# What concept does an open set axiomatise?

In the context of metric (and in general first-countable) topologies, it's reasonably clear what a closed set is: a set $F$ is closed if and only if every convergent sequence of points in $F$ converges to a point also in $F$. This naturally generalises to the definition of a closed set in an arbitrary topological space using the concept of limit points... but limit points are defined in terms of open sets, which are, to me, somewhat more mysterious than closed sets.

I was once told that ultrafilters axiomatise the concept of (sets of) big sets. I'm hoping here to find a similar conceptual picture of topologies defined by systems of open sets, preferably without reference to closed sets. One such explanation I've seen is that open set axiomatise the concept of nearness, which, I suppose, is fair enough at least for metric topologies. Indeed, if something is "happening" at points near $x \in X$, then it's usually the case that every open neighbourhood $U \subseteq X$ of $x$ contains an open subset $V \subseteq U$, $x \in V$, such that the something "happens" at all points in $V$. But what about non-metrisable topologies, particularly the coarse ones where there are no "small" open sets?

Consider, for example, the Zariski topology on affine $n$-space $\mathbb{A}^n$. $\mathbb{A}^n$ is irreducible, so every non-empty open set is dense. It seems reasonable to interpret this to mean that every non-empty open set is "large". Indeed, if we work over the complex numbers, then in the usual metric geometry, Zariski-open sets are unbounded and have full measure, so are very large indeed, so I don't think it's fair to say that open sets are capturing the notion of nearness here.

I'm also curious about the history of point set topology. When were the axioms first written down? What were the first "non-geometric" examples of topological spaces — "non-geometric" here referring to either non-metrisable topologies or topologies on sets other than sets of points of some intuitively geometric object — and were they part of the motivations for creating point set topology?

• Apr 9, 2011 at 5:27
• Zariski open sets are capturing the notion of "nearness" relative to what they are defined. Remember that a set is Zariski closed if it is the zero set of a family of polynomials; so here, "nearness" is not about metric, but about how polynomials "treat" the point. The problem in your penultimate paragraph is that you are trying to shoehorn one notion of "nearness" (the norm notion from $\mathbb{A}^n$) into a topological structure that is completely unconcerned with that notion of nearness (and instead focuses on something else). Apr 9, 2011 at 5:28
• Concerning your last paragraph: It's all in Hausdorff, Grundzüge der Mengenlehre, 1914 [rough translation of the title: basic facts on set theory]. He abstracted and cleaned up the work of Fréchet on metric spaces from about ten years before where function spaces were considered already, which I don't see as geometric objects, at least not primarily). Even if you don't read German it's worth having a look at that book.
– t.b.
Apr 9, 2011 at 5:32

Edit, 7/10/11: The idea below was recently also discussed in this blog post by Michael O'Connor.

This was thoroughly discussed on MathOverflow. The answers that I found most convincing can be summarized as follows: open sets axiomatize the notion of a semidecidable property.

That is, open sets axiomatize the notion of a condition whose truth can be verified in finite time (but whose falsehood cannot necessarily be verified in finite time). A continuous function $f : A \to B$ between two spaces is a function such that the preimage of any semidecidable subset is semidecidable, hence is a computable function in the sense that a decision procedure to verify whether $f(a) \in U \subset B$ in finite time gives a decision procedure to verify whether $a \in V \subset A$ in finite time.

To really make sense of what I just said above you should think of $A$ as the set of possible conditions of some system, $f$ as a measurement of some property of the system, and $B$ as the set of possible values of the property you're measuring. Then $f$ is computable precisely when information about $f(a)$ allows us to deduce information about $a$. In some sense it is the central premise of the scientific method that this is possible.

Note that the above description really brings out the special role of the Sierpinski space $\mathbb{S}$. Indeed, a subset of a topological space $X$ is open precisely when the indicator function $X \to \mathbb{S}$ is continuous.

Onto the axioms:

• The union of an arbitrary collection of open sets is open because any decision procedures to verify conditions $U_i, i \in I$ in finite time can be run in parallel to verify condition $\bigcup U_i$ in finite time.
• The intersection of a finite collection, but not necessarily an infinite collection, of open sets is open because a finite number of decision procedures to verify conditions $U_1, ... U_n$ in finite time can be run in parallel to verify condition $\bigcap U_i$ in finite time, but this is not necessarily true of an infinite number of decision procedures, which may take an unbounded amount of time to all complete.
• The empty set and the entire space are open because both of these conditions can be verified in zero time.

