Vtyjkyuk

Here is part of Munkres' book on topology:

I suppose the reader is supposed to use

to prove this. But I don't understand how. The lemma says that if there is a space with a fixed topology, then to show that a set forms a basis, one has to take an open set from the fixed topology and do something. But the order topology is being defined, how can I take an open set and prove that it contains a basis element if I don't know what open sets are yet?

The requirements for $mathcalBsubseteq mathscrP(X)$ to be a base for some topology on $X$ are (second edition., paragraph 13, page 78):

(1) $bigcup mathcalB = X$, or equivalently for every $x in X$ there is some $B in mathcalB$ such that $x in B$.

(2) For all $B_1, B_2 in mathcalB$, for all $x in B_1 cap B_2$: there exists some $B_3 in mathcalB$ such that $x in B_3 subseteq B_1 cap B_2$.

The lemma 13.2 is not used here: this is to characterise a base for an already given topology, while in the case at hand we want to define a topology by a suitable base.

Munkres suggests to check condition (2) by checking that the set of intervals he describes ($mathcalB$) has the property that if $B_1 , B_2 in mathcalB$ we either have $B_1 cap B_2 in mathcalB$ again (so we can take $B_3 = B_1 cap B_2$) or $emptyset$ (in which case the condition is voidly true as there are no $x in B_1 cap B_2$ to check.

$mathscr B$ doesn't define a topology on $X$. You need to take arbitrary unions and finite intersections of elements of $mathscr B$ to fulfill the definition of a topology.

$mathscr B$ is a basis for the topology it generates...

Actually, $mathscr B$ is closed under finite intersections, as Munkres notes.

But you need to throw in arbitrary unions... $mathscr B$ isn't closed under arbitrary unions.

Thus $mathscr B$ generates a topology on $X$.