Closed set
This is in Lemma 3.1 of Hartshorne's Algebraic Geometry.
"An open subset of an open subset is open" is the trick.
Proposition. Let \( Y \) be a topological space. \( Z \subset Y \) is an closed set if and only if there exists an open cover \( \{ U_\alpha \} \) such that \( Z \cap U_\alpha \) is closed in \( U_\alpha \) for all \( \alpha \).
Proof)
\( (\Rightarrow) \): Suppose \( Z \) is closed. \( Y \) is an open cover of \( Y \) and \( Z \cap Y \) is closed in \( Y \).
\( (\Leftarrow) \): If \( Z \cap U_\alpha \) is closed in \( U_\alpha \), then \( Z^c \cap U_\alpha \) is open in \( U_\alpha \). Hence \( Z^c \cap U_\alpha = Z_\alpha \cap U_\alpha \) for some \( Z_\alpha \) open in \( Y \). In particular, \( Z^c \cap U_\alpha \) is open in \( Y \).
\[ Z^c = Z^c \cap Y = Z^c \cap \left ( \bigcup U_\alpha \right ) = \bigcup ( Z^c \cap U_\alpha) \]
hence \( Z^c \) is open. Hence \( Z \) is closed.
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