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This seems rather simple, but just curious about the following definition (pulled from Lee, but definitely standard):

Given an open cover $\mathcal{U}$ of $X$, another open cover $\mathcal{V}$ is called a refinement of $\mathcal{U}$ if for each $V\in \mathcal{V}$ there exists some $U\in \mathcal{U}$ such that $V\subset U$.

Why not the following definition:

Given an open cover $\mathcal{U}$ of $X$, another open cover $\mathcal{V}$ is called a refinement of $\mathcal{U}$ if for each $U\in \mathcal{U}$ there exists some $V\in \mathcal{V}$ such that $V\subset U$.

To me this second 'definition' seems more natural, is this important? I guess these defintions are not equivalent but how bad is this definition of mine?

Squirtle
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1 Answers1

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Your definition, while initially tempting, would make $\{X,\emptyset\}$ a refinement of every open cover! Basically, we need to be sure that every piece of a refinement is small relative to the original cover, and that's what the usual definition gets that yours doesn't. On the other hand, the usual definition does permit, say, $\{X\}$ as a refinement of the cover by all opens: roughly, a refinement doesn't have to be "fine" relative to the small parts of the original cover, but just relative to the big parts. It's a general principle that the small sets in an open cover (i.e. those contained in another element of the cover) are redundant and unimportant.

Kevin Carlson
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  • Yeah, that's why I thought my definition was better, because of the ${X}$ refinement. Additionally $\mathcal{V}\subset \mathcal{U}$ is a refinement of $\mathcal{U}$ and I don't like that. In particular, $\mathcal{V}=\emptyset$ is (vacuously) a refinement... I really don't like that. Yes, a lot of it has to do with the fact that you can't spell refinement without fine. – Squirtle Jan 02 '15 at 03:01
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    No, ${\emptyset}$ is not a refinement of any cover of any nonempty space. Don't forget $\mathcal{V}$ has to be a cover-that's important! And note that ${X}$ is only a refinement of a cover that already includes $X$, so this might not be as bad as it looks. – Kevin Carlson Jan 02 '15 at 19:21
  • Very true, thanks. – Squirtle Jan 02 '15 at 19:22
  • That is two very enlightening arguments, thanks for sharing!! +1 – C-star-W-star May 29 '19 at 09:14
  • PS: Why did you consider ${X,\varnothing}$ instead of ${X}$ for the first argument? It seems $\varnothing$ plays no argumentative/demonstrative role here, or did I miss something? – C-star-W-star May 29 '19 at 09:15
  • The empty set plays an important role for ${X, \emptyset}$ to be a refinement of $\mathcal{U}$ for \textbf{any} $\mathcal{U}$ (if $X$ is an open set) according to the 2nd defiition. For the 2nd def. to be satisfied "for each $U\in\mathcal{U}$ there must exist some $V\in\mathcal{V}$ such that $V\subset U$." This holds trivially for $V=\emptyset$, but it will only hold for $V=X$ if every $U$ is a superset of $X$ – Bruno KM Feb 11 '22 at 09:09