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Unraveling the definitions, I keep getting that Cl(A) = A'.

[tex]x\in \overline{A}[/tex]

[tex]\Leftrightarrow (\forall U\in \tau)[(A\subseteq X\setminus U)\Rightarrow (x\in X\setminus U)][/tex]

[tex]\Leftrightarrow (\forall U\in \tau)[\neg (x\in X\setminus U)\Rightarrow \neg(A\subseteq X\setminus U) ][/tex]

[tex]\Leftrightarrow (\forall U\in \tau)[(x\in U)\Rightarrow (A \cap U \neq \emptyset ) ][/tex]

[tex]\Leftrightarrow x\in A'.[/tex]

(The empty set is its own closure, so if x is in A, then A is not empty.)

I suspect the problem may lie in the substitution

[tex]\neg (x\in X\setminus U) \Leftrightarrow \neg((x\in X)\& \neg(x\in U))[/tex]

[tex]\Leftrightarrow \neg(x\in X) \vee (x\in U)[/tex]

[tex]\Leftrightarrow (\forall x\in X) [x\in U].[/tex]

On it's own, the final step of deleting this "for all X" looks sound to me (we're already implicitly talking about all x in X, so why do we need to consider the possibility that x is not in X?), but in the above context, I've moved from "for all x in X, if P is true or x is in X, then ..." (which is true of all x in X) to "for all x in X, if P is true ..." (which is not necessarily true of all x in X).

Can anyone help me understand what the logical rule is here? (I.e. if this isn't a legitimate substitution, what general rule makes it illegitimate.) Is this why I'm getting the anomalous result that Cl(A) = A'?

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# Cl(A) = A' ?

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