Proving that the family of subsets of $\mathbb{R}$ containing Symmetric open intervals form a topology on $\mathbb{R}$.

Question

Proving that the family subsets of $\mathbb{R}$ containing Symmetric open intervals form a topology on $\mathbb{R}$, could anyone clarify this for me especially the step of showing that arbitrary union is in the above set?

Answer 1

Let $\tau_S$ be the symmetric open intervals, I am assuming that $\mathbb{R}$ and $\emptyset$ are included in this.

If $\{U \alpha \}_\alpha \subset \tau_S$, then $U_\alpha = (-\sup U_\alpha, \sup U_\alpha)$ for all indices $\alpha$. Then $\cup_\alpha U_\alpha = \cup_\alpha (-\sup U_\alpha, \sup U_\alpha) = (-\sup_\alpha \sup U_\alpha, \sup_\alpha \sup U_\alpha) \in \tau_S$.

Similarly, if $U_1,...,U_n \in \tau_S$, then $\cap_k U_k = (-\min_k \sup U_k,\min_k \sup U_k) \in \tau_S$.