Which of the following collections of subsets of $\mathbb{R}$ form a topology on $\mathbb{R}$? 1) $T_3 =\{ \emptyset, \mathbb{R}, [−a,a] : a \in \mathbb{R},a>0 \}$;
2) $T_4 = \{ \emptyset, \mathbb{R}, [−n,n], (−a,a) : a \in \mathbb{R}, a > 0, n \in \mathbb{N}^{>0} \}$.
I have that $T = \{ \emptyset, \mathbb{R}, (−a,a) : a \in \mathbb{R}, a>0 \}$ is a topology, but I am not sure how to continue with these 2.
To be a topology T must satisfy
(i) X and the empty set, Ø, belong to τ ,
(ii) the union of any (finite or infinite) number of sets in τ belongs to τ, and
(iii) the intersection of any two sets in τ belongs to τ
Thanks

Apologies for the blunder earlier. Hope this is correct. 
Say $A = (0, 10)$ Then, $ \bigcup_{a\in A} [-a, a] = (-10, 10) \not \in T_3$. Therefore, $T_3$ is not closed under arbitrary unions and hence does not form a Topology. 
Now to $T_4$. Consider $\bigcup_{n \in \Bbb N \setminus \{1\} } ( - \frac  2 3 + \frac 1 n, \frac 2 3  - \frac 1 n) $ which is equal to $[- \frac  2 3,  \frac  2 3]$ which is not an element of $T_4$. 
Therefore both sets do not qualify to be a Topology. 
