# Touch-axiomatization of point-set topology

This MathOverflow post presents an alternative axiomatization of point-set topology that closely aligns with the intuitive notion that a topology on a set tells you which points are "infinitesimally close" to each other. The relevant part of the post is paraphrased below.

Definition: A touch relation on a set $X$ is a binary relation $\lessdot$ between the members of $X$ and the subsets of $X$ (we read $x \lessdot A$ as "$x$ touches $A$") that satisfies the following axioms:

• No element of $X$ touches the empty set.
• If $x \in A$, then $x \lessdot A$.
• If $x \lessdot (A \cup B)$, then $x \lessdot A$ or $x \lessdot B$.
• If $x \lessdot A$ and every element of $A$ touches $B$, then $x \lessdot B$.

I would like to prove that these axioms are indeed equivalent to the usual axioms for a topology in terms of open sets. To do this, I need to exhibit for any given set $X$ a bijective correspondence between touch relations on $X$ and topologies on $X$. I believe the correct correspondence is as follows:

• Given a touch relation on $X$, we obtain a topology on $X$ by declaring $A \subseteq X$ to be open if and only if no member of $A$ touches $X \setminus A$.
• Given a topology on $X$, we obtain a touch relation on $X$ by declaring that $x \lessdot A$ if and only if $x \in \overline{A}$, where $\overline{A}$ denotes the topological clousure (intersection of all closed supersets) of $A$.

So far I have been able to show that both of these maps are well-defined (i.e., that they do produce a topology from a touch relation and vice versa). I can also show that the first map is a left inverse of the second (i.e., by starting from a topology $T$ on $X$, constructing the touch relation $\lessdot$ induced by $T$, and then constructing the topology $T'$ induced by $\lessdot$, we always have that $T = T'$). To finish off the proof, I need to show that the first map is a right inverse of the second, which I haven't yet figured out how to do.

If I've unwound the definitions correctly, showing that the touch relation $\to$ topology $\to$ touch relation round-trip is the identity map amounts to proving the following:

If $\lessdot$ is a touch relation on $X$, then $x \lessdot A$ if and only if $x$ is a member of every superset $B$ of $A$ with the property that $b \lessdot B \implies b \in B$. (Let's call such a set $B$ "touch-closed.")

The left-to-right implication is easy, but the right-to-left implication eludes me.

Question: Let $\lessdot$ be a touch relation on $X$, and let $A \subseteq X$ be given. Suppose that $x$ is a member of every touch-closed superset of $A$. How can I prove that $x \lessdot A$?

My partial proof attempt follows.

Proof: Let $C$ be the intersection of every touch-closed superset of $A$. By hypothesis, $x \in C$, so $x \lessdot C$. Using the fourth touch axiom, it suffices to show that every element of $C$ touches $A$. To do this, suppose the the sake of contradiction that some element $c \in C$ does not touch $A$. Then I claim that there exists a touch-closed superset $B$ of $A$ not containing $c$. Indeed...

• Would $B = C \setminus \{c\}$ work? – Theoretical Economist Dec 30 '17 at 21:32
• @TheoreticalEconomist Huh, I thought for some reason $B = C \setminus \{c\}$ wouldn't work, but looking at it again it definitely does. Thanks! – David Zhang Dec 30 '17 at 21:47
• I suspected it did, but didn’t have the time to work it out properly. Reading jgon’s answer below convinces me that it should though. Glad it works! Thanks for sharing this great question. – Theoretical Economist Dec 30 '17 at 21:50
• This would seem very close to the "neighborhood relation" formulation of a topology, as a relation between points and sets: $N$ is a neighborhood of $x$ if and only if $x$ does not touch the complement of $N$. I view the neighborhood formulation as being equivalent to defining some quantifier called "for close enough", satisfying: (1) if for all $y$, $\phi(y)$: then for all $x$, for $y$ close enough to $x$, $\phi(y)$; (2) if for $y$ close enough to $x$, $\phi(y)$; and for $y$ close enough to $x$, $\psi(y)$: then for $y$ close enough to $x$, $\phi(y) \wedge \psi(y)$; ... – Daniel Schepler Feb 1 at 23:31
• (3) if for $y$ close enough to $x$, $\phi(y)$; and for $y$ close enough to $x$, $\phi(y) \rightarrow \psi(y)$: then for $y$ close enough to $x$, $\psi(y)$; (4) if for $y$ close enough to $x$, $\phi(y)$, then: for $z$ close enough to $x$, for $y$ close enough to $z$, $\phi(y)$. – Daniel Schepler Feb 1 at 23:32

