Is the set of all positive real numbers dense in $\mathbb{R}$ I am working on a problem I found that asks whether the set
$S = \{x \in \mathbb{R} \mid x \ge 0\}$ is dense in $\mathbb{R}$.
The theorem I have been using states the following:
"$S$ is dense in $\mathbb{R} \iff \forall a,b \in \mathbb{R}$ with $a < b$, then $\exists x \in S$ such that $x \in (a,b)$." 
Now my logic from what I have so far is basically that for any $a$ and $b$ you give me, I can find the midpoint between $a$ and $b$, which will consistently give me a positive real number. This feels like I'm just constructing sentences and not really "proving" it, however.
I am a bit confused and would like any words of advice.
 A: In general topology, a subset $S$ of a topological space $X$ is called dense if every point of $X$ either belongs to $S$ or is a limit point of $S$. (A point $x \in X$ is a limit point of $S$ if every open neighbourhood of $x$ contains at least one point of $S$ different to $x$).
The non-negative real numbers are not dense in $\mathbb{R}$, under the metric topology, because, for example, $-1$ does not belong $S$, and $-1$ is not a limit point of $S$. (The open neighbourhood $\left(-\frac{3}{2},-\frac{1}{2}\right)$ contains $-1$ but does not contain any point of $S$.)
A: To give a third point of view:
A subset $D$ of a topological space $X$ is dense in $X$ if and only if $\overline{D} = X$ where $\overline{D}$ denotes the closure of $D$.
We also know that a set $C$ is closed if and only if $\overline{C} = C$. 
But your set $S$ is closed and a proper subset of $\mathbb R$, hence cannot be dense in $\mathbb R$.
A: That set is not dense in $\mathbb{R}$. Try picking an interval which only consists of negative numbers, does this interval contain any elements of $S$? 
Finding such a counter example is enough to prove that $S$ is not dense in $\mathbb{R}$.
