Question: Let $X$ be a continuum and $p \in X$. Under what conditions does there exist a compactification $\gamma (X-p)$ with $\gamma (X-p) - (X-p)$ connected and nondegenerate?

Throughout, $X$ is a continuum. That is to say, a compact connected Hausdorff space. (I'd be interested on any info for non-connected spaces too!)

The story is we want to remove a point $p \in X$ and replace it with a small subcontinuum. Formally we want a space $Y$ and continuous map $Y \to X$ such that all fibres are nontrivial save one, and the nontrivial fibre $F$ is a nondegenerate subcontinuum with void interior. The question is trivial if we allow $F$ to have interior as we can always take a copy of $[0,1]$ and glue it to $p$ at an endpoint.

When $X$ is metric this can always be done. For example we can resolve $X$ at the point $p$ onto the arc: consider the following subspace of $X \times [-1,1]$.

$\displaystyle Y = \Big \{ \Big (x, \sin \Big (\frac{1}{d(x,p)}\Big ) \Big): x \in X-p \Big\} \bigcup \Big ( \{p\} \times [-1,1]\Big )$.

Then the projection map onto the $x$-coordinate satisfies the above condition.

More generally we can resolve onto any locally connected continuum $F$, by recalling $F$ is the image of an arc and constructing a map from $[0,1)$ that visits each point cofinally many times.

When $X$ is not metric we cannot always do this. For example consider the long arc $[0,\omega_1]$. Removing the endpoint $p$ we get the long line whose Stone Čech compactification is known to be the same as the one-point compactification. Since the remainder of the Stone Čech maps onto the remainder of every compactification, we cannot replace $p$ with a nondegenerate continuum, as then that continuum would be the image of a point.

It is known exactly when $X-p$ admits a nontrivial remainder. Namely when $X-p$ has two disjoint closed non-compact subsets. Has it been studied whether there are any conditions for when a nontrivial connected remainder exists?

I know Smith has given a condition for when the Stone Čech remainder is connected (when $p$ fails to be a local cut point) but this is of no immediate use. For example if $X$ is a circle $(X-p)^*$ has exactly two components but the above shows we can always wrap $X-p$ around an arc. In this case we get something like a Warsaw circle.

Edit: You might try to generalise the resolution approach to something involving a gauge $\mathcal G $ of pseudometrics on $X$. But this throws up problems if $\rho(p,x)=0$ for some $x \ne p$ and $\rho \in \mathcal G$. The obvious solution is to replace $\mathcal G$ with a gauge that is positive definite at $p$ meaning $\rho(p,x)=0 \implies p=x$. But this cannot in general be done. In fact if a single $\rho$ is positive definite then $\{p\}$ is immediately a $G_\delta$ set which is not a guarantee for non-metric spaces.

  • $\begingroup$ As I understand the question, we can write $K=K(X,p)$ as the Stone-Cech remainder of $X\smallsetminus\{p\}$, and then the question boils down to understand whether $K$ admits a connected Hausdorff quotient (as a topological space) not reduced to a singleton. $\endgroup$
    – YCor
    Mar 12, 2018 at 19:42
  • $\begingroup$ $X\setminus \{p\}$ having disjoint closed sets limiting to $p$ is not enough; $X$ could be the one-point compactification of the double long-line ($X$ is like a circle). $X\setminus \{\infty\}$ has only two compactifications (one-point and two-point). $\endgroup$ Mar 12, 2018 at 20:39
  • 2
    $\begingroup$ If there are (A) infinitely many pairwise disjoint closed sets limiting to p, then the remainder of β(X∖{p}) should contain a copy of ω∗ which is certainly big enough to map onto nontrivial continua in place of p. Moreover, if (B) there is an unbounded continuous function f:X∖{p}→[0,∞), then X∖{p} has the arc as its remainder. I do not know if (A) and (B) are equivalent in this context. I would conjecture that every continuum remainder is degenerate if and only if $X\setminus \{p\}$ is pseudocompact. $\endgroup$ Mar 12, 2018 at 20:55
  • $\begingroup$ A separate stand-alone self-contained self-explanatory short formulation of the Question would be helpful. $\endgroup$
    – Wlod AA
    Mar 12, 2018 at 22:42
  • $\begingroup$ Actually I think it would be of benefit to isolate the question: which compact Hausdorff spaces admit a Hausdorff quotient topological space that is connected and not reduced to a singleton. $\endgroup$
    – YCor
    Mar 14, 2018 at 10:15

1 Answer 1


Given a compact Hausdorff space $X$ (not necessarily connected) and a point $p \in X$, the following are equivalent:

  1. $X - p$ has a compactification with non-trivial connected remainder.

  2. There is a continuous surjection $q: (X - p)^* \to I$, where $(X-p)^*$ denotes the Stone-Čech remainder of $X-p$ and $I = [0,1]$.

  3. $X-p$ has a compactification $\gamma(X-p)$ with remainder homeomorphic to $I$.

  4. There is a continuous $f: X-p \to I$ such that every point of $I$ is a cluster point of $f$ at $p$. (in other words, $\overline{f[U-p]} = I$ for every neighbourhood $U$ of $p$).

Sketch of a proof:

$(1. \implies 2.)$ $(X-p)^*$ can be mapped onto every other remainder and every non-degenerate continuum can be mapped onto $I$.

$(2. \implies 3.)$ Put $\gamma(X-p) = I \cup_q \beta(X-p)$. Since $X-p$ is open in $\beta(X-p)$, the inclusion $X-p \to \gamma(X-p)$ is a homeomorphic embedding.

$(3. \implies 4.)$ The identity mapping of $I$ has a continuous extension to $f: \gamma(X-p) \to I$. The desired property follows from the density of $X-p$ in $\gamma(X-p)$ and the correspondence between the neighbourhoods of $I \subset \gamma(X-p)$ and those of $p \in X$.

$(4. \implies 3. \implies 1.)$ The closure in $X \times I$ of the graph of $f$ is such a compactification. Clearly $I$ is connected.

Additional information on condition 1:

A compact Hausdorff space $K$ can be mapped continuously onto $I$ if and only if $K$ is not scattered. For sufficiency of this condition we need, by the Tietze-Urysohn extension theorem, only to see that there is some closed subset that can be mapped onto $I$. If $K$ is not totally disconnected, we can take this subset to be any nontrivial component. If $K$ is totally disconnected, then we can take its perfect kernel, which is zero-dimensional, compact and dense in itself and by repeated bisection find a mapping onto $\{0,1\}^\omega$, which maps onto $I$ in a straightforward way.

For necessity, observe that any continous surjection $q: K \to I$ is proper, which implies that $K$ has a closed subset $L$ such that $q|_L$ is irreducible. Since $q|_L$ is also closed, it must map every isolated point of $L$ to an isolated point of $I$, which has no isolated points. Hence $L$ is dense in itself and $K$ is not scattered.

Additional information on condition 4:

A easy sufficient, but not necessary, condition for the existence of such an $f$ is that $X-p$ is not pseudocompact. To see this, suppose $g$ is an unbounded nonnegative function on $X-p$. Then $g[X-p]$ contains an unbounded subset $W$ with order type $\omega$. Since $W$ is discrete and closed in $[0,\infty)$ there is a continuous $h: [0,\infty) \to I$ such that $h[W]$ is dense in $I$. Putting $f = h \circ g$ gives the desired result, since if $U$ is a neighbourhood of $p$, $f[U-p]$ contains all but finitely many points of $h[W]$.

An example of a pseudocompact locally compact Hausdorff space with a connected Stone-Čech remainder is easily constructed by taking the product of the long line with a non-degenerate continuum.


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