Prove that $X \setminus \operatorname{Int}A \subseteq \overline {X\setminus A} $

Question

Prove that $X \setminus \operatorname{Int}A \subseteq \overline {X\setminus A} $

$\overline {X\setminus A} =\{x \in X : \text{$x$ is adherent to $X\setminus A$}\} $

$x$ is adherent to $X\setminus A$ means that every open subset of $X$ has nonempty intersection with $X \setminus A$

$\operatorname{Int}A=\bigcup \{U : U \subset A, \text{$U$ open}\}$

$x \in X \setminus \operatorname{Int}A \implies x \notin \operatorname{Int}A $

I am trying to prove it in contrary:

$U \cap (X \setminus A) = \emptyset $

$U^{c} \cup (X \setminus A)^{c} = X$

$U^c \cup A = X$

$x \in U^c \cup A \implies x\in U^{c} \ \lor x\in A$

I am not sure if it makes sense. I don't know what to do now.

Answer 1

You can indeed prove that if $x\notin\overline{X\setminus A}$, then $x\notin X\setminus\operatorname{Int}A$ (that is, $x\in\operatorname{Int}A$).

Suppose $x\notin\overline{X\setminus A}$; then, by definition, there exists an open neighborhood $U$ of $x$ such that $U\cap(X\setminus A)=\emptyset$.

Therefore $U\subseteq A$ (this is the step you're missing to notice). Hence, by definition, $x\in\operatorname{Int}A$, as you wished to prove.

Answer 2

If $x\in X\setminus \operatorname{Int}A$ and $U$ is an open set containing $x$, then $U\subsetneq A$, so $U$ must contain a point of $X\setminus A$. Therefore $U$ has nonempty intersection with $X\setminus A$. Since $U$ was arbitrary, $x\in \overline{X\setminus A}$.

Answer 3

In fact, we have $X \setminus \operatorname{int}(A)=\operatorname{cl} (X \setminus A)$

Let $F=X \setminus G$.

$$ \begin{align*} \operatorname{int}(A) &=\bigcup \{G: G \text{ is open and } G \subseteq A \} &&\text{(By Definition of Interior)}\\ X \setminus \operatorname{int}(A)&= \bigcap \{X \setminus G: G \text{ is open and } G \subseteq A \} &&\text{(DeMorgan's Law)} \\ &= \bigcap \{X \setminus G:X \setminus G \text{ is closed and } X \setminus A \subseteq X \setminus G \} &&\text{(Relative Closure Property)} \\ &= \bigcap \{F: F \text{ is closed and } X \setminus A \subseteq F \} &&\text{(By Assumption)} \\ &= \operatorname{cl} (X \setminus A) &&\text{(By Definition of Closure)} \end{align*} $$