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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?

Gibbs
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Averroes
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1 Answers1

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

José Carlos Santos
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  • you are right. Thanks. What is the flaw in my argument though? I couldn't spot it on my own – Averroes Dec 08 '17 at 14:50
  • @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? – José Carlos Santos Dec 08 '17 at 14:53
  • Fair points. How could we prove that a construction is impossible in this case? (not possible to find an extension) – Averroes Dec 08 '17 at 14:57
  • 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 $\mathbb{R}$ – Averroes Dec 08 '17 at 15:01
  • @Averroes Because if it was possible, then $\mathbb R$ would be disconnected. But it is connected. – José Carlos Santos Dec 08 '17 at 15:01
  • Sure I know that I am stating the obvious (apologies). When I made the flaw argument I had a sphere in mind. So for which type of connected sets ,taking out a point make them non-connected? (again my guess is that somewhere the space must be the union of two connected sets plus a point of intersection, but that is just a guess) – Averroes Dec 08 '17 at 15:07
  • I also do not see how this could hold: https://math.stackexchange.com/questions/1638652/connected-spaces-minus-proper-subspaces-is-connected – Averroes Dec 08 '17 at 15:08
  • @Averroes I don't know the answer to that question. – José Carlos Santos Dec 08 '17 at 15:09
  • Let us continue this discussion in chat. – Averroes Dec 08 '17 at 15:11