$A$ is dense in $\Bbb R$ iff for every $x \in \mathbb R$ there is a sequence in $A$ that converges $x$ Prove that, for all A ⊆ R, the following are equivalent:


*

*$A$ is dense in $\Bbb R$.

*For every $x \in\Bbb R$, there is an infinite sequence $a_n$ with each $a \in A$, such that $\lim_{n\to\infty}a_n=x$.


This is double implication, so I need to prove it both ways. I'm just really struggling with manipulating the property regarding the existence of a sequence in $A$ converging to every $x \in\Bbb R$.
 A: First, assume that $A$ is dense in $\mathbb{R}$. Fix $x\in\mathbb{R}.$ For each natural number $N$, there exists a point in $A\cap B_{1/n}(x)$, where $B_{1/n}(x)$ is the ball of radius $1/n$ around $x$. (Here, the quantity $1/n$ may be playing the role of the $"\epsilon"$ that you may be used to seeing.) Let $p_n$ be any point in $A\cap B_{1/n}(x)$. Then $p_n\to x$ as $n\to\infty$. To formally argue this, you would fix $\epsilon>0,$ and choose $N$ so that $N>1/\epsilon$ by the archimedian property. Then $1/N<\epsilon$. For each $n>N$, the point $p_n$ is contained in a ball of radius $1/n$ around $x$, which is a subset of the ball of radius $1/N$ around $x$. So $|x-p_n|<1/n< 1/N<\epsilon$ for all $n>N$. This proves convergence.
For the converse, fix $x\in \mathbb{R}$. Let $(a_n)$ be a sequence of points in $A$ that converges to $x$. Any open ball around $x$ contains a point $a_n$ of the sequence (in fact all but finitely many points of the sequence), by definition of convergence, and hence every ball around $x$ intersects $A$. Since $x$ was an arbitrary point in $\mathbb{R}$, this proves that $A$ is dense. 
A: hint if $A$ is dense in $\mathbb R$ and $x \in \mathbb R$ then for each $n$ you can find some $x_n \in (x-\frac{1}{n}, x+\frac{1}{n}) \cap A$.