## 1 Answer

Let $$C = \newcommand{\set}{\left\{{#1}\right\}}\set{x:x\lessdot A}$$. By axiom (ii), this is a superset of $$A$$, by axiom (iv), it is touch closed. Clearly it is a subset of every touch closed superset of $$A$$, so it is the intersection of all touch closed supersets of $$A$$. Thus we have an equality of sets that shows the desired equivalence that $$x\lessdot A$$ if and only if $$x$$ is in every touch closed superset of $$A$$.

Edit:

My original answer to the question is above. user56834 has offered a bounty to give a precise statement and proof that the touch axiomatization is equivalent to the classical axiomatization of point set topology, and it's kind of an interesting question, which I've already partially answered, so I thought I'd complete this answer. In the interests of a self contained answer, I'll include the definition from the question.

Answer to bounty question

Definition: A touch relation on a set $$X$$ is a binary relation $$\lessdot$$ between the members of $$X$$ and the subsets of $$X$$ (we read $$x \lessdot A$$ as "$$x$$ touches $$A$$") that satisfies the following axioms:

1. No element of $$X$$ touches the empty set.
2. If $$x \in A$$, then $$x \lessdot A$$.
3. If $$x \lessdot A \cup B$$, then $$x \lessdot A$$ or $$x \lessdot B$$.
4. If $$x \lessdot A$$ and every element of $$A$$ touches $$B$$, then $$x \lessdot B$$.

The definition I will show is equivalent

There are many already well known equivalent axiomatizations for point set topology. The easiest one to connect to touch relations is the closure operator axiomatization (due to Kuratowski) see the link to Wiki for the connection between the closure operator axiomatization and the closed set axiomatization.

Definition: A closure operator on a set $$X$$ is a map $$\newcommand\calP{\mathcal{P}}\newcommand\cl{\operatorname{cl}}\cl:\calP(X)\to\calP(X)$$ such that

1. $$\cl(\varnothing)=\varnothing$$
2. $$A\subseteq \cl(A)$$
3. $$\cl(A\cup B) =\cl(A)\cup\cl(B)$$
4. $$\cl(\cl(A))=\cl(A)$$

Note that axiom 3 implies that closure operators are order preserving, since if $$A\subseteq B$$, then $$A\cup B=B$$, so $$\cl(B)=\cl(A\cup B)=\cl(A)\cup\cl(B)$$, which implies that $$\cl(A)\subseteq \cl(B)$$.

The equivalence

From comparing the axioms, it should be clear that the axioms for touch relations are a fairly direct translation of Kuratowski's axioms for closure operators. Here is the explicit correspondence.

Given a closure operator $$\cl$$, the relation $$x \lessdot A$$ if $$x\in \cl(A)$$ is a touch relation, and given a touch relation, $$\lessdot$$, the operator $$\cl(A) = \set{x\in X: x\lessdot A}$$ is a closure operator. Moreover, these correspondences are inverses of each other, yielding a bijection between touch relations and closure operators on $$X$$.

Proof:

A note on notation. I will refer to the touch axioms as T1-T4 and the closure axioms as K1-K4 (the K standing for Kuratowski).

Let $$\cl$$ be a closure operator. Then the associated touch relation satisfies

1. No $$x$$ touches $$\varnothing$$, since there are no $$x$$ in $$\cl(\varnothing)=\varnothing$$ by K1.
2. If $$x\in A$$, then $$x\lessdot A$$, since $$x\in A$$ implies $$x\in \cl(A)$$ by K2.
3. If $$x\lessdot A\cup B$$, then $$x\lessdot A$$ or $$x\lessdot B$$, since by K3, $$\cl(A\cup B)=\cl(A)\cup \cl(B)$$, so if $$x\in \cl(A\cup B)$$, then $$x\in \cl(A)$$ or $$x\in\cl(B)$$.
4. If $$x\lessdot A$$ and $$a\lessdot B$$ for all $$a\in A$$, then $$x\lessdot B$$, since if $$a\lessdot B$$ for all $$a\in A$$, then $$A\subseteq \cl(B)$$, so $$\cl(A)\subseteq \cl(\cl(B))=\cl(B)$$ since closure operators are order preserving, and by K4. Thus we have that if $$x\lessdot A$$, then $$x\lessdot B$$.

