# 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.

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