Proving a point in $\mathbb{R}$ can be reached from within a dense subset of $\mathbb{R}$ 'from any direction'. This is homework. The problem was also stated this way: 
Let A be a dense subset of $\mathbb{R}$ and let x$\in\mathbb{R}$. Prove that there exists a decreasing sequence $(a_k)$ in A that converges to x.
I know:
A dense in $\mathbb{R}$ $\Rightarrow$ every point in $\mathbb{R}$ is either in A or a limit point of A.
If x is a limit point of A, then there is a sequence in A that converges to x. 
What if $x\in A$?
Also, how can I know if the sequence is increasing or decreasing?
 A: Hint:  You have to construct your sequence so it is decreasing.  Let $a_0$ be a point in $A$ greater than $x$.  How do you know there is one?  Then let $a_1$ be a point in $A$ in $(x,a_0)$.  How do you know there is one?  Then keep going.  You also have to make sure the intervals shrink to zero length.
A: Recall that $A \subset R$ is dense if and only if every open subset of $\mathbb{R}$ contains an element of $A$. 
Let $x \in \mathbb{R}$. We will define a decreasing sequence in $A$ converging to $x$ as follows: Consider the open interval $(x + 2^{-1}, x + 2(2^{-1}))$. Since $A$ is dense, $A$ intersect this interval. Choose $a_1 \in A \cap (x + 2^{-1}, x + 2(2^{-1}))$. By recursion, suppose $a_1, ..., a_n$ has been chosen. Consider the interval $(x + 2^{-n}, x + 2(2^{-n})$. Again since $A$ is dense, there exists a $a_{n + 1} \in A \cap (x + 2^{-n}, x + 2(2^{-n})$. 
The sequence $(a_k)$ constructed in this way is decreasing and converges to $x$. 
A: Hint:  Recursively define  $a_n \in A$ such that $a_n \in (x, a_{n-1})$. 
A: There is a sequence converging to $x$, but you won't know it's decreasing - you'll have to construct it so this happens.
Try thinking about what would happen if this weren't true - then there would be an $a_0 \in A$ so that $(x,a_0) \cap A = \emptyset$... in other words this interval is full of points that are in $\mathbb{R}$, but not in $A$. Would such an $A$ still be dense?
