# A connected topological space whose points cannot be connected by irreducible components

Does there exist a topological space $$X$$ with the following properties?

1. $$X$$ is connected.
2. The set of irreducible components of $$X$$ is locally finite.
3. Not every pair of points in $$X$$ can be "connected by irreducible components", i.e., there exist points $$x,y\in X$$ such that there does not exist a finite sequence $$(Z_i)_{i=0}^n$$ of irreducible components of $$X$$ with $$x\in Z_0$$, $$y\in Z_n$$ and $$Z_i\cap Z_{i+1}\neq\emptyset$$ for every $$i\in\{0,\ldots,n-1\}$$.

Note that in such a case, there are infinitely many irreducible components that are not pairwise disjoint.

(I think that such a space must exist: We take an uncountable well-ordered set of irreducible spaces such that each of them meets "the next one" in a single point. However, while unsuccessfully trying to do this rigorously, I got the feeling that some understanding of ordinal numbers might be helpful, which I seemingly do not have; hence the corresponding tag)

This question arose while trying to understand and compare different characterisations of connectedness of topological spaces.

• If I understand the question: take X=[0,3] with usual topology on [1,2] and co-finite topology on [0,1] and [2,3]. Then [0,1] and [2,3] are irreducible but cannot be joined.
– erz
Jun 10 '20 at 15:03
• @erz: In your example, the set of irreducible components is not locally finite. Jun 10 '20 at 15:11
• What's meant by "the set of irreducible components of $X$ is locally finite"? First, I'm not sure what an irreducible component is in such a broad setting: is it a maximal irreducible closed subset? Does the condition mean that every point has a neighborhood meeting only finitely many irreducible components?
– YCor
Jun 10 '20 at 15:34
• Dear @YCor, both your guesses are correct. Jun 10 '20 at 19:10
• @YCor What is an irreducible closed subset? Jun 10 '20 at 20:21

## 1 Answer

No such space can exist.

The proof doesn't use very much about irreducible components. That is, suppose $$X$$ is connected. Let $$S$$ be any set of closed subsets of $$X$$ which exhaust $$X$$ and suppose $$S$$ is locally finite in the sense that every point $$x$$ has a neighborhood $$U_x$$ intersecting only finitely many sets $$Z_1,\ldots,Z_n$$ of $$S$$. Then every pair of points in $$X$$ can be 'connected by $$S$$-sets' à la condition (3).

The local finiteness condition can be slightly strengthened for free: every point $$x$$ of $$X$$ has a neighborhood meeting only finitely many sets in $$S$$, each of which contains $$x$$, because the intersection of $$U_x$$ with the complements of those $$Z_i$$ not containing $$x$$ is still open.

Now for $$x$$ a point of $$X$$,let $$F_x$$ be the set of all points $$y$$ in $$X$$ such that there exists a finite sequence of $$S$$-sets between $$x$$ and $$y$$ as in (3). We will show that $$F_x$$ is both open and closed, which will mean that $$F_x=X$$.

To see that $$F_x$$ is closed, let $$y$$ lie in the closure of $$F_x$$; then there exists a neighborhood $$U_y$$ of $$y$$ which intersects finitely many $$S$$-sets containing $$y$$, necessarily including one $$Z$$ which meets $$F_x$$. Then a finite sequence from $$x$$ to $$z\in Z$$ can be extended to a sequence from $$x$$ to $$y$$ by appending $$Z$$, so $$y$$ is in $$F_x$$.

To see that $$F_x$$ is open, let $$y$$ lie in $$F_x$$. Then by the same strengthening there exists a neighborhood $$U_y$$ of $$y$$ contained in the union of all $$S$$-sets containing $$y$$. Certainly each of these $$S$$-sets is in $$F_x$$ so a neighborhood of $$y$$ is contained in $$F_x$$.

• Dear @Gabriel, thank you very much! The generality of your result helps a lot in understanding "connected by certain sets"-condition. Jul 6 '20 at 5:47