Like many mathematical properties, paracompactness this is a notion of smallness. It's not about the smallness of a subset U of X, but smallness of a collection C of subsets U of X.
The definiton for a subset C of T to be locally finite says that you may consider a well choosen sample of neighborhoods (sets V∈T) and C ought to be finite with respect to that sample (i.e. finite pro V). Pic related for a concise definition.
A topologal space is paracompact if it has a cover with that local finiteness property.
The sample of V's above may be very big, so C is really only small w.r.t. the sample. In a compact space, on the other hand, the cover itself is finite (and you don't need to consider that sample).
Note that the name locally compact is already used for the situation where every point x∈X has a compact neighborhood V.
Question: What can we say, in general, about the notions of paracompact vs. locally compact ??
You talk about locally compact but your topic and the highlighted text in your pic talk about local finiteness. Which one is the property you are interested about?
John Adams
Okay, I switched up the title.
I'm interested in 1. paracompact (where the definition involves local finiteness) vs. 2. locally compact.
E.g. you you consdier the classes of paracompact spaces and the class of locally compact spaces, who do they relate to each other, which is the stronger requirement, what is to add to make on imply the other (my main interested in the quesiton)? And finally, in which cases are they the same?
Brandon Diaz
Neither condition implies the other I think. Metric implies paracompact. There are metric spaces which are not locally compact, example any infinite dimensional banach space. A locally euclidean hausdorff space is metric iff it is paracompact iff the connected components are second countable. So a locally euclidean hausdorff connected space which is not normal (example in Bredon's topology) is not paracompact. Locally compact hausdorff sigma-compact implies paracompact, though.
David Richardson
Also, I think you messed the definition, paracompact is that *any* open cover has a locally finite refinement.
Jacob Jackson
This is right. A second-countable Hausdorff space is paracompact if it's locally compact, too.
This too!
Does anyone know what it takes for a paracompact space to be locally compact? I tried googling it, but got nothing.
Joshua Roberts
>Does anyone know what it takes for a paracompact space to be locally compact? Don't think there is anything. In general locally compact 'looks' weaker, the counterexample is fairly delicate and convoluted, and a fairly tame countability condition rules it out. The counterexample for the other direction (e.g. banach spaces) is easier, and has good topological properties.
Juan Brooks
>In general locally compact 'looks' weaker I meant the opposite, it 'looks' stronger.
Adrian Carter
And this is why I was wondering if anyone knew some amplifier for paracompactness. I mean, a sequelly compact metrizable space is compact. The metrizability amplifies the notion of sequential compactness.
Still me. I think the following provides a broad class of counterexamples to one direction (if it is correct): Take any topological space X having infinitely many open sets, and with the property that any two open sets intersect in a non-trivial way. Now add a point p to X and define a subset of this new set to be open if, and only if it is an open set of X with p added to it. The assumptions I have made above are in order to force the singleton {p} to not be open. Then Xu{p} is not paracompact.
If X was compact, Xu{p} is also compact, so if X was locally compact, then Xu{p} also is.
Brody Martinez
Like in my example regarding sequentially compact spaces, how an additional assumption makes a strictly weaker property equal imply some other.
Levi Kelly
Seems to work without need for the additional point. Of course you'll have to prove something like that X exists, which is actually not that hard. But such a space is never hausdorff, though.
Elijah Ward
I don't follow. If metric is analogous to paracompact in your analogy, what is compactness and sequential compactness?
Ryan Walker
Metric is actually the thing I'm after, sequential compactness is paracompactness and compactness is local compactness. What do we need to assume to have paracompact spaces locally compact? A quick thought process reduced it to another problem: >assuming a space is Hausdorff and paracompact, what do we need to make closed sets compact? Naturally, we could assume the space itself is compact, and this would give the result. The reasoning here is based on the fact that paracompact Hausdorff spaces are normal, which is equivalent to that, every point [math]x[/math] and every open nbd [math]U[/math], there is an open nbd [math]V[/math] of [math]x[/math] such that [math]V \subset \bar{V} \subset U[/math]. Since the space would be Hausdorff, the compactness of V's closure would make it locally compact.
Brody Wilson
Seems to be the same question as here then I still think this There shouldn't be a natural one.
Thomas Scott
Yeah, there probably isn't. I just wouldn't say it so surely because of these cases like this metric and connectedness together with local connectedness implying path connectedness. I still lean to the direction that there is none.
Elijah Campbell
Right. I was thinking of something like an algebraic variety over the complex numbers, of generic dimension at least 1, with the Zariski topology should do the job. But I think we do need the additional point, so that if we need an infinite number of opens in a cover to cover X, then any refinement will cover p an infinite number of times.
Henry Lee
Are these fancy words just jargon to give unnecessarily tedious definitions to simple stuff?
David Smith
If you're not a mathematician, then yes. If you are a mathematician, sometimes you need some technical assumption to make things work. For example, usually the definition of (topological) manifold is a paracompact, Hausdorff, locally euclidean topological space.
