# “It can be checked locally that $Z$ is a closed subset”

Look at the following proposition/exercise:

A subspace $Z$ of a topological space $X$ is closed if and only if exists an open cover $\{U_\alpha\}$ of $X$ such that $Z\cap U_\alpha$ is closed in $U_\alpha$ for every $\alpha$.

Now the implication $(\Rightarrow)$ is quite trivial, infact if $Z$ is closed and we cover $X$ with some elements $U_{\alpha}$ of the basis, then by definition of topology for subspaces, we have that $Z\cap U_{\alpha}$ is closed in $U_{\alpha}$.

I have problems to write the other implication. Thanks in advance.

• In the generality of all topological spaces, the only open cover you could possibly write down for a proof is the collection of all open sets on X, and that one works. – zyx Oct 11 '13 at 16:10
• Couldn't he just choose $\{X\}$ as the open cover? – Stefan Hamcke Oct 11 '13 at 16:11
• @zyx ¿...? ${}{}{}$ – Pedro Tamaroff Oct 11 '13 at 16:13
• @PedroTamaroff, question asks for the "converse" implication that asserts an open cover exists, such that (...). There are very few open covers that provably exist on an arbitrary topological space. – zyx Oct 11 '13 at 16:15
• Also that one! @StefanH – zyx Oct 11 '13 at 16:18

Let $\mathcal U=(U_\alpha)_\alpha$ be the open cover. Let $W$ denote the complement of $Z$. We have $W=W\cap(\bigcup\mathcal U)=\bigcup_\alpha(W\cap U_\alpha)$. Now, $Z\cap U_\alpha$ is closed in $U_\alpha$, thus $U_\alpha-(Z\cap U_\alpha)$ is open in $U_\alpha$ and thus open in $X$. But $U_\alpha-(Z\cap U_\alpha)$ is just $W\cap U_\alpha$.
Essentially the same argument works if, more generally, the $U_\alpha$ are sets whose interiors cover $X$ and $Z\cap U_\alpha$ is always closed in $U_\alpha$.
If this property holds, then we say that "$X$ is coherent with the family $\mathcal U$"
Suppose that $x\notin Z$. Then $x\in U_\beta$ for some element of the cover. If $U_\beta\cap Z=\varnothing$; we're done. Else, $x\notin Z\cap U_\beta\neq \varnothing$. Since this is closed, there exists an open set $N$ such that $x\in N\cap U_\beta=N_\beta$ and $N_\beta\cap (Z\cap U_\beta)=\varnothing$. Since $U_\beta$ is open we may assume $N\subseteq U_\beta$. Then $N\cap Z=\varnothing$; $x\in N$, and we're done.