Conversely, given a touch relation $$\lessdot$$, the associated closure operator satisfies the Kuratowski axioms. Before proving that, note that T2 and T4 imply that that associated closure operator is order preserving, since if $$A\subseteq B$$, then T2 says that for all $$a\in A$$, $$a\lessdot B$$, and thus T4 says that if $$x\lessdot A$$, then $$x\lessdot B$$. Hence $$\cl(A)\subseteq \cl(B)$$ if $$A\subseteq B$$.

Now to prove that the Kuratowski axioms hold:

1. $$\cl(\varnothing)=\varnothing$$, since nothing touches the empty set by T1.
2. $$A\subset \cl(A)$$ by T2.
3. $$\cl(A\cup B)=\cl(A)\cup \cl(B)$$, since by T3, we have $$\cl(A\cup B)\subseteq \cl(A)\cup\cl(B)$$, and conversely, since the closure operator is order preserving, $$\cl(A)\subseteq \cl(A\cup B)$$ and $$\cl(B)\subseteq \cl(A\cup B)$$. Thus $$\cl(A)\cup\cl(B)\subseteq \cl(A\cup B)$$ as well.
4. $$\cl(\cl(A))=\cl(A)$$, by T4, since for all $$x\in \cl(A)$$, $$x\lessdot A$$ by definition, so if $$y\lessdot \cl(A)$$, then $$y\lessdot A$$. Hence $$\cl(\cl(A))\subseteq \cl(A)$$, and we already know $$\cl(A)\subseteq \cl(\cl(A))$$ by the proof of K2 above.

Finally, I should justify that these correspondences are inverses of each other, although it's fairly straightforward.

Starting with a touch relation, we define $$\cl(A)=\set{x\in X : x\lessdot A}$$, and then we define $$x\lessdot' A$$ by $$x\in \cl(A)$$. However $$x\in\cl(A)$$ if and only if $$x \lessdot A$$, so $$x\lessdot' A$$ if and only if $$x\lessdot A$$, so $$\lessdot'=\lessdot$$.

Conversely, starting with a closure operator, we define $$x\lessdot A$$ by $$x\in\cl(A)$$, and then a new closure operator by $$\cl'(A)=\set{x\in X : x\lessdot A}$$. But then $$x\in\cl'(A) \iff x\lessdot A \iff x\in\cl(A)$$, so $$\cl'(A)=\cl(A)$$ for all $$A$$, so $$\cl'=\cl$$.

Thus the correspondences are inverses of each other. $$\blacksquare$$

Side note

Wiki already has an article for touch relations, calling it the "closeness relation", but it doesn't note that such a relation is equivalent to a topology.

Further edit

In response to the comments below, I've decided to state and prove the direct equivalence between touch relations and open sets.

First, recall

Definition: $$\tau\subseteq \calP(X)$$ is a topology on $$X$$ and its elements are called open subsets of $$X$$ if

1. $$\varnothing,X\in \tau$$ (the null set and $$X$$ are open)
2. $$\bigcup_{\alpha\in I} U_\alpha \in \tau$$, where $$I$$ is any index set and $$U_\alpha\in \tau$$ for all $$\alpha\in I$$. (arbitrary unions of opens are open)
3. $$\bigcap_{i=1}^n U_i \in \tau$$ for all $$n\ge 0$$ when all of the $$U_i$$ are in $$\tau$$. (finite intersections of opens are open)

Note: With the axiomatization I just gave, axiom 1 is redundant, since the null set is the empty union and the whole set is the empty intersection.

