6
$\begingroup$

What's all the fuss about local bases? I find that everywhere I look people are confused of the notion of a local base, and frankly I am as well, because it seems to me it's equivalent an incredibly simple formulation, but everyone else expounds endlessly on "filters" and "filter bases" and comes up with these elaborate logical statements.

The definition of a "local base" that I've learned is that if $x\in X$ where $(X,\ \tau)$ is a topological space, and $N_x$ is the neighbourhood filter of $x$ (that is, the set of neighbourhoods of $x$), then a local base of $x$ is a subset $B_x\subset N_x$ such that $\forall A\in N_x$ there is some $B\in B_x$ such that $B\subseteq A$.

Alright, so, why don't we take $B_x$ to be the set of open sets containing $x$? That makes perfect sense to me, since every neighbourhood must contain some open set, and so the above logical formulation of a local base is satisfied. There you go, boom. Nothing special. No filters or anything. Clean and simple.

So why do topologists make such a fuss of local bases? There's all this stuff on "first countable" and "second countable" and it seems all the separation axioms of topological spaces are rooted in the idea of bases (some local, some global, from what I can tell). The definition of a locally convex TVS uses local bases. I see tons of people confused over what a local base is.

What makes local bases important? Why are people so confused, myself included over them? Why is such a complicated definition necessary? Is it because the $B_x$ I described isn't the only local base?

$\endgroup$
  • 1
    $\begingroup$ Do you understand the point of a basis for a topology? The point of a local base is pretty similar... $\endgroup$ – Eric Wofsey Apr 5 '17 at 2:51
  • $\begingroup$ We can speak of convergent sequences in $\mathbb R$ because it is a first-countable $T_1$ space. There are spaces for which non-trivial sequences don't converge.. $\endgroup$ – DanielWainfleet Apr 5 '17 at 2:59
3
$\begingroup$

Is it because the $B_x$ I described isn't the only local base?

Basically, yes. The point of talking about local bases is not to know that a local base exists: as you've observed, that is trivial. The point is to know that a nice local base exists. That is, we want to know that there exists a local base consisting of sets with certain nice properties, or a local base which itself has nice properties as a set (e.g., being countable), or a local base that is simply easy to think about. Typically the collection of all open sets containing $x$ won't be especialy nice, but there might be a smaller local base that is much nicer. And such a nicer local base $B_x$ can be used to understand the topology at $x$, since to test whether a set $U$ is a neighborhood of $x$ you just have to test whether $U$ contains some element of $B_x$.

If you look at all the applications you're referring to, none of them are talking about whether there exists an arbitrary local base. All of them are talking about whether there exists an local base that is nice in some special way. For instance, a space is first-countable if there exists a countable local base at every point. This has all sorts of powerful consequences: for instance, in a first-countable space a set $A$ is closed iff it is closed under taking limits of sequences.

$\endgroup$
4
$\begingroup$

Frequently, there are too many open sets. It is nice to be able to verify topological assertions (like continuity, for example) in terms of a smaller subclass which is more manageable.

As an analogy, consider the notion of basis of a vector space. It is quite good that we can, for instance, have a linear map entirely determined by its values on a basis. The concept of basis helps us to represent the whole structure in terms of a smaller, hopefully more manageable subclass. This is the idea.

$\endgroup$
  • $\begingroup$ This solution expresses the idea I was hoping to see expressed. Alright, so, why don't we take $B_x$ to be the set of open sets containing $x$? You could, but the whole point of reducing attention to a base is that they are economical in some way, because you can prove lots of things without thinking about too many open sets. That $B_x$ you (the OP) describe is, relatively speaking, a heavy-handed choice. You can already improve by reducing to connected open sets. (I think! Maybe I'm forgetting something.) $\endgroup$ – rschwieb Apr 5 '17 at 7:52
  • 1
    $\begingroup$ @rschwieb Also, it helps if we have a local base of connected sets (in locally connected spaces) or sets with compact closure (for locally compact Hausdorff spaces), in some arguments. And for ordered spaces we only need to conider open intervals (or half open segments). It simplifies life. Also local bases are natural filter bases for converging to a point, in some arguments. $\endgroup$ – Henno Brandsma Apr 5 '17 at 13:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.