0
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

$\endgroup$
9
  • $\begingroup$ you are right. Thanks. What is the flaw in my argument though? I couldn't spot it on my own $\endgroup$
    – Averroes
    Dec 8 '17 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 '17 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 '17 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 '17 at 15:01
  • $\begingroup$ @Averroes Because if it was possible, then $\mathbb R$ would be disconnected. But it is connected. $\endgroup$ Dec 8 '17 at 15:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.