The equivalence between touch relations and open sets

Given a touch relation $$\lessdot$$, the collection of sets $$U$$ satisfying $$x\not\lessdot U^C$$ for all $$x\in U$$ gives a topology on $$X$$. Conversely, given a topology on $$X$$, the relation $$x\lessdot A$$ if $$x\in \cl(A)$$ defines a touch relation on $$X$$ (in terms of open sets, this relation is that $$x\lessdot A$$ if there is no open $$U$$ with $$x\in U$$ and $$U\cap A =\varnothing$$). Moreover, these correspondences are inverses of each other.

Proof:

First, suppose that $$\lessdot$$ is a touch relation on $$X$$. We need to show that the definition of open given in the statement defines a topology on $$X$$.

Let $$U = \bigcup_{\alpha\in I} U_\alpha,$$ with all $$U_\alpha$$ open, and suppose $$x\lessdot U^C$$, then since $$U^C\subseteq U_\alpha^C$$ for all $$\alpha$$, by T2 and T4, we have that $$x\lessdot U_\alpha^C$$ for all $$\alpha$$, so $$x\not \in U_\alpha$$ for all $$\alpha$$, and hence $$x\not\in U$$. Thus $$U$$ is open.

For intersections, it suffices to show that the empty intersection and the binary intersection of opens are open. Now the empty intersection is $$X$$, and $$X^C=\varnothing$$, so there are no $$x$$ with $$x\lessdot X^C$$ by T1, and thus $$X$$ is open. Finally, by T3, if $$x\lessdot (U\cap V)^C = U^C\cup V^C$$, then either $$x\lessdot U^C$$ or $$X\lessdot V^C$$. Either way, when $$U$$ and $$V$$ are open, either $$x\not\in U$$ or $$x\not\in V$$, so $$x\not\in U\cap V$$. Hence if $$U$$ and $$V$$ are open, $$U\cap V$$ is open. Thus a touch relation defines a topology.

Conversely, given a topology, we know that $$x\in \cl(A)$$ defines a touch relation, by the argument I gave earlier, since $$\cl$$ is a closure operator. However, if I restrict to only talking about open sets, we can define $$x\lessdot A$$ if for all open sets $$U$$, $$x\in U$$ implies that $$U\cap A\ne \varnothing$$.

Now we'll prove that this satisfies the touch relation axioms.

1. Regardless of $$x$$, for all open sets $$U$$, $$U\cap \varnothing =\varnothing$$, so no element of $$X$$ touches the empty set.
2. If $$x\in A$$, then if $$x\in U$$, $$x\in U\cap A$$, so $$U\cap A\ne \varnothing$$, and thus if $$x\in A$$, $$x\lessdot A$$.
3. Suppose $$x\lessdot A\cup B$$, then for all open $$U$$ containing $$x$$, $$U\cap (A\cup B) = (U\cap A)\cup (U\cap B)\ne \varnothing$$. Now suppose $$x\not\lessdot A$$, so there exists an open set $$V$$ with $$V\cap A=\varnothing$$. Now for any open set $$U$$, we know that $$(U\cap V)\cap (A\cup B) = (U\cap V\cap A) \cup (U\cap V\cap B) = U\cap V\cap B \ne \varnothing,$$ and thus $$U\cap B\ne\varnothing$$. Hence if $$x\not\lessdot A$$, then $$x\lessdot B$$, and therefore T3 holds.
4. Finally, if $$x\lessdot A$$, and for all $$a\in A$$, $$a\lessdot B$$, we need to show that $$x\lessdot B$$. However, if $$x\not\lessdot B$$, then there exists an open $$U$$ with $$x\in U$$ and $$U\cap B=\varnothing$$. However, since $$x\lessdot A$$, $$U\cap A \ne \varnothing$$. Then if $$a\in U\cap A$$, we have that $$U$$ is an open set containing $$a$$ and disjoint from $$B$$, contradicting the fact that $$a\lessdot B$$. Hence T4 holds.

Finally, these should be inverse correspondences. This is immediate from the fact that the correspondences between touch relations and closure operators were inverses, since all this correspondence is is the composition of that correspondence with the correspondence between closure operators and open sets. However, working it out explicitly, we have.

