# How do I prove $U (A_1)\cap V (A_2) \neq \emptyset$? Can You help me to find where do we arrive contradiction?

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

Notations and definition in the theorem:-

$$X\sim Y$$- There is a bijection map from $$X$$ to $$Y$$.

$$X\precsim Y-$$ There is a subset $$Y'$$ of $$Y$$ such that $$X \sim Y$$

Relatively discrete. A subset $$A$$ of a topological space $$(X,\mathscr T)$$ is relatively discrete provided that for each $$a\in A$$, there exists $$U\in \mathscr T$$ such that $$U\cap A=\{x\}$$.

I am having doubt in understanding the underlined statements. I have edited and made my doubts more precise.

Doubt 1:- How do I prove $$C\setminus A$$ is closed? My attempt:- It is enough to prove that $$A$$ is an open set. Let $$x\in A\subset C$$, then there is an open set $$U\in \mathscr T: U\cap C=\{x\}.$$ So, we can write $$A=\bigcup_{x\in A}\{x\}$$ is open in $$C$$. Hence, $$A$$ is closed in $$C$$. Hence Closed in $$X$$. Am I correct?

Doubt2:-How do I prove that $$U(A_1) \cap V(A_2)\neq \emptyset$$?

What is the idea of the proof afterwards? Where do we arrive at the contradiction?

• You first suppose what you are wanting to prove. – William Elliot Feb 11 at 9:50
• I wanted to prove $(X,\mathscr T)$ is not normal. – Unknown x Feb 11 at 9:57
• I have typed the proof given in the textbook of foundation of topology by C.W Patty – Unknown x Feb 11 at 9:57
• I have doubt in two places 1. How do I prove $C \setminus A$ is closed?also How do I prove $U (A_1)\cap V (A_2) \neq \emptyset$? – Unknown x Feb 11 at 9:58
• $C$ is closed discrete in $X$, so every its subset is closed in $C$ and also in $X$. – user87690 Feb 11 at 10:32

If $$C$$ is relatively discrete, this means that all $$\{x\}$$ where $$x \in C$$ are open in $$C$$ (because there is an $$U \in \mathcal{T}$$ such that $$U \cap c=\{x\}$$). This implies, as all subsets are unions of singletons trivially, that all subsets of $$C$$ are open in $$C$$. As every subset of $$C$$ also has an open-in-$$C$$ complement (by the previous) it is in fact itself closed in $$C$$ and so closed in $$X$$.

Now by normality, we can find for every non-empty $$A \subseteq C$$ disjoint open subsets $$U(A)$$ and $$V(A)$$ such that $$A \subseteq U(A)$$ and $$C-A \subseteq V(A)$$, as $$A$$ and $$C-A$$ are disjoint closed subsets of $$C$$ and thus of $$X$$.

The aim is now to show that the function $$f(A)=U(A) \cap D \in \mathscr{P}(D)$$ from $$\mathscr{P}(C)$$ is 1-1 (injective)(you can add $$f(\emptyset)=\emptyset$$ to make it defined on the whole powerset, if you like). I'll rephrase the quoted argument slightly:

Suppose $$f(A_1) = f(A_2)$$ while $$A_1 \neq A_2$$.

$$A_1 \neq A_2$$ means either $$A_1 - A_2 \neq \emptyset$$ or $$A_2 - A_1 \neq \emptyset$$.

Say the former holds. Then we have some $$p \in A_1, p \notin A_2$$. Then $$p \in U(A_1) \cap V(A_2)$$ by how the sets are chosen, and so we have a non-empty open set, which thus intersects the dense set $$D$$.

But $$V(A_2) \cap f(A_1)= U(A_1) \cap D \cap V(A_2)\neq \emptyset$$ while $$V(A_2) \cap f(A_2)=V(A_2) \cap U(A_2) \cap D =\emptyset$$ as $$U(A_2)$$ and $$V(A_2)$$ are disjoint. This contradicts $$f(A_1)=f(A_2)$$: the same set intersected with the same $$V(A_2)$$ cannot be both empty and non-empty. Contradiction 1.

We get another contradiction if $$A_2 - A_1 \neq \emptyset$$, using $$V(A_1)$$ intersections this time (check this, the text skipped this due to symmetry considerations, but I mention it for completeness).

These contradictions show that $$f(A_1)=f(A_2)$$ must imply $$A_1=A_2$$.

The last part is set theory: by Cantor's theorem we have that $$\mathscr{P}(C)$$ does not inject into $$C$$, or $$\mathscr{P}(C) \not\precsim C$$ .

But by the map $$f$$, $$\mathscr{P}(C) \precsim \mathscr{P}(D) \precsim C$$ (the last by the theorem's assumption). So combining: $$\mathscr{P}(C) \precsim C$$: a contradiction with Cantor's theorem. QED

Regarding your first question: the earlier part of the first paragraph showed that if $$B \subsetneq C$$, then $$B$$ is closed in $$C$$. If we have a set $$A \subsetneq C$$ with $$A \neq \varnothing$$, then setting $$B = A$$ shows that $$A$$ is closed in $$C$$ (and hence closed in $$X$$), whereas setting $$B = C-A$$ shows that $$C-A$$ is closed in $$C$$ (and hence closed in $$X$$).

Regarding your second question: the assumption that $$A_1 - A_2 \neq \varnothing$$ implies that $$A_1 \cap (C-A_2) \neq \varnothing$$. Since $$A_1 \cap (C-A_2) \subset U(A_1) \cap V(A_2)$$, this means that $$U(A_1) \cap V(A_2) \neq \varnothing$$.

Regarding the final contradiction: the second paragraph shows that there is an injection from $$\mathcal{P}(C)$$ into $$\mathcal{P}(D)$$. If $$\mathcal{P}(D) \subset C$$, then $$\mathcal{P}(C)$$ would be in bijection with a subset of $$C$$, which would contradict Cantor’s theorem.

• Why do they have assumption $A_1\setminus A_2\neq \emptyset$? From there what does he want to prove? – Unknown x Feb 11 at 14:56
• He wants to prove that if $\varnothing \neq A \subsetneq C$ and we pick $U(A), V(A) \subset C$ with $A \subset U(A)$ and $C-A \subset V(A)$, then the map $A \mapsto U(A)$ is injective. – Jordan Green Feb 11 at 15:04