To prove open sets form topology Question:
Let $X=[0,\infty)$ and
$$
\Omega =  \lbrace \varnothing \rbrace \cup \lbrace (a,\infty) \mid a \geq0 , a \in \mathbb{R} \rbrace
$$
Is $\Omega$ a topological structure?
My Answer:
I know that $\varnothing , X \in \Omega$. For proving arbitrary unions I have if $U_i := (a_i,\infty) \forall i \in I$ and thus $\cup_{i\in I} U_i = (\inf a_i,\infty)$ and thus it follows similarly for intersection it will be $\sup a_i$.
Is this answer correct ? 
How should I prove the statement that$\cup_{i\in I} U_i = (\inf a_i,\infty)$?
 A: Yes, I think your answer is partially correct. First, it is correct to say that $\bigcup_{i \in I} (a_i, \infty) = (\inf_{i \in I} a_i, \infty)$. It is important here that $a \geq 0$, otherwise one could take $\bigcup_{n \geq 1}(-n, \infty) \notin \Omega$ and this would not be a topological structure. 
However, it is not correct to say that $\bigcap_{i \in I} (a_i, \infty) = (\sup_{i \in I} a_i, \infty)$. Counterexample: $\bigcap_{n \geq 1} (1 - \frac{1}{n}, \infty) = [1, \infty)$, which is not an open set above. The set $\Omega$ is closed under finite intersections: $\bigcap_{i = 1}^n (a_i, \infty) = (\max_{i = 1, ..., n} a_i, \infty)$. 
A: First of all, you have a problem in that it is not clear to me why you say $X \in \Omega$. This can easily be fixed by adding $X$ explicitly in the definition, which is something you really should do.
For intersection, remember that you only need finite intersections, so instead of the supremum, you will have the maximum. In fact, $\Omega$ is not closed under arbitrary intersections; for example, let $(b_i)_{i \in \mathbb N}$ be an increasing sequence of real numbers $0 < b_i < 1$ converging to $1$, then $$\bigcap_{i \in \mathbb N} (b_i,\infty) = [1,\infty),$$ since $1 = \sup_{i \in \mathbb N} b_i$ and $1 \in (b_i,\infty)$ for all $i$.
The easiest way to prove $\bigcup_{i \in I} U_i = (\inf a_i,\infty)$ is by chasing elements and using properties of the infimum.
