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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 $X$.

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  • $\begingroup$ 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 Aug 19 '16 at 8:52
  • $\begingroup$ @ForeverMozart Thanks, but the property must be true for any $a \in X$ $\endgroup$ – alex alexeq Aug 19 '16 at 8:55
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    $\begingroup$ 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 Aug 19 '16 at 11:45
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    $\begingroup$ 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 Aug 19 '16 at 15:42
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    $\begingroup$ @JoelDavidHamkins you sunk my battleship :( $\endgroup$ – Forever Mozart Aug 20 '16 at 5:38
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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).

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    $\begingroup$ 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 Aug 19 '16 at 16:54
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    $\begingroup$ @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 Aug 19 '16 at 17:28
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    $\begingroup$ I believe this comment should be part of the answer. $\endgroup$ – მამუკა ჯიბლაძე Aug 19 '16 at 19:09
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    $\begingroup$ @მამუკა ჯიბლაძე: Point taken -- I've added the proof to my answer. $\endgroup$ – Will Brian Aug 19 '16 at 19:32
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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.)

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  • $\begingroup$ 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 Aug 19 '16 at 15:03
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    $\begingroup$ @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 Aug 19 '16 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 Aug 19 '16 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 Aug 19 '16 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 Aug 19 '16 at 17:23

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