Munkres' Topology

• Aug 4th 2012, 11:41 PM
Diamondlance
Munkres' Topology
Hello,

I'm looking at Munkres' Topology (2nd edition), trying to refresh myself on some of the basics. I'm having an issue with an early Lemma of his (Lemma 13.1), which states:

Let X be a set, and let B be a basis for a topology T on X. Then T equals the collection of all unions of elements of B.

My issue with this is that any topology contains the empty set, but nothing about the definition* of a basis for a topology requires the empty set to be in B. Indeed, Munkres' proof of this Lemma ignores the empty set in T altogether. For this being such a well reputed book, I'm wondering if I'm missing something. Any thoughts?

*Definition of a basis in Munkres:
If X is a set, a basis for a topology on X is a collection B of subsets of X such that:
-For each x in X, there is at least one basis element U in B containing X.
-If x belongs to the intersection of two basis elements U and V, then there is a basis element W containing x such that W is contained in the intersection of U and V.
• Aug 5th 2012, 05:49 AM
Plato
Re: Munkres' Topology
Quote:

Originally Posted by Diamondlance
I'm looking at Munkres' Topology (2nd edition), trying to refresh myself on some of the basics. I'm having an issue with an early Lemma of his (Lemma 13.1), which states: Let X be a set, and let B be a basis for a topology T on X. Then T equals the collection of all unions of elements of B.
My issue with this is that any topology contains the empty set, but nothing about the definition* of a basis for a topology requires the empty set to be in B. Indeed, Munkres' proof of this Lemma ignores the empty set in T altogether. For this being such a well reputed book, I'm wondering if I'm missing something. Any thoughts?

I do not have a copy of that text. That said, there is a notion of an empty union using an empty indexing set.
That may be what the author has in mind. But that is a guess.
• Aug 5th 2012, 06:31 AM
GJA
Re: Munkres' Topology
When I studied topology my professor gave the same explanation Plato just did for this.
• Aug 5th 2012, 09:33 AM
Diamondlance
Re: Munkres' Topology
Okay, I think that makes sense. So basically the Lemma is saying that $\displaystyle T=\bigcup_{A\subset B}\bigcup_{U\in A}U$, where A may well be empty, even though the empty set may not be in B.
• Aug 5th 2012, 12:43 PM
Deveno
Re: Munkres' Topology
don't get lost in the formalism. a base for a topology is just something we can recover the topology by "unioning arbitrary elements of". for example, a base for the standard topology on the real line is the set of all open intervals B = {(a,b): a,b in R}. note that this example includes the empty interval (a,a). you can think of a base as being sort of a generating set for a topology.

note that if we speak of "the topology generated by (a base) B", then whether or not B includes the empty set is largely irrelevant, as every topology includes the empty set.

if B is closed under (finite) intersections, it's clear we automatically get a topology. but B need not be closed under intersections. if not, we have, for any U1,U2 in B:

U1 ∩ U2 = U {Ux in B : Ux ⊂ U1 ∩ U2, x in U1 ∩ U2},

and the definition of a base guarantees the existence of such Ux, for every x in the intersection of U1 and U2, which shows that the intersection of any two elements of B is contained in the set of unions of elements of B.

if U1 and U2 are disjoint, then we indeed have an "empty union" as referred to in the previous posts (because we don't have any x's).