Im picking up from this question Verifying that a certain collection of intervals of $\mathbb R$ forms a topology We need to prove that "Let ℝ be the set of all real numbers. Prove that each of the following collections of subsets of ℝ is a topology.
a. τ1 consists of ℝ , ∅ and every interval (−n, n) for n any positive integer
Assuming the following
$$ 1)\ A_n = (-n,n) \in \tau_1, \ \forall n \in \Bbb N\\ 2)\ A_n \lt A_{n+1}, \ \forall n \in \Bbb N\\ $$
Two cases for intersection and union of finite sets have already been proven in the reference above. For the infinite case i.e. When $n$ is not bounded above by the Archimedean Property $\bigcup_{n=1}^\infty (-n, n) = \Bbb R \in \tau_1 \\$ hence proved AND We can also prove that the $\bigcup_{n=1}^\infty (-n, n)$ is not bounded above by the Boundedness of Convergent Sequence Theorem
I've tried mailing Dr. Morris and dont have a facebook account. Can someone please tell me if I've been thinking about this correctly?
Also please let me know if the formatting of the text doesnt seem right.
Let $\emptyset \ne S\subset \Bbb Z^+$ where $S$ has no largest member. Then for every $n\in \Bbb Z^+$ there exists $m\in S$ with $n<m.$ For any $r\in \Bbb R$ there exists $n\in \Bbb Z^+$ with $|r|<n$ and hence $r\in (-n,n),$ and there exists $m\in S$ with $n<m,$ so $r\in (-n,n)\subset (-m,m)\subset \cup_{m'\in S}(-m',m').$ Therefore $\Bbb R=\cup_{m'\in S}(-m',m').$