# Problem in topology

Can a connected space have a countable disjoint cover of closed subsets with at least two elements?

Office_Shredder
Staff Emeritus
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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 R2, 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

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 R2, 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 pick an irrational number r between 0 and 1, wouldn't R x (r,1) U R x (0,r) induce a separation of the comb space? It is by the way impossible to cover R (or any linear continuum) in the order topology with disjoint closed intervals in this way. I have trouble generalizing this to arbitrary connected spaces and arbitrary disjoint closed sets. It might be easier to prove with path-connected spaces as you mentioned though.

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Office_Shredder
Staff Emeritus
Gold Member
You're right, I was too hasty and didn't construct the space properly.

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?

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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

I don't get this, wouldn't the inverse image of a function be in the space, and not in [0,1]?

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?

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.

So the question becomes can a union of path connected components be a closed subset of a connected space?

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.

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Office_Shredder
Staff Emeritus
Gold Member
I don't get this, wouldn't the inverse image of a function be in the space, and not in [0,1]?

If f:[0,1] to X is a path, and X is the union of closed disjiont sets Ai, then let Bi=f-1(Ai). Each Bi 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

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.

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

If you have a countable number of points xi in a path connected space, you can always find a path between all of them simultaneously: the first half of your path is from x1 to x2, the next quarter is from x2 to x3, the next eight from x3 to x4 etc. This requires there are only countably many points.

So now given a general connected space, which is partitioned by closed sets Ai, we can consider how each Ai 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 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.

The path connected components aren't closed though, because the y-axis contains limit points of the sine curve portion

If f:[0,1] to X is a path, and X is the union of closed disjiont sets Ai, then let Bi=f-1(Ai). Each Bi 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
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.

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

Certainly, but I didn't see how it followed since I didn't see where you used that there was only a countable amount of closed subsets. The point is that to me it seemed your argument as it were would apply to an uncountable cover of closed subsets.

So now given a general connected space, which is partitioned by closed sets Ai, we can consider how each Ai 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

Ok, but how do you prove that each path-connected component must be contained in some closed set in the partition? The closed sets need not path-connected or even connected, so I don't see how it follows. I might be missing something here.

The path connected components aren't closed though, because the y-axis contains limit points of the sine curve portion

I am aware of that, it was just a comment to how you had reduced the problem, since you didn't mention that they had to be closed. Adding the criterion that the path-connected components themselves are closed fixes this, and I guess it was implied in your sentence "So the question becomes can a union of path connected components be a closed subset of a connected space?".

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lavinia
Gold Member
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.

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?

Can't you just take Q^2 (Q=the rational numbers) with the regular topology and take all horizontal lines?

lavinia
Gold Member
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.

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.

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.

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Can't you just take Q^2 (Q=the rational numbers) with the regular topology and take all horizontal lines?

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.

lavinia
Gold Member
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.

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.

Surely a linear continuum L is hausdorrf. If x,y is in L with x<y, there is a z such that x<z<y, so $$x \in (-\infty,z)$$ and $$y \in (z,\infty)$$ and these subsets are open and disjoint. Or did you refer to something else?

EDIT: I realize you probably meant the last counter-example you brought up. You may be right, connected and hausdorff might be enough to ensure that no such cover exists.

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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.

Duhh of course. Sorry, I forgot to include the connected part :/