This is almost certainly not new, but it’s something I’ve been thinking about recently and haven’t seen before (it’s pretty close to a treatment in terms of closure operators, which I have seen before but not as a pedagogical approach).

Normally a topology is defined as a family of open sets. This is unfortunate, as open sets are in many ways the least intuitive topological object when you first encounter them. I was thinking in terms of what a more intuitive way of defining topology would be and I hit upon the following approach.

Given a set X, a topology is a binary relationship between points in X and subsets of X. We will call this relation “touches”.

The intuitive idea is that x touches A if A contains points “arbitrarily close to” x, a term which we will leave undefined for the moment and which is only intended to be an intuitive justification. For example, if you consider the open interval \((0, 1) \subseteq \mathbb{R}\), this touches the points 0 and 1 despite not containing them, but it does not touch the point 2.

We require this relationship to obey the following axioms:

1. Nothing touches the empty set
2. If \(x \in A\) then \(x\) touches \(A\)
3. If \(x\) touches \(A\) and \(A \subseteq B\) then \(x\) touches \(B\)
4. If x touches \(A \cup B\) then \(x\) touches \(A\) or \(B\)
5. If every point \(x \in B\) touches \(A\) and \(y\) touches \(B\) then \(y\) touches \(A\)

The last two are hopefully the only ones that aren’t immediately obvious. I don’t really have a good justification for 4 at the moment, but hopefully it’s not too unreasonable. The reasoning behind 5 is to exclude things where you can “make the set much larger” by adding in nearby points.

For example, you could consider \(X = \mathbb{N}\), and that \(x\) touches \(A\) if it’s within distance 1 of some element of \(A\). Then by repeatedly adding in points you would eventually fill up the whole set. This is something we want to avoid.

A topological space is then a set X together with this touches relation. We will tend to abuse notation by referring to the set as identical with the set and letting the topology be assumed.

I’m now going to show how this definition relates to the normal ones.

Let \(X\) be a topological space defined as above. Say that a set \(A \subseteq X\) is closed if \(x \in A\) whenever \(x\) touches \(A\)

We’ll prove the following properties of closed sets:

Proposition: \(\emptyset\) and \(X\) are both closed.

This is true almost by definition. \(X\) certainly contains every point in the space, so contains every point it touches. No points touch the empty set and thus every point it touches is contained in it.

Proposition: If \(\{ A_i : i \in \mathcal{I} \}\) are all closed, then so is \(\bigcap_{i \in \mathcal{I}} A_i \)

Proof: Suppose \(x\) touches \(\bigcap_{i \in \mathcal{I}} A_i \). For any \(i\) this set is a subset of \(A_i\), and thus by axiom 3, \(x\) also touches \(A_i\). Thus \(\forall i. x \in A_i \). Therefore \(x \in \bigcap_{i \in \mathcal{I}} A_i \).

Proposition: If \(A, B\) are closed then so is \(A \cup B\).

Proof: If \(x\) touches \(A \cup B\) then by axiom 4 it touches at least one of \(A\) or \(B\) and thus because they are closed is contained in at least one of \(A\) or \(B\). Therefore \(x \in A \cup B\).

So our closed sets satisfy all the usual conditions for closed sets of a normally defined topological space, thus their complements satisfy all the axioms for open sets in a topological space.

Now how do we recover the “touches” relationship from a normal topological space?

Well if we define the closure of a set as normal as the smallest closed set containing it, i.e.

\[ cl(A) = \bigcap \{ B \supseteq A : B \text{ is closed } \} \]

Then we can define a touches relationship as \(x\) touches \(A\) if \(x \in cl(A)\).

Does this satisfy all our axioms?

First we must prove the following properties of closure operators:

1. \(A = cl(A)\) iff \(A\) is closed
2. \(A \subseteq cl(A) \)
3. \(A \subseteq B \) implies \(cl(A) \subseteq cl(B)\)
4. \(cl(A \cup B = cl(A) \cup cl(B)\)
5. \(cl(cl(A)) = cl(A)\)

The first three properties are pretty trivial consequences of the definitions and the properties of closed sets.

To prove 4: First note that \(cl(A) \cup cl(B)\) is a closed set containing \(A \cup B\), so certainly \(cl(A \cup B) \subseteq cl(A) \cup cl(B)\). Further because \(A, B \subseteq A \cup B\) we have \(cl(A), cl(B), \subseteq cl(A \cup B)\), so \(cl(A) \cup cl(B) \subseteq cl(A \cup B)\). So we’ve proved the inclusion both ways and the two are equal.