Suppose we start with a touch relation $$\lessdot$$. Then we obtain a topology $$\tau$$ and a new touch relation $$\lessdot'$$ in the manner described above. We need to show that $$x\lessdot A$$ if and only if $$x\lessdot' A$$.

Suppose $$x\not \lessdot A$$, then define $$U=\set{x\in X : x\not\lessdot A}$$. $$U\cap A=\varnothing$$, by T2, and $$x\in U$$ by assumption, and $$U$$ is open, since if $$x\lessdot U^C$$, then $$x\lessdot A$$ by T4, so $$x\not\in U$$. This $$U$$ shows that $$x\not\lessdot' A$$.

Conversely, if $$x\lessdot A$$, then if $$U$$ is open and $$x\in U$$, then supposing that for all $$a\in A$$, $$a\in U^C$$, then by T4, we'd have $$x\lessdot U^C$$, contradicting that $$x\in U$$ and $$U$$ is open. Hence for some $$a\in A$$, $$a\in U$$. Since $$U\cap A\ne \varnothing$$ for all open $$U$$ containing $$x$$, we have $$x\lessdot' A$$. Thus $$\lessdot=\lessdot'$$ as desired.

On the other hand, suppose we start with a topology $$\tau$$. Let $$\lessdot$$ be the associated touch relation, and $$\tau'$$ the topology associated to this touch relation. We need to show that $$\tau=\tau'$$.

If $$U\in\tau$$, then for all $$x\in U$$, $$x\not\lessdot U^C$$, since $$U$$ is an open set containing $$x$$ with $$U\cap U^C=\varnothing$$. Thus $$U\in \tau'$$, and $$\tau\subseteq \tau'$$.

On the other hand, if $$U\in \tau'$$, then for all $$x\in U$$, $$x\not\lessdot U^C$$, so there exists $$V_x\in\tau$$ with $$x\in V_x$$, and $$V_x\cap U^C=\varnothing$$. Then $$U=\bigcup_{x\in U} V_x,$$ so $$U\in \tau$$. Hence $$\tau=\tau'$$ as desired. $$\blacksquare$$

Side comment

After some more thinking, I realized that the touch relation fits neatly into a rough characterization of the common axiomatizations of point set topology.

Ignoring the "partial" definitions (i.e. things like bases and nhood bases), we can categorize them in the following way: $$\begin{array}{ccc} \text{global} & \text{operator} & \text{local} \\ \hline \text{open sets} & \text{interior operator} & \text{neighborhoods} \\ \text{closed sets} & \text{closure operator} & \text{touch relation} \\ \end{array}$$

The parallel between neighborhoods and touch relations is that we can think of specifying the neighborhoods of a point as giving a relation $$x \prec N$$ if $$N$$ is a neighborhood of $$x$$. Then to convert between $$\prec$$ and $$\lessdot$$, we say $$x\lessdot A$$ if $$x\not\prec A^C$$ and $$x\prec N$$ if $$x\not \lessdot N^C$$.

• I feel somewhat conflicted about this answer, because on the one hand it's great (it states everything clearly and makes a clear connection), but on the other hand, by "the classical axiomatization" I really meant the one based on open sets. (though I admit that that was ambiguous). – user56834 Feb 2 at 7:24
• Though the one based on neighborhoods would also satisfy what I meant by "classical axiomatization" (but again, I admit this was not clear from my bounty description). – user56834 Feb 2 at 7:42
• @user56834 Edited to add the equivalence to the open set axiomatization. – jgon Feb 2 at 16:04
• amazing. I will award the bounty in one hour (can't do it now). – user56834 Feb 2 at 16:17
• I have re-read the answer, and I suspect this part may be wrong: In the proof that the closure operator induced by a touch relation satisfies axiom $K3$: "and conversely, since the closure operator is order preserving, ..." My problem with this is, that this assumes that the closure operator induced by a touch relation is actually a closure operator (since you deduced that the closure operator is order preserving from $K3$ itself). So it's circular reasoning. I suspect that maybe axiom $T3$ is wrong, and should be "x touches $A\cup B$ if and only if $x$ touches $A$ and $x$ touches $B$". – user56834 Feb 8 at 10:27