Is $\mathcal{T}$ a topology on $\mathbb{R}$? Let $a \in \mathbb{R}$, we have $V_a=(a,+\infty)$, $F_a=[a,+\infty)$.
1.Prove that $\mathcal{T}=\left\lbrace \emptyset,\mathbb{R},V_a:a\in\mathbb{R}\right\rbrace$ is a topology over $\mathbb{R}$.
2.Is $\mathcal{T'}=\left\lbrace \mathbb{\emptyset}, \mathbb{R}, F_a:a \in \mathbb{R}\right\rbrace$ is a topology on $\mathbb{R}$ ?
For question 1, I have already proved that $\emptyset,\mathbb{R} \in \mathcal{T}$ and if $G_1,G_2 \in \mathcal{T} $ then $G_1 \cap G_2 \in \mathcal{T}$ but I stuck at the codition about the union because it has so many cases which made me lost.
For question 2, I prove it is a topo the same as question 1, but I think it's wrong, I think it is not a topology ? 
Can you help me with 2 these questions please ? 
 A: For the unions, let $A \subseteq \mathbb{R}$ nonempty. We have
$$\bigcup_{a \in A} V_a = \begin{cases} (\inf A, +\infty), &\text{ if $A$ is bounded from below}\\
\mathbb{R}, &\text{ otherwise} \end{cases}$$
so $\bigcup_{a \in A} V_a \in \mathcal{T}$. If we add $\emptyset$ to this union it remains the same, and if we add $\mathbb{R}$ the result is $\mathbb{R}$. In any case, $\mathcal{T}$ is closed under unions.
$\mathcal{T'}$ is not a topology since
$$\bigcup_{n \in \mathbb{N}} F_{1/n} = \bigcup_{n\in\mathbb{N}}\left[\frac1n,+\infty\right) = (0,+\infty) \notin \mathcal{T}'$$
so $\mathcal{T}'$ is not closed under unions.
A: For question $1$ assume we have a non empty set $I\subseteq\mathbb{R}$. We want to prove that $\cup_{a\in I} V_a\in\mathcal{T}$. Let $b=\inf I$. If $b=-\infty$ then $\cup_{a\in I} V_a=\mathbb{R}\in\mathcal{T}$. If $b>-\infty$ then you can check that $\cup_{a\in I} V_a=V_b\in\mathcal{T}$. 
As for question $2$-it is not a topology.Take the set $I=\{\frac{1}{n}:n\in\mathbb{N}\}$. Then $\cup_{a\in I}F_a=(0,\infty)\notin\mathcal{T'}$. 
A: Crutch:  In the "usual" metric topology the sets $(a,\infty)$ are open so its reasonable to think this will be compatible with being open.  In the "usual" $[a,\infty)$ are closed so that's a big red flag this will never work.
But... crutches are bad; dont use them.
But you can use them as motivation for insight.
The reason any union of open intervals fits into the topology of disjoint union of open intervals is because if any $(a_i, b_i)$ intervals overlap their union is $(\inf a_i, \sup b_i)$ an open interval.
For $V_a$ it's even easier.  Let $E\subset \mathbb R$.  If $x \le \inf E$ then there is no $a\in E$ so that $a < x$ so $x \in (a,\infty)=V_a$ for any $a\in E$ so $x \not \in \cup_{e\in E} V_e$.  
And if $x > \inf E$ there is an $a\in E$ so that $a < x$ and $x\in(a,\infty)=V_a \subset \cup_{e\in E}V_e$.  So $\cup_{e\in E}V_e = (\inf E,\infty) \in \mathcal {T}$.
But what if $E$ isn't bounded below?  Well, then for all $x \in \mathbb R$ there is an $a\in E$ so that $a <x$ and $x \in V_a\subset\cup_{e\in E}V_e$ and so $\cup_{e\in E}V_e =\mathbb R \in \mathcal{T}$.
And the reason we can't do the same for closed intervals is because it is possible a set of closed intervals $[a_i, b_i]$ will be such that $\inf a_i$ exists (or $\sup b_i$) but isn't one of the $a_i$ and so $\inf a_i \not \in \cup [a_i, b_i]$ and the union will have an "open" not "closed" endpoint.  
The same is true for unions of $F_a$.  If $E$ is a set that is bounded below but $\inf E \not \in E$ then for any $x \le \inf E$ we have $x$ is not in any $F_a$.  But if $x > \inf E$ there is an $a< x$ so $x \in F_a$.  So $\cup_{e\in E}F_e = (\inf E, \infty) \not \in \mathcal {T'}$.
So $\mathcal{T'}$ is not a topology.
Similar logic for intersections.  
The reason we need a finite intersection of open intervals is that if you have an infinite intersections of $(a_i, b_i)$ and all $a_i < \sup a_i$ then the $\sup a_i$ is in every interval but nothing smaller is.  So $\sup a_i$ would be a "closed" endpoint. 
But if we have finite intersection then the $a_i$ are finite and a max exist and it is one of the $a_i$ and we have an open interval.
So if $E$ is a finite set then $\max E$ exist and $\max E \in E$.  If $x \le \max E$ then $x \not \in (\max E, \infty) =V_{\max E}$ so $x \not \in \cap_{e\in E}V_e$.  But if $x > \max E$ then $x > e$ for all $e \in E$ and $x \in (e,\infty)$.  So $\cap_{e\in E}V_e = (\max E,\infty)\in \mathcal {T}$.
So $\mathcal{T}$ is a topology.
(We can note, but we don't have to, that if $E$ is infinite and bounded above by $\sup E$ and $\sup E \not \in E$ then we'd have $\cap_{e\in E}V_e = [\max E,\infty)\not \in \mathcal {T}$.  But infinite intersections don't count.
We've already shown $\mathcal{T'}$ is not a topology.  But to hone our skills we can try to calculate what $\cap_{e\in E} F_e$ are.
It actually doesn't matter whether $E$ is finite or not.
If $E$ is bounded above then for any $x \ge \sup E$ we have $x \in [e\infty)$ for all $e \in E$ so $x \in \cap_{e\in E} F_e$.  If $x < \sup E$ then there is an $a > x$ and $x \not \in F_a$ so $x \not \in \cap_{e\in E} F_e$.  So $\cap_{e\in E} F_e = [\sup E, \infty)\in \mathcal {T'}$.
If $E$ is not bounded above then for any $x$ there is an $a \in E$ so that $a > x$ and $x \in \in F_a$  so $\cap_{e\in E} F_e = \emptyset \in \mathcal{T'}$.
So $\mathcal{T'}$ would satisfy the intersection condition.  But it already failed the union condition.
