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$$X \setminus ( X\setminus A) =A$$


doing some self studying on Croom's book on principles of topology. I am able to prove it using sets but not element argument on left way

Easy prove using sets and thm 1.2 if $A \subset B $ it is equivalient to $A \cap B =A$

$$\begin{aligned} X \setminus (X \setminus A) &=X \cap (X \setminus A)^c \\ &= X \cap( X^c \cup A) \\ &= (X \cap X^c) \cup (X \cap A) \\ & = \emptyset \cup (X \cap A) \\ & = (X \cap A) \\ &=A &&\text{since } A \subset X \Leftrightarrow A \cap X = A \end{aligned} $$

proving using set element notation

$\Rightarrow$] ($X \setminus (X \setminus A) \subset A$)

assume that $x \in X \setminus (X \setminus A)$ so $x\in X \wedge x\not \in X \setminus A \Leftrightarrow x\in X \wedge (x\not \in X \vee x\in A)$ by disjoing syllogims $x\in A$

$\Leftarrow ]$ could use help here thanks! Using element argument ofcourse

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    $\begingroup$ Suppose $x\in A$. Then $x\notin X\setminus A$. Since $x\in A$ it follows that $x\in X$ as well. Therefore since $x\in X$ and $x\notin X\setminus A$... $\endgroup$ – JMoravitz May 16 '17 at 2:17
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    $\begingroup$ Nitpick. You never stated that X was the universal set or that A was a subset of X. This isn't true if A is not a subset of X. $\endgroup$ – fleablood May 16 '17 at 2:24
  • $\begingroup$ @fleablood. Very good nitpick. Hidden assumptions, no matter how small, are the bane of all attempts at reasoning. $\endgroup$ – DanielWainfleet May 16 '17 at 21:16
  • $\begingroup$ @fleablood Unwilling to upvote your comment because you say it is a nitpick. It is not. $\endgroup$ – Git Gud Feb 18 at 21:04
  • $\begingroup$ It's a nitpick about the post. Not about the math. It is essential that someone indicated that $X$ is a universal or at least a superset. And I assume someone did. This is a nitpick for the poster for (probably-- I'm giving the benefit of the doubt) assuming it was self-evident and not bothering to point it out. But you are right. The necessity is not a nitpick. $\endgroup$ – fleablood Feb 18 at 22:17

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