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Can a connected space have a countable disjoint cover of closed subsets with at least two elements?
Yes. If you take a comb space which consists of precisely the lines in the plane satisfying the equations y=q for q a rational number between 0 and 1, then the space is connected if you give it the induced topology from R^{2}, and the set of lines y=q gives a countable disjoint cover of closed subsets.
At first glance it appears to be impossible for path connected spaces
If you have a path connected space, start by picking one point in each of the closed subsets. Construct a path connecting them all. The inverse image of the function is going to give a countable set of closed subsets of [0,1] which are disjoint and cover it. I guess you actually have to pick two points in each of the subsets in order to satisfy that condition
Now if you just have a general connected space that is partitioned into closed subsets, each path connected component is going to be partitioned as well based on which subsets it lies in. It's impossible to do this so each path connected component must be contained entirely inside of a single closed set. Therefore each closed set is the union of path connected subsets of our connected space. So the question becomes can a union of path connected components be a closed subset of a connected space?
So the question becomes can a union of path connected components be a closed subset of a connected space?
I don't get this, wouldn't the inverse image of a function be in the space, and not in [0,1]?
Are you sure of this? R has a partition of the closed singleton sets, but path components of R are never contained in such a set. You don't seem to be using the countability criterion.
The topologists sine curve is an example of this, and that space is connected. I don't know if this is a counter-example to your argument though, I hope you could explain it a bit further.
Of course, you are right. Bad error on my part. This proves that path-connectedness is sufficient to disprove the possibility of a countable cover, since it implies a countable closed cover of [0,1], given that it is impossible. It's impossibility is demonstrated given that your second argument is sound, but I still have some question about it underneath.If f:[0,1] to X is a path, and X is the union of closed disjiont sets A_{i}, then let B_{i}=f^{-1}(A_{i}). Each B_{i} is closed, and they are disjoint, and they cover [0,1]. Since [0,1] can't be covered, X can't have been covered to begin with
When I said the closed subsets have to contain the path connected subsets, the closed subsets I was referring to have to satisfy the condition in your OP. The set of singletons in R doesn't
So now given a general connected space, which is partitioned by closed sets A_{i}, we can consider how each A_{i} intersects a path connected component. These intersections partition the path connected component, but we know there aren't many ways that this partitioning is possible
The path connected components aren't closed though, because the y-axis contains limit points of the sine curve portion
I think this works.
On a countably infinite set put the topology whose open sets are the entire set minus only finitely many points. This space is connected and each singleton is closed. The whole space is the union of all of the singletons, and so is a union of countably many closed sets.
Excellent, nice example. Now, what if we restrict ourselves to uncountable connected spaces? Can a countable disjoint union of at least two closed subsets exist?
Well this isn't really what you want but I am not too proud to say it anyway. Take a countable infinite collection of disjoint closed disks and do the same thing - the open sets are the union of all of the disks minus finitely many of them.
Can't you just take Q^2 (Q=the rational numbers) with the regular topology and take all horizontal lines?
Ah, yes, of course. This is exactly what I want, nice and easy counter-examples. I guess this does not generalize neatly at all to general connected spaces. But I think it still might work in a linear continuum. Closed disjoint intervals cannot cover such a space, but whether a countable union of disjoint closed subsets in general can I am not sure of.
The problem I see with this example is that it is no longer a Hausdorff space. Let's add that condition and see what happens.
By the regular topology I guess you mean the subspace topology from R? In that case Q^2 is not connected. The same applies for the order topology induced from Q.