# Why is the bounded functions not a sheaf?

I know that the presheaf of bounded functions is not a sheaf but I don't see why. I checked in wikipedia they say that this presheaf does not verify the axiom of "Glue". For me it verifies this axiom. Indeed, if U and V are open sets, and f and g are bounded functions on U and V respectively, and they agree on the intersection, then combining them in the obvious way - let h(x) = f(x) if x is in U, g(x) if x is in V, is a bounded function with the bound being max(|f|, |g|). Do you think I don't understand well what gluing means?

Thanks.

• The functions may glue together to form an unbounded function. – Lord Shark the Unknown Oct 7 '18 at 19:34
• The covering needs not be finite, so your $\max(\lvert f\rvert,\lvert g\rvert)$ is not really a bound. – Saucy O'Path Oct 7 '18 at 19:35
• I don't see why? If you define $h$ (by gluing) as I did I don't see why it's unbounded? @LordSharktheUnknown – algebra1112 Oct 7 '18 at 19:36
• Because $\max\{1,2,3,4,5,6,7,\cdots\}=\boxed{??}$. – Saucy O'Path Oct 7 '18 at 19:39
• You may not be glueing two functions, you may be glueing infinitely many functions. – Lord Shark the Unknown Oct 7 '18 at 20:04

Let us take the $$\mathcal{F}$$ presheaf of bounded real functions on the real line $$\mathbb{R}$$. Then for each $$U\subset \mathbb{R}$$ open we have $$U \mapsto \mathcal{F}(U) = \{ f\colon U \longrightarrow \mathbb{R} \mid \sup_U |f| < \infty \}$$ It is clearly a presheaf. Now let's see the sheaf requirements.
Fix $$U\subset \mathbb{R}$$ and an open covering $$U=\cup_i U_i$$.
1. For $$s,t \in \mathcal{F}(U)$$, we need that $$s|_{U_i} = t|_{U_i}, \forall i \Rightarrow s=t$$
2. For a family $$\{s_i \in \mathcal{F}(U_i)\}$$ we need that $$s_i|_{U_i\cap U_j} = s_j|_{U_i\cap U_j}, \forall i,j \Rightarrow \exists s \in \mathcal{F}(U) \mid s|_{U_i}= s_i$$
It is not hard to see that 1. holds. Now for 2. take $$U_i = (i-2,i+2)$$ an open interval for each $$i\in \mathbb{Z}$$. We have $$\mathbb{R} = \cup U_i$$. Define $$s_i\colon U_i \longrightarrow \mathbb{R}, \, s_i(t) = t$$ Then $$\sup_{U_i} |s_i| = \max\{|i-2|,|i+2|\}$$ and $$s_i \in \mathcal{F}(U_i)$$. Suppose that there exists $$s\in\mathcal{F}(\mathbb{R})$$ such that $$s|_{U_i}= s_i$$. Let $$N \in \mathbb{Z}$$ be such that $$N> \sup_{\mathbb{R}} |s|$$. Then we have an absurd since $$s(N) = s_N(N) = N > \sup_{\mathbb{R}} |s|.$$