Topology question open sets A subset of $\mathbb{R}$ is open if it is either $\mathbb{R}$, $\emptyset$, or an interval of the form $(a,\infty)$ where $a$ is any real number.
How do I show that this choice of open sets defines a topology on $\mathbb{R}$?
 A: Call $\tau\subseteq P(\mathbb{R})$ the collection of sets as you said. Clearly $\mathbb{R}$ and $\emptyset\in \tau$. Now consider an arbitrary union $\bigcup_{T\in I} T$ where each $T\in I$ belongs to $\tau$. If one of the $T$'s is $\mathbb{R}$, this union is $\mathbb{R}$ and thus the union belongs to $\tau$. Of one of the $T$'s is $\emptyset$, we can simply drop them from the union. Hence we may assume that each $T\in I$ is of the form $(a_T,\infty)$.
Note that $\bigcup_{T\in I}(a_T,\infty)=(a,\infty)$ where $a:=\inf_{T\in I}a_T$. This shows that arbitrary unions of things in $\tau$ again belongs to $\tau$.
It remains to show that finite intersection of elements in $\tau$ belongs to $\tau$. I'll leave that case to you.
A: There are three axioms that must be met for a collection of subsets to form a topology on $\mathbb{R}$.
The first one is that $\emptyset, \mathbb{R}$ must be contained in the collection of subsets. Looks like you have that one covered.

The second axiom is that given an arbitrary collection of open sets, their union must also be open. Let $\{(a_i,\infty)\}_{i \in I}$ be a collection of open sets indexed by some set $I$ (note that I may be uncountable for all we know!).
There are two cases to consider:
case 1: The set $\{a_i\}_{i \in I}$ has no lower bound.
Thus, for every $n \in \mathbb{N}$, $\exists i \in I$ s.t. $n < a_i$, and therefore $\bigcup_{i \in I} (a_i,\infty) = \mathbb{R}$.
case 2: The set $\{a_i\}_{i \in I}$ has a lower bound.
In this case, let $\tilde{a} = inf_{i \in I} a_i$.
Then we have $\bigcup_{i \in I} (a_i,\infty) = (\tilde{a},\infty)$.

The third axiom is given any FINITE collection of subsets, their intersection is an open. So, let $\{(a_i,\infty)\}_{i \in I}$ be a collection of subsets where the cardinality of $I$ is finite.
Since this collection is finite, let $\tilde{a} = max\{a_i\}_{i \in I}$
Then we can see that $\bigcap_{i \in I} (a_i,\infty) = (\tilde{a},\infty)$.

Thus, the three axioms for a collection of subsets to be a topology have been met!!!
