Uncovering the Logic Behind Cl(A) = A

[Rasalhague]

Cl(A) = A' ??

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

$$x\in \overline{A}$$

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

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

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

$$\Leftrightarrow x\in A'.$$

(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

$$\neg (x\in X\setminus U) \Leftrightarrow \neg((x\in X)\& \neg(x\in U))$$

$$\Leftrightarrow \neg(x\in X) \vee (x\in U)$$

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

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'?

 

[micromass]

What is the definition of A' ??

 

[Rasalhague]

Ooh, ooh, I see the mistake! I was garbling the definition of A':

$$\left \{ x | (\forall U)[(x\in U)\Rightarrow (U\setminus \left \{ x \right \} \cap A \neq \emptyset)] \right \},$$

rather than simply

$$\left \{ x | (\forall U)[(x\in U)\Rightarrow (U \cap A \neq \emptyset)] \right \}.$$

So actually $\overline{A}=A\cup A'$. If x is in the closure of A, either x is a limit point of A, or x belongs to A.

 

[micromass]

Aah yes, I found your A' a bit weird in the OP Good you found the mistake!

 

[Rasalhague]

It dawned on me just before I read your hint! Thanks, micromass - ever ready to spring to my rescue : )

Did my question about the substitution make sense?

 

[micromass]

Yeah. Your substitution looks ok to me.

 

[Rasalhague]

And I see why now, at last! The statement is actually of the form $P\vee Q \Rightarrow R$, where P is not true. That being the case, the antecedent is equivalent to Q. I think it was the double negation that confused me.