# is the preorder of locally closed immersions complete?

Let $X$ be a scheme. Consider the preorder of locally closed immersions into $X$ (considered, say, as a full subcategory of $Sch/X$). Is it complete or cocomplete? That is, are there infima or suprema?

Ok it's easy to see that every nontrivial locally closed subscheme of $\mathbb{A}^1_\mathbb{C}$ contains only finitely or cofinitely many rational points of $\mathbb{A}^1_{C}$. Therefore we should restrict to finite infima or suprema.

I guess everything works fine for infima if we restrict to seperated schemes. Take the ideal sheaves in the open subschemes which correspond to the closed immersions, restrict them to their intersection, take the smallest quasi coherent ideal which contains them and consider the corresponding closed subscheme of the intersection.

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I would find this question easier to read if you stated explicitly which way the order relation goes. There may be a standard convention in Sch/X, but not everyone who might find the question interesting is necessarily familiar with the convention. –  Charles Staats Apr 10 '10 at 15:37
As I said, the preorder is a full subcategory of $Sch/X$. Thus for locally closed immersions $i : U \to X, j : V \to X$, we have $i \leq j$ iff there is a $X$-morphism $U \to V$ (which is unique since $j$ is mono). –  Martin Brandenburg Apr 10 '10 at 16:36
Also it can be proven without much difficulty that this category is "essentially-small" in the sense that there is a small skeleton. Thus for many purposes it makes sense to regard this category as a small preorder. –  Martin Brandenburg Apr 10 '10 at 17:22

It does not have suprema. For example, consider the affine plane $\mathbb A^2_{\mathbb C}$, and let $\mathbb A^1_{\mathbb C}$ by the line $y=0$. Consider the locally closed subschemes $U := \mathbb A^2_{\mathbb C} \smallsetminus \mathbb A^1_{\mathbb C}$ and $V := \{(0,0)\}$ with the reduced scheme structure. The locally closed subschemes that contain $U$ and $V$ are the complements of finite set of points in $\mathbb A^2_{\mathbb C} \smallsetminus \{(0,0)\}$; so there is no supremum.