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Let $X_1,X_2$ be two topological space, assume we have open subset $U_1\subset X_1, U_2\subset X_2$ identified together via isomorphism $\phi:U_1\to U_2$. therefore $X_1,X_2$ gluing together to $X= X_1\cup_{\phi}X_2$

Assume the closed point set are $C_i = \{p\in X_i\mid \bar{\{p\}} = \{p\}\}$ for $i = 1,2$ .

I want to prove the set of closed point in $X$ is indeed the $C_1\cup C_2$ via the identification $\phi$.

To be more precise we have $j_i:X_i \to X$ is a topological embedding of this two pieces , I want to show the set of closed points in $X$, denotes it $C = j_1(C_1)\cup j_2(C_2)$.

my attempt: first show $ C\subset j_1(C_1)\cup j_2(C_2) $ i.e. needs to show closed point in $X_i$ are also closed point in $X$, denote the closure of a set $S$ in the topolgoical space $Y$ as $\text{cl}_Y(S)$. Therefore $p\in C_i$ iff $\text{cl}_{X_i}(p) = p$ in the mean time:

$$\text{cl}_X(p)\cap X_i = \text{cl}_{X_i}(p) = p$$ and $$\text{cl}_X(p) = \text{cl}_{X}(p)\cap(X_1\cup X_2) = \text{cl}_{X_1}(p)\cup \text{cl}_{X_2}(p)$$

therefore if $\text{cl}_X(p) = p$ then $\text{cl}_{X_1}(p) = p$ or $\text{cl}_{X_2}(p) = p$

I'm not sure how to prove the other direction?

yi li
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1 Answers1

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You can't prove the other direction because it's false. Consider the following counterexample: let $X=\{0,1\}$ with the open sets being $\{\emptyset,\{0\},\{0,1\}\}$ (this is the Sierpinski space, known better in algebraic geometry as the spectrum of a DVR). Let $X_1=\{0,1\}$ and $X_2=\{0\}$. Then $C_1=\{1\}$ and $C_2=\{0\}$, so $C=X$ but the only closed point of $X$ is $1$.

KReiser
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  • thank you @KReiser , I was working with the projective space $\Bbb{P}{k}^n$ which covered by $n+1$ affine scheme $U_i = \operatorname{Spec} k[x{0/i},...,x_{n/i}]/(x_{i/i} - 1)$ and the closed point of $U_i$ is clear, in this specific example, does the other direction holds ( I guess it's true)? – yi li Jan 08 '23 at 04:45
  • where $k$ is algebraic closed. – yi li Jan 08 '23 at 05:04
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    Yes, it does. One way to prove it in your case is that for a scheme of finite type over a field $k$, a point is closed iff its residue field is a finite extension of $k$. This criteria lets you show that a point is closed iff it's closed as a point in any open subscheme. – KReiser Jan 08 '23 at 05:04
  • Perfect solution thank you KReiser, related question in the comment (If someone needed) here:https://math.stackexchange.com/a/1108119/360262 – yi li Jan 08 '23 at 08:04