To prove 5: This is a simple consequence of the fact that the closure of a set is closed.

We can now show that the touches relationship we’ve defined obeys all the properties we asked for.

1. \(\emptyset\) is closed, so \(cl(\emptyset) = \emptyset\) and no points touch it.
2. \(A \subseteq cl(A)\), so every point in \(A\) touches \(A\)
3. \(cl(A) \subseteq cl(B)\) so if \(x \in cl(A)\) then \(x \in cl(B)\)
4. If \(x \in cl(A \cup B\)\) then \(x \in cl(A) \cup cl(B)\) so \(x \in cl(A)\) or \(x \in cl(B)\) so \(x\) touches \(A\) or \(B\)
5. If every \(x \in B \) touches \(A\) then \(B \subseteq cl
   (A)\) so \(cl(B) \subseteq cl(cl(A)) = cl(A)\). So if \(y\) touches \(B\) then \(y \in cl(B)\) and so \(y\) touches \(A\)

So we get a proper touches relation back as we wanted.

Now the next question is: Do these define the same thing? If we start from a touches relation and go to a set of closed sets then back again, do we get the same relation back? And vice versa?

The answer is still yes. Time for more definition chasing!

We’ve already seen that a set is closed iff it is equal to its closure, and that we can construct a touches relation from the closure. So if we show that the touches relation also uniquely defines the closure operator then we’re done.

So for this we need a theorem:

Given the closed sets constructed from our touches operator, \(cl(A) = \{ x : x \text{ touches } A \} \)

Proof:

This is where we will finally have to use our fifth axiom for touching (we haven’t so far). In order to see that it’s necessary, consider our example with \(\mathbb{N}\) and touching meaning “within distance 1”. Then in this case the closed sets we will generate will be only the sets \(\emptyset\) and \(\mathbb{N}\), so the touches relationship we will get back is that every point touches every set.

Now, suppose \(x\) touches \(A\). Then it touches every closed set containing \(A\), and thus it touches the closure which is the intersection of those sets. So \(cl(A) \supseteq \{ x : x \text{ touches } A \} \). We will now show the right hand side is closed, and thus contains the closure and we will be done.

But this is almost immediate from the 5th axiom: Suppose \(y\) touches \( \{ x : x \text{ touches } A \}\). Then certainly every element of this set touches \(A\) (by definition). Therefore \(y\) touches \(A\). Therefore \(y \in \{ x : x \text{ touches } A \}\). So the set contains every element that touches it, and the result is proved.

So this means that \(x\) touches \(A\) if and only if \(x \in cl(A)\) with the closure operator we’ve defined, and for every closure operator we can define a touches relationship which satisfies this. So we’re done.

We can also nicely define continuity in terms of the touches relationship.

Given two topological spaces \(X\) and \(Y\), we can then define a function \(f : X \to Y \) to be continuous if whenever \(x \in X, A \subseteq X\) with \(x\) touching \(A\), we have \(f(x)\) touching \(f(A)\). i.e. a function is continuous if it doesn’t pull points away from sets.

How does this connect to the normal definition of continuity?

First, if \(f\) is continuous under this definition, let \(A \subseteq Y\) be closed. Then let \(x\) be close to \(f^{-1}(A)\). Then we must have \(f(x)\) close to \(f(f^{-1}(A)\) \subseteq A\). So \(f(x)\) is close to \(A\) and thus in \(A\) (because it is closed), so \(x \in f^{-1}(A)\).

Therefore if \(f\) is continuous under this definition then whenever \(A\) is closed then \(f^{-1}(A)\) is also closed.

Now to show the converse. Suppose that whenever \(A\) is closed then \(f^{-1}(A)\) is also closed, we will show \(f\) is continuous.

Suppose that \(x\) is close to \(A\) but \(f(x)\) is not close to \(f(A)\). Let \(B = cl(f(A))\). Then \(f(x) \not\in B\). But \(f^{-1}(B)\) is a closed set, and it contains \(A\). Therefore it must contain \(cl(A)\). But \(cl(A) \ni x\), so we must have \(x \in f^{-1}(B)\), contradicting our hypothesis. QED