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I am reading the answer in the post Continuity of Rational Maps in Zariski Topology, where the answerer mentioned the following lemma (sounds like an elementary fact but I fail to prove it):

Lemma: Let $S\subseteq X$ and $\{U_i\}_{i\in I}$ be an open covering of $X$, then $S$ is closed in $X$ iff $S\cap U_i$ is closed in each $U_i$.

My attempt: When $S\cap U_i$ is closed in $U_i$, we get: $$ S=\bigcup_{i\in I} (S_i\cap U_i)$$ where $S_i$ is closed in $X$. But how to proceed?

Any hints or references are welcomed. Thanks!

Zoudelong
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1 Answers1

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Suppose that $S \cap U_i$ is closed in $U_i$ for every $i$. Equivalently, $U_i \setminus S$ is open in $U_i$ and in particular open in $X$. Then, $X \setminus S = \bigcup_{i} (U_i \setminus S)$ is open, so $S$ is closed in $X$.

A good intuition is to think about convergence. A convergent sequence eventually should be a local phenomenon, i.e. eventually concentrated in some $U_i$, and here we can check closeness. It's probably a good exercise to make this intuition precise leading to another proof.

Qi Zhu
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