Continuous functions are those which preserve approximate measurements

I like the point-set topology definition of continuous function. It’s elegant, generalises well, and I think puts a bunch of things on firmer foundations than epsilon-delta definitions.

But it also confusing to some people. Why are open sets? Why is it that the pre image of an open set under a continuous function is open rather than the image?

One way to fix this is to start with different but equivalent definitions of topological spaces. This is fine, but it’s a little unsatisfying. The open set formulation is widely used because it’s quite powerful. It would be nice to be able to make intuitive sense of it. Additionally, the same sort of definition crops up elsewhere – e.g. a measurable function is one where the pre-image of measurable sets are measurable.

So I’d like to give you some intuition as to why open sets make sense and why given that intuition the definition of the continuous function is the “obvious” one.

I suggest that the intuitive concept you should attach to an open set is that and open set is an approximate measurement.

What does this mean?

Well, first let me pin down what I mean by the words individually.

A “measurement” does not here mean something like “this rod is exactly 1.23 meters long”. “This rod is less than a mile long” or “this rod is between 1 and 2 meters long” are also measurements. “The length of this rod is no more than 100 times its diameter” is also a measurement. A measurement in this case is anything that helps you pin down the range of possible objects.

And “approximate” does not mean “I guessed”. It means “you do not need to know the exact value arbitrarily well in order to validate this measurement”. You can easily validate that the rod is between 1 and 2 meters long with a tape measure. You can’t validate that it’s exactly 1.23 meters long with a tape measure (but you can validate that it’s not).

An approximate measurement is, more or less, one where you only need a finite amount of information to validate it.

Note that you might need an infinite amount of information to refute it. If I tell you that the rod is less than one meter long and it turns out that the rod is exactly one meter long down to such a subquantum scale that it turns out we’re all living in a simulation of a platonic euclidean universe then you need to measure its length infinitely precisely in order to tell me I’m wrong – even if you measure it down to the nearest micron it might be half a micron short of one meter.

So this is our intuitive and imprecise definition of an open set: An open set is one where for any member of the set we can prove that it’s a member of that set with a finite amount of information.

This is of course nonsense. How does this give rise to different topologies? And what constitutes information?

What those questions are then determines our topology. They don’t need to actually correspond to any notion of finiteness (for example we could simply define the discrete topology in which all of the questions “Is it this point?” are permitted), but many classic ones do: e.g. You only need to evaluate a real number to a finite number of decimal places to prove that it’s in an open set.

Essentially these two resolve themselves together: Topologies correspond to different sorts of questions we can ask, and then “finite amount of information” just means that for every member we can prove that it’s a member by only asking a finite number of those questions.

This intuition corresponds nicely to the topology axioms: You only need 0 questions to determine if a member of the whole set is a member of the whole set, the empty set satisfies the property vacuously. If you have an arbitrary union \(\bigcup U_i\) then for \(x \in \bigcup U_i\), \(x \in U_j\) for some \(j\) and you only need a finite set of questions to prove that. If \(x \in U \cap V\) then you can take the finite proof that \(x \in U\) and the finite proof that \(x \in V\) and union them together.

You can make all this formal and get yet another characterisation of topological spaces but it’s not very interesting and ends up mostly corresponding to existing notions.

With that notion of approximate measures hand waved, we can now hand wave our notion of a continuous function:

If you apply a continuous function to some input and make an approximate measurement of the result, this gives you an approximate measurement of the input.

So for example if we just consider the length of a rod and make an approximate measurement of that, this gives us an approximate measurement of the whole rod: It still constrains the space of possible objects in a way we only need to ask finitely many questions to answer.

And this is precisely what “the preimage of an open set is open” means: If we make some measurement \(V\) and constrain \(f(x) \in V\) then this precisely corresponds to \(x \in f^{-1}(V)\). So “an approximate measurement of the result of a continuous function gives an approximate measurement of its input” is exactly “The preimage of an open set under a continuous function is open”.

But why does that match what we would intuitively think of as “continuity”?

Well, in some cases it doesn’t really, but that’s OK. For examples where we have more intuition about what continuous should mean it matches quite nicely:

Consider e.g. \(f\) with \(f(0) = 1\) and \(f(x) = 0\) otherwise. Now consider the measurement \(f(x) > \frac{1}{2}\). In order to know whether this holds for \(x\) we’re back in the “this rod is exactly one meter long” territory – no matter how precisely you measure \(x\) it might be just a bit closer to zero than that but still non-zero.

This works in more generality: At any point of discontinuity \(x\) you will find open sets that you need to know \(y\) arbitrarily well to distinguish it from \(x\) in order to determine membership.

Note also that an approximate measurement of the input to a continuous function does not give you an approximate measurement to the output. Consider e.g. the constant function \(f(x) = 1\). Then given some open set \(U\), in order to determine if \(y \in f(U)\) we need to test if \(y = 1\). This requires infinitely many decimal points of \(y\) and thus is not an approximate measurement.

Anyway, that’s enough hand waving. I don’t know if this actually clears things up for anyone (I figured this representation out long after I’d already internalized the rules of topology), but hopefully it’s given a different perspective on it.

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