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Stream: learning: questions

Topic: When the presheaf of bounded functions is a sheaf

Peva Blanchard (Apr 01 2024 at 23:54):

I'm continuing here a side quest started on the topic started by @David Egolf

We practiced basic exercises in sheaf theory, showing that the presheaf $\mathcal{F}$ of bounded functions on $\mathbb{R}$ is not a sheaf. Then I was wondering what could be the conditions on a topological space $X$ for the presheaf of bounded functions to be a sheaf.

Peva Blanchard (Apr 01 2024 at 23:56):

Here are some examples or candidates:

• any topological space $X$ where every open subset is compact
• any topological space $X$ with a finite number of open sets
• any topological space $X$ with the trivial topology (i.e., $\emptyset$ and $X$ are the only open sets)
• any finite topological space $X$

Peva Blanchard (Apr 02 2024 at 00:08):

Let's assume that $\mathcal{F}$ is a sheaf on $X$.

A first reduction. We can decompose $X$ into connected components. There must be a finite number of connected components, as otherwise we could build an unbounded function on $X$ defined as a constant on each connected component.

So we can focus on the case where $X$ is connected.

Morgan Rogers (he/him) (Apr 02 2024 at 08:14):

Suppose there exists an infinite open cover of $X$ with no finite subcover; i.e. suppose $X$ is not compact. Then we may construct a countable sequence $\{U_n\}_{n=0}^{\omega}$ which is strictly increasing, in the sense that $U_k \not\subseteq \bigcup_{n=1}^{k-1} U_n$. Under some further conditions (to be determined), could one construct bounded functions which hit a maximal value of $k$ in each $U_k$?

Peva Blanchard (Apr 02 2024 at 08:17):

I was thinking about partitions of unity.

If $X$ admits a partition of unity for the open cover you suggest, then I think we can build an unbounded function.

Chris Grossack (they/them) (Apr 02 2024 at 20:04):

I mean, as soon as you have countably many disjoint open sets you're screwed, right? So if you want the presheaf of bounded functions to be a sheaf you need to know that there's no countable family of disjoint opens.

But such spaces have to be highly non separated. For instance, this excludes all (infinite) hausdorff spaces

Chris Grossack (they/them) (Apr 02 2024 at 20:09):

I think the converse might be true as well? If $X$ has no countable family of disjoint opens, can you show that every family of opens is a finite union of "sunflowers" or similar? If yes, then each of the finitely many sunflowers would get a real, but the whole function would stay bounded

Chris Grossack (they/them) (Apr 02 2024 at 20:27):

Yup! This sounds right. Some quick googling for antichains in the lattice of open sets led me to Jean Goubault-Larrecq's Spaces with No Infinite Discrete Subspace (available here). Right above Definition 2.2, we read

Perhaps the closest result in the literature, except for the preorder-theoretic results mentioned in the introduction, is due to A. H. Stone [11, Theorem 2]: a space X has no infinite discrete subspace if and only if every open cover of every subspace A of X has a finite subfamily whose union is dense in A, if and only if every continuous real-valued function on every subspace of X is bounded

So it looks like "X has no infinite discrete subspace" is the right condition, and it's already known to be equivalent to the statement that the presheaf of bounded real functions is a sheaf (that's basically the last sentence) ^_^

Peva Blanchard (Apr 02 2024 at 21:12):

Cool, thanks!

Peva Blanchard (Apr 02 2024 at 21:20):

The cited paper, by A. H. Stone, is called "Hereditarily compact spaces", so I was not too far away. I don't have access to the paper, but I would be interested in A.H.Stone's proof technique (that relates having no infinite discrete subspaces and the boundedness of continuous functions).

Chris Grossack (they/them) (Apr 02 2024 at 22:41):

Here you go ^_^
stone1960.pdf

Chris Grossack (they/them) (Apr 03 2024 at 03:43):

Also, @Peva Blanchard, your "every open subset is compact" conjecture was very close! In fact it's literally correct if you moreover assume that $X$ is $T_1$! (This is in Stone's paper)