Basic proof that $T$ is a topology. 
Question:
Let
  $T = \{U \subseteq \mathbb{R} : U = \emptyset \text{ or } U = \mathbb{R}\text{ or } U = (−\infty, a) \text{ for some } a \in\mathbb{R}\}$.
Prove that $T$ is a topology on $\mathbb{R}$.

I know the axioms are:


*

*The empty set and $X$ are in $T$.

*The intersection of any finite collection of subsets of $X$ in $T$ is also in $T$.

*The union of any collection of sets in $T$ is also in $T$. 


and can prove the 1st axiom easily but I am struggling to understand how to go about actually showing that the 2nd and 3rd axioms of topologies hold for any example question I attempt, not just this one.
I would appreciate if someone could give me the basics of the process involved in proving axiom 2 and 3 hold.
 A: What are your axioms 2 and 3?
You need to show:
$\ \ \ $1) $X$ and $\emptyset$ are in $T$.
$\ \ \ $2) Any union of elements of $T$ belongs to $T$.
$\ \ \ $3) Any finite intersection of elements of $T$ belongs to $T$

1) holds by the definition of $T$.

For 2):
Let $ \{ U_\alpha\mid \alpha\in I\}$ be a non-empty collection of elements of $T$. You need to verify that $O=\bigcup\limits_{\alpha\in I} U_\alpha\in T$. This is  perhaps best done by considering cases. 
If one of the $U_\alpha$ is $\Bbb R$, so is the union, and then $O\in T$. 
If all $U_\alpha=\emptyset$, then $O=\emptyset\in T$.  
Otherwise, let $\beta=\sup\{\alpha\mid\alpha\in I\}$.  If $\beta=\infty$, you can show $O=\Bbb R\in T$.  If $\beta$ is finite, you can show that $O=(-\infty,\beta)\in T$.

3) is easier to verify: 
For the nontrivial case where no member of the finite collection of sets is empty, given a finite collection $\{(-\infty,\alpha_1),\ldots, (-\infty,\alpha_k)\}$ of elements in $T$
(here $a_k$ is allowed to be infinity), the intersection is $(-\infty, \min\{\alpha_1,\ldots,\alpha_k \})\in T$.
A: Hint: Use the fact that $\mathbb R$ is linearly ordered and order complete to deduce that:


*

*$(-\infty,a)\cap(-\infty,b)=(-\infty,\min\{a,b\})$, and 

*that the union of intervals $(-\infty,a_i)$ for $i\in I$ is $(-\infty,\sup\{a_i\mid i\in I\})$.

A: I guess that Axiom 2 is for you the fact the any union of open sets is an open set, while Axiom 3 is the same for the intersection of any two open sets. Well, in general there is no recipe. In your example, let's take the intersection of two open sets. If one of them is $\emptyset$, the intersection is empty. If one of them is $\mathbb{R}$, the intersection is the other open set. If both have the form $(-\infty,a)$ and $(-\infty,b)$, then their intersection is $(-\infty,\min\{a,b\})$. In any case, the Axiom holds true. The proof for an arbitrary union is slightly more involved, but not impossible.