Finally, note that it is intuitively possible to verify whether a point in a metric space lies in an open ball in finite time by showing that it lies in an even smaller open ball (which can be done using finite-precision computations), but it is not necessarily possible to do the same for a closed ball because the point may lie on the boundary and we cannot make arbitrarily precise measurements in finite time.

In the particular case of the Zariski topology, it is always possible to verify in finite time whether a polynomial is nonzero at a point by computing with sufficient precision, but without additional information it is not necessarily possible to verify in finite time whether a polynomial is zero at a point.

• Note that this gives a natural interpretation to clopen sets: they're precisely the properties whose truth and falsehood can both be verified in finite time. In my opinion this makes things like Stone's representation theorem (en.wikipedia.org/wiki/…) a little more natural. Apr 9, 2011 at 17:18
• Sorry to resurrect an old question, but I just realised that the analogy is not quite so clear. It's reasonable enough to say that the union of a countable collection of semidecidable sets is again semidecidable, by a standard interleaving argument... but it's not clear that an uncountable union of semidecidable sets should be semidecidable. Jun 2, 2011 at 17:21
• @Zhen: you can run an arbitrary number of decision procedures in parallel. (This is a fairly idealized model of computation.) I'm not sure what you mean by "a standard interleaving argument." Jun 2, 2011 at 17:23
• Hmmm. It's more idealised that Turing machines, that's for sure! By interleaving argument, I mean the following: Let $P_1, P_2, P_3, \ldots$ be decision procedures. Then we can construct a single decision procedure $P$ which halts when some $P_n$ halts in this fashion: First, run the step $1$ of $P_1$, then run step $1$ of $P_2$ and step $2$ of $P_1$, and then run step $1$ of $P_3$, step $2$ of $P_2$, and step $3$ of $P_1$, and so on. I'm not entirely certain this is permitted in the Turing machine model, but it seems reasonable enough. Jun 2, 2011 at 17:30
• @Zhen: yes, this isn't a Turing machine model. Jun 2, 2011 at 18:28

I just want to give an example to the Qiaochu's beautiful answer.

In Jet Nestruev's book there is following physical interpretation of structures on smooth manifold (or rather how physical system can be modeled with smooth manifold):

• physical system = manifold $M$
• state of system = point of $M,$ $x\in M$
• measuring device = function on $M,$ $f:M\to\mathbb{R}$ (smooth, but stick to continuous ones)

• condition $C$ on system such that verifying if state satisfies $C$ only requires finite amount of measuring devices = open set in $M,$ $U\subset M.$

I think it's a mistake to look for meaning in open sets. They are simply an artifact of how topological spaces are commonly axiomatized.

To motivate this axiomatization, consider the high school definition of a continuous function from $\mathbb{R}$ to $\mathbb{R}$: $f$ is continuous at $x$ if for all $\varepsilon > 0$, $\exists \delta > 0$ such that if $x'$ is within $\delta$ of $x$, $f(x')$ is within $\varepsilon$ of $f(x)$. In other words, defining a ball centered at $x$ of radius $r > 0$, $B(x, r)$, to be the points (strictly) within $r$ of $x$, the definition reduces to this: for each ball $B := B(f(x), \varepsilon)$ there is a ball $B(x, \delta)$ that $f$ maps to $B$.

At this point we may see that there is no technical reason to require balls to consist of points within a certain distance of a center. We may simply define the balls in a space explicitly and carry over the definition of continuity to our new notion of spaces. The resulting notion of continuity is a strict generalization of the original definition and turns out to be very useful, allowing all sorts of spaces to be reasoned about similarly.

In other words, open sets make topological spaces work, and topological spaces are a very general notion of a set on which there is a notion of continuous functions.

