# Connected space minus a singleton remains connected

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?

But the statement is false! $\mathbb{R}$ is connected, but $\mathbb{R}\setminus\{0\}$ isn't.
• @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? Dec 8, 2017 at 14:53
• 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 $\mathbb{R}$ Dec 8, 2017 at 15:01
• @Averroes Because if it was possible, then $\mathbb R$ would be disconnected. But it is connected. Dec 8, 2017 at 15:01