# Showing that $\mathbb{R},\mathcal{T} = \{]-a,a[ \mid a > 0\} \cup \{\emptyset,\mathbb{R}\}$ is a topology

I have to show that $\mathbb{R},\mathcal{T}$ is a topology, where $\mathcal{T} = \{]-a,a[ \mid a > 0\} \cup \{\emptyset,\mathbb{R}\}$.

However, $$\cup_{n \in \mathbb{N}} ]-2 + \frac{1}{n} , 2 -\frac{1}{n}[ = [-2,2]$$which clearly is not an element in $\mathcal{T}$. Is the exercise faulty? Or am I missing something?

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The union of the open sets $(-2+1/n, 2-1/n)$ is the open interval $(-2,2)$. (sorry, but I cannot bring myself to use the notation $]a,b[$ for open intervals). – Shaun Ault Nov 16 '12 at 2:04
@Shaun: And another +1 for the parenthetical remark! – Brian M. Scott Nov 16 '12 at 2:07

In order to show that $\mathcal T$ really is a topology on $\Bbb R$, you need to prove the following result:
Let $A$ be any non-empty set of positive real numbers. Then $$\bigcup_{a\in A}(-a,a)=\begin{cases} \Bbb R,&\text{if }A\text{ is unbounded}\\ (-u,u),&\text{if }\sup A=u\in\Bbb R\;. \end{cases}$$
Note that in the second case $u$ is not in the union.