Does any relation $\propto\,\subseteq X\times \mathcal P(X)$ that extends $'\!\!\in'$ in the way that:

  1. $x\in M\Rightarrow x\propto M$
  2. $\neg\exists x\in X:x\propto\emptyset$
  3. $x\propto A \subseteq B \Rightarrow x\propto B$
  4. $x\propto A\cup B\Rightarrow x\propto A \vee x\propto B$

defines a closure operation on subsets of $X$ by $x\in \overline M \Leftrightarrow x\propto M$?

My approach is: all the sets $\overline M$ satisfies the axioms for closed sets:

  • $\emptyset$ is closed since $x\in\overline\emptyset\Leftrightarrow x\propto\emptyset \Leftrightarrow x\in\emptyset$
  • $X$ is closed since $x\in \overline X\Leftrightarrow x\propto X\Leftrightarrow x\in X$
  • $x\in \overline{A\cup B}\Leftrightarrow x\propto A\cup B\Leftrightarrow x\propto A \vee x\propto B\Leftrightarrow x\in\overline A\cup\overline B\quad$ (3 & 4)
  • $\displaystyle x\in\overline{\bigcap_iM_i}\Leftrightarrow x\propto\bigcap_iM_i\Leftrightarrow\forall i: x\propto M_i\Leftrightarrow\forall i: x\in \overline{M}_i\Leftrightarrow x\in\bigcap_i\overline{M}_i\quad$ (3)$\;\square$

I become unsure of the method when Hagen von Eitzen showed that a closure operation on $\mathcal P(X)$ I had defined with this method was faulty.

  • 1
    $\begingroup$ There are some known axioms for closure operators and idempotent closure operators. Is it $\overline A\subseteq B$ $\Rightarrow$ $\overline A\subseteq\overline B$? I think your conditions 1, 2, 4 can be expressed as $M\subseteq\overline M$, $\overline\emptyset=\emptyset$, $\overline{A\cup B}\subseteq \overline A\cup\overline B$. I am not sure how to rephrase the condition 3 in a way which does not use $\propto$ but is expressed only using the closure operator. $\endgroup$ – Martin Sleziak Jan 15 '15 at 11:33
  • $\begingroup$ Ok, 3 seems to be $\overline A\subseteq B$ $\Rightarrow$ $A\subseteq B$. $\endgroup$ – Martin Sleziak Jan 15 '15 at 14:42
  • $\begingroup$ May I ask how do we know that $(\forall i) x \propto M_i$ implies $x\propto \bigcap M_i$? (I see that the other implication follows from the 3rd condition, but I do not see how to get this one. This is part of equivalence in the last line of your proof.) $\endgroup$ – Martin Sleziak Jan 15 '15 at 14:47
  • $\begingroup$ @MartinSleziak: Thank you! It is a faulty equivalence - the equivalence of the lazy... I've become convinced that the four axioms of mine isn't enough to prove $\bar{\bar M}\subseteq \bar M$. There must be a fifth axiom - at least. $\endgroup$ – Lehs Jan 15 '15 at 17:31
  • $\begingroup$ @MartinSleziak: I'm unsure about your 3'th. As I read my axiom 3 it could be reformulated as $x\in\bar A\wedge A\subseteq B\implies x\in \bar B$. $\endgroup$ – Lehs Jan 15 '15 at 19:42

Your fourth argument is incorrect. As Martin Sleziak mentioned in the comments, $x\propto M_i$ for each $i\in I$ does not imply that $x\propto\bigcap_iM_i$. Consider the relation $\propto$ defined on $\Bbb R\times\wp(\Bbb R)$ by $x\propto A$ iff $x\in\operatorname{cl}A$, where the closure is taken in the usual Euclidean topology; this relation evidently satisfies all of your axioms. However, if we set $M_0=(0,1)$ and $M_1=(1,2)$, then $1\propto M_0$ and $1\propto M_1$, but $1\not\propto\varnothing=M_0\cap M_1$.

Note that this doesn’t mean that $M\mapsto\overline M$ isn’t a closure operator: the ordinary closure operator in $\Bbb R$ doesn’t satisfy your fourth condition, either. The real question is whether your axioms imply that if $x\propto\overline M$, then $x\in\overline M$, so that $M\mapsto\overline M$ is idempotent. Unfortunately, they don’t.

Define $\propto\subseteq\Bbb N\times\wp(\Bbb N)$ as follows: $n\propto A$ if and only if either

  • $n\in A$, or
  • $A\ne\varnothing$ and $n=\min(\Bbb N\setminus A)$.

It’s straightforward to check that $\propto$ satisfies conditions (1)-(4). However, $\overline{\{0\}}=\{0,1\}$, but $\overline{\{0,1\}\}}=\{0,1,2\}\ne\{0,1\}$: $2\propto\overline{\{0\}}$, but $2\notin\overline{\{0\}}$.

  • $\begingroup$ Yes I agree that the axioms isn't enough. But if I add the axiom $x,y\propto M\implies x\propto M\cup\{y\}$? Or did I misunderstand? $\endgroup$ – Lehs Jan 15 '15 at 17:26
  • $\begingroup$ @Lehs: That’s won’t work: ordinary closure doesn’t satisfy it (take $M=(0,1)$, $x=0$, $y=1$), and my example with $\Bbb N$ does. $\endgroup$ – Brian M. Scott Jan 15 '15 at 17:35
  • $\begingroup$ @Lehs: I was talking about two different examples. My first example, using ordinary closure in $\Bbb R$, does not satisfy your suggested property. My second example, using $\Bbb N$, does satisfy it, even though it doesn’t give you a closure operator. $\endgroup$ – Brian M. Scott Jan 15 '15 at 17:48
  • $\begingroup$ Yes I saw that. I must check this and see if I can get it. Thanks a lot! $\endgroup$ – Lehs Jan 15 '15 at 17:50
  • $\begingroup$ @Lehs: You’re welcome! $\endgroup$ – Brian M. Scott Jan 15 '15 at 17:51

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