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This question already has an answer here:

Let $X$ be a topological space, and let $\{U_i\}$ be an open cover. If $Y$ is subset of $X$ such that $Y\cap U_i$ is closed in $U_i$ (for each $i$), does this imply that $Y$ is closed in $X$?

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marked as duplicate by user2345215, Shaun, user99914, Lord_Farin, Moishe Kohan Jan 31 '15 at 10:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Note that $Y^c\cap U_i = U_i \setminus Y \cap U_i$ is open in $U_i$. Therefore it is open in $X$. Now, since $\bigcup U_i = X$, $Y^c = \bigcup (Y^c \cap U_i)$ is open in $X$. Hence $Y$ is closed.

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  • $\begingroup$ ahh i didn't think about complements. cheers for the rapid answer. $\endgroup$ – M Davolo Aug 25 '12 at 18:34
  • $\begingroup$ You're welcome :) Would you mind accepting it as an answer? Just click the check mark next to it... $\endgroup$ – ronno Aug 25 '12 at 18:39
  • $\begingroup$ I had to wait for 10 minutes before I could accept it due to the speed of your answer lol $\endgroup$ – M Davolo Aug 26 '12 at 16:13

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