# Is this a sufficient condition for a subset of a topological space to be closed?

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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possible duplicate of Is whether a set is closed or not a local property? –  Hagen von Eitzen Nov 12 '12 at 16:17

## 1 Answer

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