{a_n\} \in C$ sucht that ${a_n} \to x$.

I have no clue to even start this problem.

All I know so far is that $\Bbb R$ is a set of equivalent classes of the limit of rational Cauchy sequences.

Can anyone give me a proof or a hint ?

I am new to analysis, so it would be nice if you could tell me the rationale of this problem as well.

hyg17
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4 Answers4

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I’m going to write this answer on the assumption that you’re working through a fairly rigorous construction of the reals; if you’re not, it’s more technical than you actually need.

Let $x\in\Bbb R$; then by the definition with which you’re working, $x$ is an equivalence class of Cauchy sequences of rational numbers. Let $\sigma=\langle p_k:k\in\Bbb N\rangle$ be one of those Cauchy sequences. The whole idea of this construction of the real numbers is that $x$ should be the limit of the sequence $\sigma$. There’s a technical problem, however: the rational numbers $p_k$ aren’t actually in $\Bbb R$ when $\Bbb R$ is viewed as a set of equivalence classes of Cauchy sequences of rationals. However, if you’ve been shown this construction, you should have been shown that this $\Bbb R$ contains a copy of the rationals.

Specifically, for each $k\in\Bbb N$ let $\pi_k$ be the constant sequence $\langle p_k,p_k,p_k,\ldots\rangle$; then $\bar\pi_k$ is the real number that corresponds to the rational $p_k$, where I write $\bar\tau$ for the equivalence class of a sequence $\tau$ of rationals. Thus, we expect that $\langle\bar\pi_k:k\in\Bbb N\rangle$ converges to $x=\bar\sigma$ in $\Bbb R$. To show this, we must show that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N$ such that $\bar d(\bar\pi_k,\bar\sigma)<\epsilon$ for each $k\ge m_\epsilon$, where $\bar d$ is the metric in $\Bbb R$. If you’ve done the rigorous development, you know that

$$\bar d(\bar\pi_k,\bar\sigma)=\lim_{n\to\infty}|p_k-p_n|\;.$$

Since $\sigma$ is a Cauchy sequence, there is an $m_\epsilon\in\Bbb N$ such that $|p_\ell-p_n|<\frac{\epsilon}2$ whenever $\ell,n\ge m_\epsilon$.

  • Now show that $\bar d(\bar\pi_k,\bar\sigma)\le\frac{\epsilon}2<\epsilon$ whenever $k\ge m_\epsilon$ and conclude that $\langle\bar\pi_k:k\in\Bbb N\rangle$ is a sequence of rationals in $\Bbb R$ that converges to $x=\bar\sigma$.
Brian M. Scott
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    Do infinite decimals provide a rigorous construction of the reals? – Mikhail Katz Jun 14 '13 at 08:45
  • @user72694: One could use them as the basis for a rigorous construction, but it would involve a lot of work; off the top of my head I suspect that overall it would be a bit more work than either the construction using Dedekind cuts or the construction using equivalence classes of Cauchy sequences. – Brian M. Scott Jun 14 '13 at 08:49
  • There are textbooks that use specifically the infinite decimal construction. I am not claiming this approach is preferable to the others, but that it is feasible, and in certain cases preferable (as the case in point). My former colleague D. Barlet (the famous complex analyst) preferred this construction over the others. Notice that the idea of representing all numbers by infinite decimals is older than people generally believe. In fact, it is due to Stevin, over 300 years before Cantor, Dedekind, and Weierstrass. I have a paper on the subject; if you drop me an email, I can send you a copy – Mikhail Katz Jun 14 '13 at 08:58
  • OMG I am so intimidated by what you guys are saying. I really appreciate your help, but I feel as though I am not at that level where I can understand you at all. – hyg17 Jun 14 '13 at 18:54
  • @hyg17: What text are you working from? (It’ll be at least half a day before I get back to this, but I will check.) – Brian M. Scott Jun 14 '13 at 20:19
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Consider an infinite decimal expansion of $x$, and truncate it at rank $n$. This gives a sequence $(a_n)$ that tends to $x$.

Mikhail Katz
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  • It was mentioned to me that "Any Goedel-style enumeration can cover only those real numbers that are definable by formulas (in some fixed language). Thus, if in one's set theory, there are uncountably many real numbers, then some of these numbers must be undefinable (by formulas)". My comment is: A real number x cannot therefore, in general, be given - and therefore cannot be truncated. This can only be done for a countable subset of the real numbers, such as Q. – Mikael Jensen Jun 15 '13 at 17:12
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    Every real number has a decimal expansion, whether or not it is definable. If the number is not definable, then its string of decimal digits will not be definable, but it certainly "exists" in the usual mathematical sense. In fact, the reals can be constructed using infinite decimals, see http://en.wikipedia.org/wiki/Construction_of_real_numbers#Stevin.27s_construction – Mikhail Katz Jun 17 '13 at 11:51
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Let $x\in Q$ Then define the sequence of rationals by $x_n=x-\frac{1}{n}$.(Note that $x_n\in R$) Then $x_n\to x$ as $x\to \infty.$

Let $x\in Q^c$.Let $A_n=(x-\frac{1}{n},x+\frac{1}{n})$.There exists $x_n\in Q\cap A_n$ using the claim 1 below. So we have $x_n\to x$ as $n\to \infty$.

Claim 1:Let $p\ne q$ and $p,q\in R$. I will prove that $\exists x\in Q$ with $p<x<q$.

By Archimedean principle $\exists m\in R$ such that $m(q-p)>1\Rightarrow mq-mp>1$

Now as $mq-mp>1$ so $\exists n\in N$ such that $mp<n<mq\Rightarrow p<\frac{n}{m}<q$ and as $\frac{n}{m}\in Q$ so $x=\frac{n}{m}$ satisfies the claim.

Proof of Archimedean property,

Let $a>0$ and $b\in R$ and $S=\{n|n\in N\text{ and }na\le b\}$ We know that as $a>0$ so $na\to \infty $ as $n\to \infty$ so $|S|<\infty$ i.e. $S$ is finite.And as $S$ is finite so $\sup S$ exists. Now by choosing $m\in N, m>\sup S$ we have a natural no. such that $ma>b$ . This proves the archimedean property.

Abhra Abir Kundu
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The real numbers are, or can be, defined using the type limit of limit you mention so there is nothing to prove, except that axiom A implies A.

Mikael Jensen
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