Consider a set $S \subseteq \mathbb{R}$. I have learned that $S$ is an open set if
$(\forall a \in S)(\exists \epsilon \in \mathbb{R}^+)(\forall b \in \mathbb{R})(|a - b| < \epsilon \implies b \in S)$
That is, $S$ is an open set if for every point $a$ in $S$, you can choose some positive real $\epsilon$ such that all of $(a - \epsilon, a + \epsilon) \in S$.
Now, I want to show that if $S \subseteq \mathbb{R}$ is a open set, and we have the points $x, y \in S$, then there exists a point $z \in S$ where $z$ is strictly between $x$ and $y$.
Here is some work I've done so far:
- Without loss of generality, let $x < y$. Then we want to show that there exists some $z \in S$ where $x < z < y$.
- Because $S$ is given to be an open set, we know that there exists some $\epsilon \in \mathbb{R}^+$ such that $(x - \epsilon, x + \epsilon) \in S$.
- Here is where my logic gets fuzzy. I think that if there is a "gap" after $x$, then no positive real $\epsilon$ will satisfy the requirement for $S$ to be an open set, because the real numbers in $S$ would be "cut off" at $x$. Thus, I think $S$ must also include the range $(x, k)$ for some $k > x$. This would show that there exists some point between $x$ and $y$, as desired.
I'm not sure if this is valid (and if it is, I need help in formalizing my argument). Can someone provide any hints or insights to prove this basic property of open sets?