How do I prove $C\setminus A$ is closed subset of $X$?

by Unknown x   Last Updated February 11, 2019 09:20 AM - source

Let $(X,\mathscr T)$ be topological space with dense subset $D$ and a closed,relatively discrete subset $C$ such that There is a bijection from $\mathscr{P}(D)$ to a subset of $C.$ Then $(X,\mathscr T)$ is not normal.

Proof. Suppose $(X,\mathscr T)$ is not normal subspace of $X$. Let $A$ be a nonempty proper subset of $C$. Let $x\in C \setminus A$. Then there exists $U\in \mathscr T$ such that $U \cap C=\{x\}$. which is open. So, $C\setminus A=\bigcup_{x\in C\setminus A}\{x\}$ is open in $A$. So $A$ is closed in $C$. So,$A$ is closed in $X$. But

I am not able to prove that $C \setminus A$ is a closed subset of $X$.

By the normality, there exists a disjoint open subsets $U(A) $ and $V(A) $ such that $A \subset U(A) $ and $C \setminus A \subset V(A)$....................................................(I)

Now suppose $A_1$ and $A_2 $ are nonempty proper subsets of $C$ Such that $A_1\subset A_2 \neq \emptyset.$ Apply (I) to $A_1$ and $A_2 .$ We get open subsets $U(A_1) $ and $V(A_1) $ such that $A_1 \subset U(A_1) $ and $C \setminus A_1 \subset V(A_1) $ and open subsets $U(A_2) $ and $V(A_2) $ such that $A_2 \subset U(A_2) $ and $C \setminus A_2\subset V(A_2) $. Then the author wrote $U (A_1)\cap V (A_2) \neq \emptyset.$ How do I prove $U (A_1)\cap V (A_2) \neq \emptyset$? Since $D$ is a dense subset in $X,$ $U(A_1) \cap V(A_2) \cap D\neq \emptyset.$ But $U(A_1) \cap V(A_2) \cap D \subset U(A_1) \cap D$ and $U(A_2) \cap V(A_2) \cap D \not\subset U(A_2) \cap D.$ Since, Thus if $A_1$ and $A_2$ are two distinct subset of $X$. We get $U(A_1) \cap D$ and $U(A_2) \cap D$ are distinct subset of $D.$ Therefore there is a subset of $\mathscr{P}(D)$ which is bijective to $\mathscr{P}( C). $ Also $C$ is not bijective to $\mathscr{P}(C)$ and $C$ is bijective to some subset of $\mathscr{P}(C)$. How this contradict to the assumption There is a bijection from $\mathscr{P}(D)$ to a subset of $C.?$ Please help me to understand the proof.

Related Questions

If J is uncountable, then $R^J$ is not normal.

Updated April 25, 2017 12:20 PM

Product of $T_1$ spaces is $T_1$

Updated March 11, 2019 03:20 AM