Talk:Praclosure operator

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[edit] Topology or not

The collection of all open sets generated by the praclosure operator is a topology. Really? If you mean that this collection of 'open sets' is automatically closed under unions, I don't see why. It will be closed under binary intersection, from the axioms. Charles Matthews 14:49, 26 November 2006 (UTC)

I am not sure I know what you're getting at. The article defines a set A to be closed if and only if

A = [A] p

and it defines it to be open if and only if $X\setminus A$ is closed.

To demonstrate a topology, I need to show that if U, V are open by the above definition, then $U\cup V$ and $U\cap V$ are also open. This is more or less straightforward, (I did verify this, as I was concerned, and can reproduce the proof here if it helps). The one part that I did not do is to verify that, even under an infinite bunch of unions, the result is still an open set. I am not sure I know how to do that. Is that the objection?

FWIW, I cribbed most of the article while sitting with the given reference in my lap; that book makes the claim that the collection of all open sets generated by the praclosure operator is a topology. That this collides with the article on pretopological space is what concerned me. linas 02:58, 28 November 2006 (UTC)

[edit] Proof

I'm going to write this down before chucking my scrap of paper in the trash, because I'd be in pain if I had to re-invent this. If U, V are open, then proving that $U\cap V$ is open is easy (so I won't write it here). The proof that $U\cup V$ is open is a little bit harder. It depends on the following lemmas:

Lemma 1: One has $[A\cap B] \subseteq [A]\cap [B]$ which holds for any A, B.

Proof: Note that

$[A]=[A]\cup [A\cap B]$

and

$[B]=[B]\cup [A\cap B]$

Right? Yes, since one has

$\begin{align} \;[A]\cup [A\cap B] &= [A\cup (A\cap B)] \\ &= [(A\cup A) \cap (A\cup B)] \\ &= [A \cap (A\cup B)] \\ &= [A] \end{align}$

Soo ... Combining and distributing:

$\begin{align} \;[A]\cap [B] &= ([A]\cup [A\cap B]) \cap ([B]\cup [A\cap B]) \\ &= ([A]\cap [B]) \cup ([A]\cap [A\cap B])\cup ([B]\cap [A\cap B]) \cup ([A\cap B]\cap [A\cap B]) \\ &= ([A]\cap [B]) \cup [A\cap B] \end{align}$