Subbasis for an order topology

algebraic-topologygeneral-topology

I'm a little confused about how Munkres' defines subbasis for a topology. It says

The topology generated by the subbasis $\mathcal{S}$ is defined to be the collection $\mathcal{T}$ of all unions of finite intersections of elements of $\mathcal{S}$

So given a topology and a collection of sets, to show that the given collection is a subbasis for the topology, isn't it sufficient to show that the collection formed by taking finite intersections of elements of the given collection forms a basis for the topology?

But while proving that for an order topology, the subbasis is the collection of all open rays, Munkres does the following:

Because the open rays are open in the order topology, the topology they generate is contained in the order topology. On the other hand, every basis element for
the order topology equals a finite intersection of open rays; the interval $(a, b)$ equals
the intersection of $(−\infty, b)$ and $(a, +\infty)$, while $[a_0, b)$ and $(a, b_0]$, if they exist, are
themselves open rays. Hence the topology generated by the open rays contains the
order topology.

I do not understand why the two-way proof is required. Isn't just showing that any possible finite intersections of the open rays will be basis elements for the order topology enough for the proof?

Best Answer

Munkres's argues in two steps:

  1. Because the open rays are open in the order topology, the topology they generate is contained in the order topology.

  2. Every basis element for the order topology equals a finite intersection of open rays; the interval $(a, b)$ equals the intersection of $(−\infty, b)$ and $(a, +\infty)$, while $[a_0, b)$ and $(a, b_0]$, if they exist, are themselves open rays. Hence the topology generated by the open rays contains the order topology.

You are right, one could alternatively argue that the set $\mathcal I$ of all finite intersections of open rays agrees with the set $\mathcal B$ of intervals used as a basis for the order topology.

In step 2 Munkres proved $\mathcal B \subset \mathcal I$. But what about $\mathcal I \subset \mathcal B$? Either you brush it aside as "trivial" or you need a formal argument. This involves induction on the number of elements in the intersection and to consider a number of cases. I don't think this is easier than step 1.

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