Verifying that a certain collection of intervals of $\mathbb R$ forms a topology

Solution 1:

With closure under unions, you've only shown a certain (finite) union is contained in $\tau_1$, not that an arbitrary union is (which is what the definition of a topology requires).

Let's say $U_\alpha \in \tau_1$, where $\alpha \in A$ and $A$ is some arbitrary indexing set. Let's look at $$U = \bigcup_{\alpha \in A}U_\alpha$$

What can we say about $U$ if $U_\alpha = \mathbb{R}$ for some $\alpha \in A$? What if no $U_\alpha$ is equal to $\mathbb{R}$? (There are a couple cases: all $U_\alpha = \emptyset$, some $U_\alpha \not= \emptyset$ and the set of all $n$ such that $U_\alpha = (-n,n)$ is bounded above, and some $U_\alpha \not= \emptyset$ and the set of all $n$ such that $U_\alpha = (-n,n)$ is not bounded above).

This same method will work for $\tau_2$, as well.