Intuition behind the $cl(cl(A))=cl(A)$ Kuratowski closure axiom in general topology

general-topologyintuition

If we use the real numbers as a model of topology (with the standard topology), then it is obvious why $cl(cl(A))=cl(A)$ should hold, given the definition of "closure of a set" in the context of the real numbers.

But what a good intuition for why this axiom should hold in the context of general topology?

I have an intuition for all 3 other axioms:

  1. $cl(\emptyset)=\emptyset$ (intuition: no objects touch the empty set, because you can't touch a set consisting of no objects)

  2. $A\subseteq cl(A)$ (intuition: every element of $A$ touches $A$, because elements "touch themselves").

  3. $cl(A\cup B) = cl(A)\cup cl(B)$ (intuition: if an object touches a set consisting of two parts, then it must at least touch one of those parts, because if you don't touch either of two things, you can't suddenly touch the combination of the two).

  4. $cl(cl(A))=cl(A)$ (intuition: if a set of objects $B$ all touch $A$, then any object that touches $B$ must also touch $A$, but this doesn't feel right to me. it feels too restrictive).

What is a good intuition for why nr 4 should hold for topologies, given that topologies formalize the notion of "touch-relations"?

Best Answer

$\DeclareMathOperator\cl{cl}$In terms of metric spaces, the statement "$x\in\cl(A)$" that you want to think of as "$x$ touches $A$" becomes "the distance of $x$ to $A$ is $0$". Here the distance $x$ has to $A$ is defined as $$ \operatorname{dist}(x,A) = \inf_{a\in A} d(x,a). $$ Now take $x\in\cl(\cl(A))$ and let's figure out what this means in terms of distance. The point $x$ has a distance of $0$ to the set of all points having distance $0$ to $A$. Now in general for three points $x,y,z$ we have $$ d(x,z) \le d(x,y) + d(x,z). $$ So if $x$ has distance $0$ to all points $y$ having distance $0$ to $A$, these distances will at most sum up and $x$ has distance at most $0+0=0$ to $A$. We conclude that $x$ is indeed touching $A$ and hence $$ \cl(\cl(A)) \subseteq \cl(A). $$ The other inclusion is given by property 2.

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