# Why are topological spaces defined in terms of open sets, and not in terms of connected open sets?

I'm starting to learn about point-set topology, and I find the definition of topological spaces and open sets to be very weird. Since we care about continuity in topology, it's odd that topological spaces are defined based on open sets, which can be disconnected. It makes more sense that they should be built on connected sets.

Below is an alternative formulation I made of topological spaces:

A topological space is a set $$X$$ of points, and the set $$\tau$$ of all "bunches" in $$X$$. All bunches are subsets of $$X$$. A topological space must have the following properties:

• The empty set and $$X$$ are both bunches.
• Given a non-empty collection of bunches, if their intersection is non-empty, then their union is also a bunch.
• Given two bunches, their intersection is also a bunch.

My definition of bunch is intended to be equivalent to connected open sets. Open sets can be then defined as the union of disjoint bunches.

Is my alternative formulation equivalent to the usual formulation? If yes, why is this formulation not used? If no, then what's the difference?

Note: I'm told that my formulation seems related to bases, although I'm not sure what the exact relationship is.

Note 2: My initial intuition was to define continuous functions as functions that sends bunches to bunches, but that doesn't quite work.

• Why do you think "it makes more sense" to define topology on base of "connected sets"? And unions of connected sets are usually not connected... May 15, 2021 at 10:14
• @DonAntonio Connected sets are a way to try to capture the idea of "nearness" without using distances. And yes, unions of connected sets are usually not connected, which is why I have the "if their intersection is non-empty" clause, which is the only difference between my axioms for bunches and the usual axioms for open sets. May 15, 2021 at 10:18
• The whole Euclidean space $\;\Bbb R^n\;$ is connected. How does that give us a "sense of nearness"? And again: in what sense does your definition make more sense than the usual one? May 15, 2021 at 10:20
• @DonAntonio Maybe it's subjective whether this formulation is more intuitive. I find it to me more intuitive though, so I kind of want to know whether it's equivalent to the normal one. May 15, 2021 at 10:30
• The intersection of two connected open sets need not be connected. For that reason, at least, your idea needs more work. That's not to say you should give up on it; but I used to worry about this kind of thing, and got nowhere, so I know there is a risk of getting bogged down in attempts to make an established rigorous theory comply with one's intuitions. May 15, 2021 at 14:26

## 3 Answers

One is of course free to define everything, but I do not think that the concept of a "bunch topology" is useful. It seems that you imagine a bunch to be a a connected open set. Since you require $$X$$ to be a bunch, you would restrict to connected spaces, but why not?

Let us now have a look at $$X = \mathbb R^2$$: What would be a bunch topology on $$X$$?

I think to obtain something useful we should regard each open disk $$B_r(x)$$ with radius $$r$$ around $$x \in X$$ as a bunch. Then $$U = B_{1+\epsilon}(0,0) \cup B_{1+\epsilon}(0,2)$$ is a bunch for each $$\epsilon > 0$$, similarly $$V = B_{1+\epsilon}(2,0) \cup B_{1+\epsilon}(2,2)$$ is a bunch. Hence also $$U \cap V$$ is a bunch. But for sufficiently small $$\epsilon$$ this set is not connected. This shows that your axioms do not produce the result you want.

• Now I feel stupid lol May 16, 2021 at 3:06
• @TheemathasChirananthavat There is no reason to feel stupid because of an idea. Without ideas things stand idle. May 16, 2021 at 8:22

I think you have hit the issue spot on when you attempt to deal with continuity. With open sets a continuous function is one where the inverse image of an open set is open. Changing that definition to accommodate your connected open sets is trickier than you might imagine.

For example, if you think about a function as simple as $$f(x)=x^2$$ and a small connected open interval around $$f(x)=4$$. There are two components to the inverse image - intervals around $$x=+2$$ and $$x=-2$$ and to establish the continuity at $$f(x)=4$$ you need to deal with both. Conventional topology wraps these up together, and you would have to deal with cases. It is pretty obvious that the inverse image of an open set can be a whole lot more complicated than that, so the machinery for dealing with the cases becomes complicated too.

On the image of an open set being open, of course the constant function is a challenge, because you would then need singleton sets to be open. So that definition of continuity won't ever work as you want it to.

• Yeah, a friend of mine gave me an identical example (f(x) = x^2). I would have thought that at the very least my formulation could just define open sets in terms of bunches, then define continuity in terms of open sets, right? May 15, 2021 at 10:45
• @TheemathasChirananthavat The idea you have that this might be related to the base of a topology is correct. In the article you linked under bases the first example is that open intervals (connected sets) form a base for the topology on $\mathbb R$. The first open sets we learn about are open intervals, and that is how we form our intuitions - but one aspect of mathematical development is that ideas find applications beyond what we first thought of. The point of learning topology is to extend mathematical range, so expect the ideas to be unfamiliar at first. May 15, 2021 at 11:01
• So are you saying that my formulation is equivalent to the usual formulation, just that people don't like it? Is it possible to prove that they're equivalent? May 15, 2021 at 13:12
• @TheemathasChirananthavat Your formulation would create the base for a topology, but you need unrestricted unions (and therefore potentially open sets which are not connected) to give the usual definition. Don't forget that these definitions have been forged in the crucible of proof, counter-example and generalisation so may well have been tested in circumstances we don't at first imagine. May 15, 2021 at 14:39

There are very useful topological spaces which do not have any non-trivial connected subspaces, just $$\emptyset$$ and all sets $$\{x\}, x \in X$$. Nevertheless we want to talk about continuity on such spaces and if we'd have defined that in terms of connected subsets we'd either have all maps continuous or maybe none, which is not what we'd want.

Moreover, once we have open sets we can talk about connectedness and compactness, continuity etc. These notions turn out to be useful. But if we start with a set and an abstract collection of "connected" sets, it's unclear how to then define continuity and also what the basic axioms for a structure would be. For convex sets there is a general theory along these lines. But that is much more retrictive than just connected sets.

• Can you give an example of such a topological space? May 15, 2021 at 10:39
• The rationals $\Bbb Q$ and the Cantor set $\{0,1\}^{\Bbb N}$ etc. @TheemathasChirananthavat May 15, 2021 at 10:55
• If we redefine "connected" to mean "connected, if we treat the numbers outside the topological spaces as nonexistent", then I think we can consistently treat the set of rationals between 0 and 1 as a "connected" set. And my formulation should be able to handle this case fine, right? May 15, 2021 at 13:14
• @TheemathasChirananthavat No, that doesn't solve it, as connectedness is already intrinsic (it doesn't refer to points outside). There is no way I could conceive $\Bbb Q$ as "connected". May 15, 2021 at 13:53