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

Question

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.

Answer 1

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.

Answer 2

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.

Answer 3

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.