Basic topological space proof

Question

$T = \{(-n,n):n\in{\rm I\!R}\backslash\{\pm\infty\}\}$

Show that $T$ is not a topology on ${\rm I\!R}$.

The question was that (-n,n) is bounded. My professor said both (1) "$\phi,X\in T$" and (2) "The arbitrary union is in $T$" fail. I found (1) since (-n,n) is bounded. But I don't know why (2) fails. I thought the arbitrary union of intervals of the form (-n,n) is the largest interval in $T$.

Answer 1

Let $G_n = (-n,n), n\in \mathbf Z^+$. Then $$\bigcup_{n=1}^{\infty}{G_n}=(-\infty, \infty)\notin T$$

That is, $T$ is not closed under arbitrary unions.

Answer 2

$\varnothing$ is actually in $T$ because the set $$(-0,0) = \{x \in \Bbb{R} : 0 < x < 0\}$$ is clearly empty. Still, $X = \Bbb{R}$ is not in $T$ because $\Bbb{R}$ is not bounded, while such is every set of $T$.

$T$ is also not closed for arbitrary infinite unions: $$ A = \bigcup_{n \in \Bbb{N}} (-n, n) \not\in T. $$ If you added $\Bbb{R}$ to $T$, then that would turn it into a topology.