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.



A connected topological space whose points cannot be connected by irreducible components
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 $F\_x$.


