Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

On page 15, in Lemma 3.1 it is claimed: "A subset Z of a topological space Y is closed if and only if Y can be covered by open subsets U such that Z $\cap$ U is closed in U for each U."

How do I prove this?

share|cite|improve this question
up vote 6 down vote accepted

For the $(\Leftarrow)$ direction, let $Z$ be a subset of $Y$ and we will show that $V := Y \setminus Z$ is open. The given covering of $Y$ has the property that $V \cap U$ is open in $U$ for all $U$. But $U$ is open in $Y$, so $V \cap U$ is open in $Y$. And $V = \cup (V \cap U)$ since the $U$'s cover $Y$, so $V$ is open in $Y$, as desired.

share|cite|improve this answer

For the first direction, if $Z$ is closed, then $Y$ is open and $\{Y\}$ is an open cover of $Y$ such that $Z\cap Y$ is closed in $Y$.

For the converse direction, if $Z$ is not closed, then there is some $z\in \partial Z \cap Z'$. Let $\{U\}$ be an open cover of $Y$ such that $Z\cap U$ is closed in each $U$. Consider $U_z \ni z$. Then in $U_z$, $z\in\partial (Z\cap U)$ but $z\notin (Z\cap U)$, so $Z\cap U$ cannot be closed in $U$ after all.

share|cite|improve this answer

If $Z$ is closed, consider the points of $Y-Z$. For each such point, take an open set containing it disjoint of $Z$. Also, for each point of $Z$, consider any open set containing it. Then you'll have open sets covering $Y$ satisfying the conditions required on the intersections. This proves $(\Rightarrow)$.

share|cite|improve this answer
For the $(\Rightarrow)$ direction, you can take the covering consisting of the single open set $Y$. – Michael Joyce Oct 21 '12 at 15:55

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.