Assume $X$ is $T_1$-space. prove that, $X$ is normal if and only if for every closed set $C\subset X$ and open set $U\subset X$ such that $C\subset U$ there exist an open set $V\subset X$ such that $C\subset V\subset \overline{V} \subset U$.

My Solution:

Part a) Assume $X$ normal I want prove $V\subset X$ such that $C\subset V\subset \overline{V} \subset U$. So I have to prove three relation: First, $C\subset V$. Second, $V\subset \overline{V}$. Third, $\overline{V} \subset U$.

First, based on property assumption for every closed set $C$ and open set $V$ , $C\subset V$.

Second one is obvious, For every set V we have $V\subset \overline{V}$.

Third, $X$ is normal, let given $C$ and given neighborhood of $C$ called $U$. We know $C$ is closed, Let $B=X-U$, as $U$ is open $B$ is closed. As $X$ is normal I used definition of normal so $C$ , $B$ are closed and they are disjoint. So there is two open set $V$ , $W$ such that $V$ contain $C$ and $W$ contain $B$, and $V\cap W = \varnothing$ , I cant understand why $\overline{V}$ is disjoint from $B$.and get stuck here although I know if $y\in B$ there is exist nieghborhood $W$ of $y$ such that $W$ is disjoint from $V$.

Part b) We want to prove if the question condition hold, $X$ is Normal. we have $X$ is $T_1$ and so it is enough to show that for every two closed subset that are disjoint there is two open set disjoint that contain evrey of those. So we have Two closed sets $C\subset X$ and $Y=X-U \subset X$, $V$ is open set contain $C$. If let $W=X-\overline{V}$, it is a open set contain $Y$ and disjoint from $V$. So we defint two closed set and two open set contain them that have not intersection. but I do not sure it is correct and enough.

every correction and suggestion would be helpful.

  • 2
    $\begingroup$ Since $V$ and $W$ are open, and $V\cap W = \varnothing$, you even have $\overline{V}\cap W = \varnothing$. Perhaps that makes it easier to see. $\endgroup$ – Daniel Fischer Nov 27 '13 at 22:56
  • $\begingroup$ How you know $\overline{V} \cap W =\varnothing$, Is it properties? $\endgroup$ – Zeezee Nov 27 '13 at 23:05
  • $\begingroup$ $W$ is open, so its complement is closed. $V$ is contained in the complement of $W$. $\endgroup$ – Daniel Fischer Nov 27 '13 at 23:07

Proof: Since $C \subset U$, it follows that $C \cap (X\setminus U) = \emptyset$. Then there are two disjoint open sets $V$ and $W$ of $X$ such that $C \subset V$, $(X\setminus U) \subset W$ and $V \cap W=\emptyset$. As pointed out by Daniel Fischer, we have $\overline{V} \cap W=\emptyset$, and hence $C \subset V \subset \overline{V} \subset X\setminus W \subset U $. This completes the proof.


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