1. **Problem Statement:** Prove that the open interval $(a,b)$ is an open subset of the real numbers $\mathbb{R}$. 2. **Definition of an Open Set:** A subset $U$ of $\mathbb{R}$ is open if for every point $x \in U$, there exists an $\epsilon > 0$ such that the interval $(x-\epsilon, x+\epsilon)$ is contained entirely within $U$. 3. **Goal:** Show that for any $x \in (a,b)$, we can find such an $\epsilon$ so that $(x-\epsilon, x+\epsilon) \subseteq (a,b)$. 4. **Proof:** Let $x \in (a,b)$. Since $x$ is strictly between $a$ and $b$, the distances $x - a$ and $b - x$ are both positive. Define $$\epsilon = \min(x - a, b - x).$$ 5. By this choice, $\epsilon > 0$. Consider the interval $(x - \epsilon, x + \epsilon)$. 6. Since $\epsilon \leq x - a$, we have $x - \epsilon \geq a$. Similarly, since $\epsilon \leq b - x$, we have $x + \epsilon \leq b$. 7. Therefore, $(x - \epsilon, x + \epsilon) \subseteq (a,b)$. This shows that every point $x$ in $(a,b)$ has a neighborhood contained entirely in $(a,b)$. 8. **Conclusion:** By the definition of open sets, $(a,b)$ is an open subset of $\mathbb{R}$.