Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $X$?

share|improve this question
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 1

up vote 13 down vote accepted

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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.