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.

  • $\begingroup$ 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. $\endgroup$
    – erz
    Jun 10 '20 at 15:03
  • $\begingroup$ @erz: In your example, the set of irreducible components is not locally finite. $\endgroup$ Jun 10 '20 at 15:11
  • 1
    $\begingroup$ 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? $\endgroup$
    – YCor
    Jun 10 '20 at 15:34
  • $\begingroup$ Dear @YCor, both your guesses are correct. $\endgroup$ Jun 10 '20 at 19:10
  • $\begingroup$ @YCor What is an irreducible closed subset? $\endgroup$ Jun 10 '20 at 20:21

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 $F_x$.

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

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