how do you prove the set of accumulation points of Q is R. 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 $0<|x-s|<y$.
 A: 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.
A: 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.
A: Well, that depends on your definition of the reals: the thing you want to prove is one possible definition. Two other possibilities:


*

*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 $1$.

*Reals defined using Dedekind cuts.

*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.
