Prove that $\mathcal{T}=\left\{\emptyset, \mathbb{R},[n,\infty):n\in \mathbb{N}\right\}$ is a topology on $\mathbb{R}$.
Here is my attempt:
Clearly, $\emptyset, \mathbb{R}\in \mathcal{T}$. For the union closure, it is clear that $\mathbb{R}\cup [n,\infty)=\mathbb{R}\in \mathcal{T}$ and $\emptyset \cup [n,\infty)=[n,\infty)\in \mathcal{T}$ for all $n\in \mathbb{N}$. Next, if there exists a finite collection of intervals $[n,\infty)$ in $\mathcal{T}$, then there exists a minimum $n$ such that the union will be this interval, which is assumed to be in $\mathcal{T}$.
I am having trouble for when the collection of intervals is infinite. Do I use the well-ordering principle for the left endpoints? For example, if $[n_{1},\infty),[n_{2},\infty),...\in \mathcal{T}$, can I guarantee a minimum left endpoint?
Next, the intersection of a finite collection of intervals is clearly in $\mathcal{T}$, since there exists a maximum left endpoint that will cause the intersection to be that interval, which is assumed in $\mathcal{T}$.
Any advice is appreciated. :~)