7
$\begingroup$

Is there a Tychonoff space $X$ without isolated points with the following property:

For any $a\in X$ and any function $f : X\longrightarrow \mathbb{R}$, if $f$ is continuous on $X\backslash \{a\}$ then we can redefine $f$ at $x=a$ such that (the new) $f$ is continuous on $X$.

Improve this question
$\endgroup$
6
  • $\begingroup$ Yes. $X=\omega_1+1$ with order topology is such a space ($a=\omega_1$). If you want no isolated points, use the Long Line. $\endgroup$
    – Forever Mozart
    Commented Aug 19, 2016 at 8:52
  • $\begingroup$ @ForeverMozart Thanks, but the property must be true for any $a \in X$ $\endgroup$
    – alex alexeq
    Commented Aug 19, 2016 at 8:55
  • 3
    $\begingroup$ No linear order topology can have the requested property, since for any $a$ we can let $f(x)=0$ for $x<a$ and $f(x)=1$ for $x\geq a$. Similarly, if deleting any point disconnects the space (with the point in the closure of two components), the same idea works. $\endgroup$
    – Joel David Hamkins
    Commented Aug 19, 2016 at 11:45
  • 2
    $\begingroup$ Such $X$ also cannot be metrizable. If you can extend $f(x):=\frac1{d(x,a)}$ with some $f(a)$, then in a small enough punctured neighborhood of $a$, $f(a)-\varepsilon<\frac1{d(x,a)}<f(a)+\varepsilon$, so that $f(a)$ is positive, and moreover $d(x,a)>\frac1{f(a)+\varepsilon}$ for all $x\ne a$ in a neighborhood of $a$. Whereas if $a$ is not isolated then in all of its punctured neighborhoods this distance will take values arbitrarily close to zero. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 19, 2016 at 15:42
  • 1
    $\begingroup$ @JoelDavidHamkins you sunk my battleship :( $\endgroup$
    – Forever Mozart
    Commented Aug 20, 2016 at 5:38

2 Answers 2

Reset to default
4
$\begingroup$

Consistently, $\omega^*$ has this property. In the paper

E. van Douwen, K. Kunen, and J. van Mill, "There can be $C^*$-embedded dense proper subspaces of $\beta\omega - \omega$," Proc. Amer. Math. Soc. 105 (1989), pp. 462-470, available here.

it was shown to be consistent that, for every $p \in \omega^*$, every bounded continuous real-valued function on $\omega^* - \{p\}$ can be extended to $p$. This almost gives a consistent answer to the OP's question, the only problem being that the property in the question does not mention boundedness. However, in the context of $\omega^*$ it turns out that adding in boundedness does not hurt anything:

Observation: If $p \in \omega^*$, then every real-valued continuous function on $\omega^* - \{p\}$ is bounded.

Proof: Suppose $f$ is an unbounded continuous real-valued function on $\omega^* - \{p\}$. Pick a sequence $\langle x_n : n < \omega \rangle$ of points in $\omega^* - \{p\}$ such that $f(x_n)$ converges to infinity. It is well known that $\omega^*$ is compact and contains no nontrivial convergent sequences. Thus there must be a point $q \in \omega^*$ with $q \neq p$ at which the $x_n$ cluster. Since $f(q)$ is some (finite) real number but $f(x_n)$ goes to infinity, this contradicts the continuity of $f$. $\qquad$ QED

On the other hand, Fine and Gillman proved that $\omega^*$ consistently fails to have the OP's property (CH implies that it does not).

I do not know whether your question has a positive answer in ZFC (but I suspect that it does).

Here is a link to a related paper that you might find interesting (it's where I learned about the two results I mention above).

Improve this answer
$\endgroup$
4
  • 1
    $\begingroup$ Thanks. Please correct me if I am wrong: It is proved (in Douwen's paper) that for and $p \in \omega^*$, $\omega^*\backslash p$ is $C^*$-embedded in $\omega^*$ but my question is about $C$-embedded property. $\endgroup$
    – alex alexeq
    Commented Aug 19, 2016 at 16:54
  • 1
    $\begingroup$ @alexalexeq: Yes, but it amounts to the same thing in this context because every continuous real-valued function on $\omega^* - \{p\}$ is bounded. To see this, suppose it is not the case and $f: \omega^* - \{p\} \rightarrow \mathbb{R}$ is unbounded. Pick a sequence $x_n$ of points in $\omega^* - \{p\}$ such that $f(x_n)$ converges to infinity. It is well known that $\omega^*$ is compact and contains no nontrivial convergent sequences. Thus there must be a $q \neq p$ at which the $x_n$ cluster. Since $f(q)$ is some (finite) real number while $f(x_n)$ diverges, this contradicts $f$'s continuity. $\endgroup$
    – Will Brian
    Commented Aug 19, 2016 at 17:28
  • 1
    $\begingroup$ I believe this comment should be part of the answer. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 19, 2016 at 19:09
  • 1
    $\begingroup$ @მამუკა ჯიბლაძე: Point taken -- I've added the proof to my answer. $\endgroup$
    – Will Brian
    Commented Aug 19, 2016 at 19:32
2
$\begingroup$

Any extremally disconnected space without isolated points has this property because a space is extremally disconnected if and only if every dense subset is C*-embedded. (This result is in problem 6M of Gillman and Jerison.)

Improve this answer
$\endgroup$
5
  • $\begingroup$ This is probably a stupid question but can an extremally disconnected space without isolated points be Tychonoff? The examples I know are not even Hausdorff. $\endgroup$
    – benblumsmith
    Commented Aug 19, 2016 at 15:03
  • 1
    $\begingroup$ @benblumsmith For example, compact extremally disconnected spaces are precisely Stone spaces of complete Boolean algebras. They have no isolated points iff the algebra is atomless. E. g. the algebra of regular open (or of regular closed) subsets of a space without isolated points will do. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Aug 19, 2016 at 15:06
  • $\begingroup$ My previous answer is not correct because it assumes that continuous functions on certain dense subsets are bounded. However, it is the case that any example must be extremally disconnected. $\endgroup$
    – Anonymous
    Commented Aug 19, 2016 at 15:12
  • $\begingroup$ Actually, even that is not obvious. It is not necessary that all dense subsets be C*-embedded, but only certain ones. I apologize. $\endgroup$
    – Anonymous
    Commented Aug 19, 2016 at 15:15
  • $\begingroup$ In fact, any compact extremally disconnected space without isolated points (such as the absolute of the interval [0,1]) has this property. The reason is that in an extremally disconnected space every countable subset is C*-embedded. It follows that if a point is removed from such a space, what is left is countably compact and therefore pseudocompact. Now use the fact in my first attempted answer. $\endgroup$
    – Anonymous
    Commented Aug 19, 2016 at 17:23

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .