Dense set in $(0,\infty)$ If 0 is a limit point of a subset $A$ of $(0,\infty)$, then prove that the set of all $x$ in $(0,\infty)$ that can be expressed as a sum of (not necessarily distinct) elements of $A$ is dense in $(0,\infty)$. 
I must prove that for every positive real number $x$, there's a sequence in $A$ converging to $x$. Help me with the convergence part please. 
 A: First of all, I think you've got wrong what you have to prove, correct me if my interpretation is wrong. If the first part is correct, then what you have to prove is that for every positive real number $x$, there's a sequence in the set of sums of elements of $A$, converging to $x$. Not for $A$, $A$ is not the dense set here, it's the set of the sums of elements of $A$.
I would prove that by contradiction. Let's call $B$ this set of the sums of not necessarily distinct elements from $A$. Let's assume $B$ is not dense, that means there is a point $x\in(0, \infty)$ with no sequence from $B$ converging to $x$, this implies that there exists $\epsilon > 0$ such that $(x-\epsilon, x+\epsilon) \cap B = \emptyset$.
Now, for the intuition of the proof, the main problem is that if you take each point $y\in A$ and you take the set $ Z = \{ y\cdot n \hspace{.2cm} | n\in \mathbb{N} \}$ we know that $Z$ is a subset of $B$ (because it's just the sum of $n$ times of $y$), and you'll notice that every interval in $(0, \infty)$ with length larger than $y$ will have non-empty intersection with $Z$ (if you have problems visualizing this fact, I recommend you to draw this $Z$ set for $y=1$ and for $y=0.1$). The proof of this is just constructing the element of $Z$ in an interval $(a,b)$, with $a, b > 0$, assuming that $b-a > y$. This element could be $b-y = (b/y-1)\cdot y$ if $b/y$ is a natural, or $\text{floor}(b/y) \cdot y$ if $b/y$ it's not a natural number. In both cases we have that the element is less than $b$ and greater than $a$, and it's in $Z$ because of definition of $Z$.
There is a sequence from $A$ convergent to $0$ so there exists $y\in A$ with $0 < y < \epsilon$. Take $Z = \{ y\cdot n \hspace{.2cm} | n\in \mathbb{N} \}$, $Z$ has non-empty intersection with $(x-\epsilon, x+\epsilon)$ because this interval has length $2\epsilon > y$. But that means there is a point in $(x-\epsilon, x+\epsilon) \cap Z$ which contradicts our first assumption that $(x-\epsilon, x+\epsilon) \cap B = \emptyset$.
We got a contradiction, which means that $B$ was dense.
A: The basic idea behind the proof should be that one can get arbitrarily close to any positive real $x$ because if $0$ is a limit point of $A$ then there are arbitrarily small positive reals in $A$. For instance, if $$A = \{10^{-n} \mid n \in \mathbb{Z}_{\geq 0} \}$$ then it's clear any positive real $x$ can be approximated via its decimal expansion. 
Let $x$ be given. Consider the open balls $B_0(10^{-n})$, for $n \in \mathbb{Z}_{\geq 0}$. (I say open balls, but of course as subsets of $\mathbb{R}_{>0}$ the negative elements are omitted.) By virtue of $0$ being a limit point of $A$, there is $y_n \in B_0(10^{-n}) \cap A$. Define the sequence 
$$x_n = \left\lfloor \frac{x}{y_n} \right\rfloor y_n.$$
Note $x_n$ is a sum of not necessarily distinct elements of $A$; it's $y_n$ summed $\lfloor x_n/y_n \rfloor$ times. One way to proceed is to show if $y \rightarrow 0$ then $\lfloor x/y \rfloor \cdot y \rightarrow x$. (We are done, as $y_n \rightarrow 0$ as $n \rightarrow \infty$.) Alternatively, note $$|x_n - x| < y_n < 10^{-n} \xrightarrow{n \to \infty} 0.$$ 
EDIT: I should add you can change this argument slightly; it might be cleaner to argue for $\epsilon > 0$ arbitrarily small that there exists a sum of not necessarily distinct elements of $A$, say $y$, such that $y \in B_x(\epsilon)$. (This of course says $x$ is a limit point of the set of sum of not necessarily distinct elements of $A$.) Then take $y = \lfloor x/z \rfloor \cdot z$ where $z \in B_0(\epsilon) \cap A$. The previous proof is basically this with $\epsilon = \epsilon_n = 10^{-n}$ rather than $\epsilon$ arbitrarily small.  