• +1 I don't necessarily agree with this answer, but I still think it well thought-out and justified. Mar 17, 2017 at 20:18

• The union of an arbitrary collection of open sets is open because any decision procedures to verify conditions Ui,i∈I U i , i ∈ I in finite time can be run in parallel to verify condition ⋃Ui ⋃ U i in finite time.
• The intersection of a finite collection, but not necessarily an infinite collection, of open sets is open because a finite number of decision procedures to verify conditions U1,...Un U 1 , . . . U n in finite time can be run in parallel to verify condition ⋂Ui ⋂ U i in finite time, but this is not necessarily true of an infinite number of decision procedures, which may take an unbounded amount of time to all complete.
• The empty set and the entire space are open because both of these conditions can be verified in zero time.

In response to Zhen Lin's comment:

Sorry to resurrect an old question, but I just realised that the analogy is not quite so clear. It's reasonable enough to say that the union of a countable collection of semidecidable sets is again semidecidable, by a standard interleaving argument... but it's not clear that an uncountable union of semidecidable sets should be semidecidable. – Zhen Lin Jun 2 '11 at 17:21

@Zhen: you can run an arbitrary number of decision procedures in parallel. (This is a fairly idealized model of computation.) I'm not sure what you mean by "a standard interleaving argument." – Qiaochu Yuan Jun 2 '11 at 17:23 6

Hmmm. It's more idealised that Turing machines, that's for sure! By interleaving argument, I mean the following: Let P1,P2,P3,… P 1 , P 2 , P 3 , … be decision procedures. Then we can construct a single decision procedure P P which halts when some Pn P n halts in this fashion: First, run the step 1 1 of P1 P 1 , then run step 1 1 of P2 P 2 and step 2 2 of P1 P 1 , and then run step 1 1 of P3 P 3 , step 2 2 of P2 P 2 , and step 3 3 of P1 P 1 , and so on. I'm not entirely certain this is permitted in the Turing machine model, but it seems reasonable enough. – Zhen Lin Jun 2 '11 at 17:30 1

@Zhen: yes, this isn't a Turing machine model. – Qiaochu Yuan Jun 2 '11 at 18:28

Like Qiachou Yuan said, you can verify the openness of the union of arbitrary open sets in finite time by running the decision procedures for each Ui,i∈I in parallel (the total time would be equal to the slowest verification procedure).

However, this is not the case for an arbitrary intersection of open sets. This is because any decision procedure for verifying an intersection requires at least 2 open sets (to calculate the overlap). You could try to run this in parallel by computing the intersections of pairs of open sets, and then computing the intersections of the intersections, and so on. However, for an infinite collection of open sets, this process would have to continue indefinitely (to visualize this, remember that counting the integers by 2's - 0, 2, 4, ... - still produces a countably infinite set).

It's okay to run infinitely many decision processes in parallel (because we are already dealing with infinite spaces), but for infinite intersections even the parallel process wouldn't terminate.

I initially think there're some unclear points in the semi-decidability arguments, so I tried to post a question about them. However, I think I somehow resolved them as I was composing the question. So I will just post my questions as well as my answers here in case someone else runs in to the confusion.

Question 1: To compute the result of disjunction property, if we only need to witness the 'yes' answer from one procedure (the one disjunct that causes the whole disjunction to be true), why do other disjuncts even need to be semi-decidable? Can't they just be undecidable?

Answer 1: The semi-decidability assertion concerns only with the truth value of the entire property, rather than the truth values of individual components of the entire property, or how individual components are configured to make the entire property true. That is to say, the semi-decidability has to work with any configuration of the individual property. If one of the disjunct is undecidable, it may entirely be possible that the same disjunct is the only one that causes the whole disjunction to be true. As a consequence, our entire parallel-running procedure cannot yield 'yes' and because that undecidable sub-procedure gets stuck at finding the corresponding disjunct is actually true.

Question 2: If we can run infinite amount of procedure in parallel, why can't we do the same for the conjunction case? If all sub-procedure finishes in finite time, can't we just sum up their results?

Answer 2: Even if all of the sub-procedure terminates in finite time, the collection of infinite number of them doesn't need to terminate in finite time. Imagine there're $$\mathbb{N}$$ number of procedures, each finishing in $$n \in \mathbb{N}$$ time. To wait for all of them terminate the overall time is unbounded. Because as long as we say we can finish the observation in $$n$$ time, there will always be another procedure that requires $$n + 1$$ time to finish. This is different from the disjunction case, as in that case we only need to wait one of the procedure terminates.