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I know that the set of accumulation points for the rational numbers is the real numbers, but I'm not sure how to prove this.

I need to use the definition: $x$ is an accumulation point of $S$ if, for every $y>0$, there exists a point $s$ in $S$ such that $0<|x-s|<y$.

Chappers
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Magen
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    This follows from the density of $\mathbb{Q}$ on $\mathbb{R}$. Do you know that? Or is that what you want to prove? it is usually done by using the Archimedean property – Alonso Delfín May 03 '15 at 01:57
  • Btw, by density of $\mathbb{Q}$ on $\mathbb{R}$ I mean that for every $r_1< r_2 \in \mathbb{R}$ there exist a $ q \in \mathbb{Q}$ such that $$ r_1< q <r_2 $$ – Alonso Delfín May 03 '15 at 02:05

3 Answers3

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Take an interval $\langle a-\varepsilon,a+\varepsilon\rangle$, by Archimedean property, there exists some $ n_{0} \in\mathbb{N}$ such: $$\frac{1}{n_0}<(a+\varepsilon)-(a-\varepsilon)=2\varepsilon$$

Then consider the set $A$ defined by: $$A=\{m\in\mathbb{N}:\frac{m}{n_0}\geq2\varepsilon\}$$ This is notempty because $\lceil 2\varepsilon n_0 \rceil \in A$, and $1\notin A$. For the Well-Order-Principle there exists some $m_0\in A$ the minimun element. Since $m_0\neq 1$ there exists $m_0-1\in\mathbb{N}$ and $m_0-1\notin A$

Is easy prove that $ a-\varepsilon<\dfrac{m_0-1}{n_0}$ and by contradiction prove $\dfrac{m_0-1}{n_0}<a+\varepsilon$ (remember that if we have positive numbers $a,b,c,d$ such $a\leq b\leq c\leq d$ then $d-a\geq c-b$)

So, set $r=\dfrac{m_0-1}{n_0}\in\mathbb{Q}$ and the rest is yours.

L F
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Before proceeding to answer the question it is better to revise the definition of accumulation point in plain English.

A number $x$ is said to be an accumulation point of a non-empty set $A\subseteq\mathbb{R}$ if every neighborhood of $x$ contains at least one member of $A$ which is different from $x$.

A neighborhood of $x$ is any open interval which contains $x$.

In this question we have $A = \mathbb{Q}$ and we need to show if $x$ is any real number then $x$ is an accumulation point of $\mathbb{Q}$. This is almost obvious because if $x$ is any specific real number then any neighborhood $B$ of $x$ contains infinitely many rational numbers (and hence at least one of them is different from $x$ itself).

The fundamental property which we are using here is the following:

If $a < b$ are two real numbers then there is a rational $x$ with $a < x < b$ and an irrational number $y$ with $a < y < b$.

This above fact implies that there are infinitely many rational and irrational numbers between $a$ and $b$. In other words any interval $(a, b)$ contains infinitely many rational and irrational numbers. The neighborhood $B$ in my answer above is an interval of this type and hence contains many rational numbers.

Paramanand Singh
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Well, that depends on your definition of the reals: the thing you want to prove is one possible definition. Two other possibilities:

  1. Reals defined as base-$b$ expansions or continued fractions, with some way of dealing with recurring $(b-1)$s or the last number in the fraction not being $1$.

  2. Reals defined using Dedekind cuts.

  3. Reals defined using the "increasing sequences that are bounded above converge" property.

In all of these cases, it's a matter of inventing a construction that represents the Cauchy sequence you take in terms of the basic reals construction.

Chappers
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