My (likely flawed) argument is that a continuous function (wrt to the subspace topology: IMPORTANT premise) from $C \setminus \{x\}$ to a discrete set (say $\{0,1\}$) is constant, if it were not then by adding x again we would find a continuous function from $C$ to $\mathbb{N}$ wich is not constant.

My guess that MAYBE not all continuous function on $C \setminus \{x\}$ to $\{0,1\}$ could be extended to a continuous function on $C$. Is it so?


1 Answer 1


But the statement is false! $\mathbb{R}$ is connected, but $\mathbb{R}\setminus\{0\}$ isn't.

  • $\begingroup$ you are right. Thanks. What is the flaw in my argument though? I couldn't spot it on my own $\endgroup$
    – Averroes
    Dec 8, 2017 at 14:50
  • $\begingroup$ @Averroes I don't understand your argument. What do you mean by “adding $x$ again”? If the domain of a function is $C\setminus\{x\}$ you can't just add $x$. And if you extend the function to $C$, what makes you think that the extension is continuous? $\endgroup$ Dec 8, 2017 at 14:53
  • $\begingroup$ Fair points. How could we prove that a construction is impossible in this case? (not possible to find an extension) $\endgroup$
    – Averroes
    Dec 8, 2017 at 14:57
  • $\begingroup$ In your example I can see that whatever value I would give to $f^*(0)$ $f^*$ being any extension of a continuous function $f$ defined on $\mathbb{R}\setminus\{0\}$ with two distinct discreet values. $f^*$ would fail to be continuous wrt to the usual topology on $\mathbb{R}$ $\endgroup$
    – Averroes
    Dec 8, 2017 at 15:01
  • $\begingroup$ @Averroes Because if it was possible, then $\mathbb R$ would be disconnected. But it is connected. $\endgroup$ Dec 8, 2017 at 15:01

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