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I'm reading the following proof here on Mathstack Exchange:

Let $Y$ be a topological space that can be partitioned into dense subsets $D$ and $E$. If $X \subset Y$ is closed, then there is a $V \subset X$ such that $\operatorname{Fr} V = X$.

Take $V = X \setminus (D \cap \operatorname{Int} X)$. Then since $V \subset X$ we have $\operatorname{Cl} V \subset \operatorname{Cl} X$, and $\operatorname{Fr} V \subset X$ and $E \cap \operatorname{Int} X$ dense in $\operatorname{Int} X$, therefore $\operatorname{Cl} V = X$. On the other hand $Y \setminus X$ is dense in $Y \setminus \operatorname{Int} X$ and $D \cap \operatorname{Int} X$ is dense in $\operatorname{Int} X$, therefore $\operatorname{Int} V = \emptyset$. It follows that $ \operatorname{Fr} V = X$.

My question: I would like to know how to prove that $\operatorname{Cl} V=X$ and $\operatorname{Int} V=\emptyset$.

I tried to use the relation $Y-\operatorname{Int}(E)=\operatorname{Cl}(Y-E)$ and $Y-\operatorname{Cl}(E)=\operatorname{Int}(Y-E)$ but without success.

I had the following idea:

\begin{align*} Y-\operatorname{Int} V=& \operatorname{Cl}(Y-V)\\ =& \operatorname{Cl}(Y \cap V^c)\\ =& \operatorname{Cl}(Y \cap (X \cap (D \cap \operatorname{Int} X)^c)^c)\\ =& \operatorname{Cl}(Y \cap (X^c \cup (D \cap \operatorname{Int} X))\\ =&\operatorname{Cl}(Y \backslash X \cup (D \cap \operatorname{Int} X))\\ =&\operatorname{Cl}(Y \backslash X) \cup \operatorname{Cl} (D \cap \operatorname{Int} X)\\ \supset & (Y \backslash \operatorname{Int} X) \cup \operatorname{Int X}\\ =& Y. \end{align*} Therefore, $\operatorname{Int} V=\emptyset.$

Is this correct?

With these informations we have that $\operatorname{Fr}(V)=\operatorname{Cl}(V)\cap \operatorname{Cl}(Y\backslash V)=\operatorname{Cl}(V) \cap (Y \backslash \operatorname{Int} V)=X\cap Y=X$.

Math
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1 Answers1

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The solution you are referring to states that $E\cap \operatorname{Int}(X)$ is dense in $\operatorname{Int}(X)$, which is a rather obvious fact. This can be written as $$\operatorname{Cl}(E\cap\operatorname{Int}(X))\cap \operatorname{Int}(X)=\operatorname{Int}(X)$$

  • Notice that $$V:=X\setminus(D\cap\operatorname{Int}(X))=\big(X\cap E\big)\cup\big(X\setminus\operatorname{Int}(X)\big)=(X\cap E)\cup \partial(X)$$ Since $\partial X$ is closed, we have that \begin{align} \operatorname{Cl}(V)&=\operatorname{Cl}(X\cap E)\cup\operatorname{Cl}(\partial X)\\ &\supset\big(\operatorname{Cl}(\operatorname{Int}(X)\cap E)\cap\operatorname{Int}(X)\big)\cup\partial X\\ &=\operatorname{int}(X)\cup\partial X\\ &=\operatorname{Cl}(X)=X \end{align} Since $V\subset X$, $\operatorname{Cl}(V)\subset \operatorname{Cl}(X)=X$, and the conclusion of your first question follows.

  • for you second question, here is a more pedestrian approach: If $W\neq\emptyset$ is an open set in $Y$ and $W\subset V$, then $W\subset \operatorname{Int}(X)$ and $W\cap D\neq\emptyset$. Consequently, $W\cap \big(D\cap \operatorname{Int}(X)\big)\neq\emptyset$, but $W\subset V\subset Y\setminus(D\cap\operatorname{Int}(X))$, which is a contradiction. Hence, $\operatorname{Int}(V)=\emptyset$.

Mittens
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