Basis for a topology on $\mathbb{R}$ Let $B = \{[a, b] \mid \forall\, a, b \in \mathbb{R}, a < b\}$. Then $B$ a basis for some topology on $\mathbb{R}$.
Is it true that the set of all closed subsets in $\mathbb{R}$ generate a basis for some topology on $\mathbb{R}$? I know from previous assignments that the lower limit topology with half open intervals generates a topology but I doubt that the closed intervals would do the same. However I am not sure how to prove this.
 A: As you have defined it, $B$ does not form a basis for a topology on $\mathbb{R}$.  For $a<b<c$, we have $[a,b],[b,c]\in B$, but $\{b\}=[a,b]\cap[b,c]$, and no basis element both contains $b$ and belongs to $\{b\}$
A: Let $X$ be a set and $\mathcal B$ some family of subsets of $X$. Then there exists unique topology with basis $\mathcal B$ if and only if:
$(1)$ $\bigcup\mathcal B = X$
$(2)$ $(\forall B_1,B_2\in\mathcal B)(\forall x\in B_1\cap B_2)(\exists C_x\in\mathcal B)\ x\in C_x\ \wedge\ C\subseteq B_1\cap B_2$
Proof. One direction is trivial since any open set is union of elements of basis. Assume that $X$ is a set and $\mathcal B$ is family of subsets of $X$ satisfying $(1)$ and $(2)$. Define $\tau = \{ \bigcup S\mid S\subseteq \mathcal B\}$. Then, obviously $\emptyset$ and $X$ are in $\tau$ and $\tau$ is closed under arbitrary unions. To see that it is closed under finite intersections, it is enough to prove that for any $B_1, B_2\in\mathcal B$ we have $B_1\cap B_2\in\tau$. But this is a direct consequence of $(2)$, since $B_1\cap B_2=\bigcup_{x\in B_1\cap B_2} C_x$. I will leave the uniqueness to you (either show that two topologies with the same basis are the same or that $\tau$ defined above is the intersection of all topologies on $X$ containing $\mathcal B$).$\tag*{$\square$}$
Your family $B$ satisfies $(1)$, but not $(2)$. Counterexample is already given in the answer by Aweygan, and I will not repeat it. I will just add that any counterexample is of that form, since the other cases are $[a,b]\cap[c,d] = [c,b]$ and $[a,b]\cap[c,d] = \emptyset$ neither of which contradicts $(2)$.
What this means is that all you need to change in the definition of $B$ is "$a < b$" to "$a\leq b$" to get basis for topology. But that would mean that $B$ contains all singletons and thus topology generated by $B$ would be discrete topology on $X$, as noted by Alex Kruckman in the comments (actually, one only needs singletons to generate discrete topology, other segements are superflous).
