It's well-known that the space of connected components of a Hausdorff space is T1 because connected components are closed and so points of the quotient space are also closed.

I'm wondering if there is an example of a compact Hausdorff space $X$ such that it's space of connected components $X/{\sim}$ is not also Hausdorff.

The space of path-connected components can easily be non-Hausdorff, as for example the closed topologist's sine curve shows, but unfortunately such examples seem harder to construct when only quotienting by connectedness.

If there are perhaps also any mild criteria such as metrisability on $X$ that ensure Hausdorffness, that would also be appreciated.


1 Answer 1


This can't happen for compact Hausdorff spaces. If $X$ is compact Hausdorff, then the components of $X$ are the same as the quasicomponents of $X$. In other words, any two distinct components are separated by clopen sets of $X$. Since a clopen set is a union of components (as is its complement), its image in $X/{\sim}$ is also clopen, and so $X/{\sim}$ is not just Hausdorff but totally separated (any two points are separated by clopen sets).

Without compactness, though, this can fail even for otherwise very nice spaces. For instance, let $X=\{1,1/2,1/3,\dots\}\times[0,1]\cup\{0\}\times\{0,1\}\subset\mathbb{R}^2$. Then $\{(0,0)\}$ and $\{(0,1)\}$ are both components of $X$, but any neighborhood of either intersects all but finitely many of the components $\{1/n\}\times[0,1]$ and so they do not have disjoint neighborhoods in $X/{\sim}$.

  • $\begingroup$ Brilliant, thanks for the swift answer. $\endgroup$
    – Dan Rust
    Jun 28, 2020 at 17:20

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