0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – zyx Oct 11 '13 at 16:10
  • $\begingroup$ Couldn't he just choose $\{X\}$ as the open cover? $\endgroup$ – Stefan Hamcke Oct 11 '13 at 16:11
  • $\begingroup$ @zyx ¿...? ${}{}{}$ $\endgroup$ – Pedro Tamaroff Oct 11 '13 at 16:13
  • $\begingroup$ @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. $\endgroup$ – zyx Oct 11 '13 at 16:15
  • $\begingroup$ Also that one! @StefanH $\endgroup$ – zyx Oct 11 '13 at 16:18
3
$\begingroup$

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 $\mathcal U$"

$\endgroup$
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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