I have found two different definitions of the topology term "subbasis" and I am not sure why they (apparently) are equivalent:

  1. "A subbasis $S$ for a topology on $X$ is a collection of subsets of $X$ whose union equals $X$"

  2. "$S \subset \tau$ is a subbasis for a topology $\tau$ if $\forall U \in \tau$ and $\forall x \in U$, $\exists \{S_1, \ldots, S_n\} \subset S$ such that $x \in (S_1 \cap \ldots \cap S_n) \subset U$. That is, finite intersections of sets in $S$ form a basis."

These two definitions seem to have nothing to do with each other. The first definition is from "Topology" by Munkres and the second is from lecture notes by my professor. Are these two definitions the same?


The second definition defines what it means for a collection $S$ to be a subbase for a given topology $\tau$ on a set $X.$

The first "definition" or rather characterization answers the question: under what conditions is a given collection $S$ a subbase (in the sense of the second definition) for some topology on the set $X$? The answer is that, if $\bigcup S=X,$ then there is a unique topology on $X$ for which $S$ is a subbase; it is called the topology generated by $S.$


These are not definitions of the same thing:

  • The first definition is for "$S$ is a subbasis"
  • The second definition is for "$S$ is a subbasis for $\tau$"

However, there is an equivalence between them, in the sense that:

  • If "$S$ is a subbasis", then there exists a unique topology $\tau$ such that "$S$ is a subbasis for $\tau$".
  • If "$S$ is a subbasis for $\tau$", then "$S$ is a subbasis"

and you also have

  • For every topology $\tau$, there is an $S$ such that "$S$ is a subbasis for $\tau$"

If you have a specific topology for which you're interested in finding a subbasis for, you want to use the second definition. The first definition, however, lets you recognize what constitutes a subbasis without having to already know the topology it generates.

(disclaimer: I have not gone over the details finely enough to prove the statements above, so I may be in error)


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