# Prove that $\Bbb R^2\setminus G$ is connected

If $$f:\Bbb R\to [-1,1]$$, let $$D$$ be it's graph $$D=\{(x,f(x))\in \Bbb R^2:x\in\Bbb R\}$$.

Show that if $$f$$ is discontinuous, then $$\Bbb R^2\setminus D$$ is connected.

If $$f$$ is not continuous, then there exists $$x_0\in \Bbb R$$ such that $$x_n\to x_0$$ but $$f(x_n)\to y\neq f(x_0).$$ With the help of sequential definition of continuity I want to solve this but I am really stuck.

• It might be easier to prove the contrapositive: If $\mathbb{R}^2-D$ is disconnected, then $f$ is continuous. Commented May 21, 2021 at 14:54
• Also, discontinuous allows for $f(x_n)$ to not converge at all, not just that $f(x_n)$ converges to some $y\neq f(x_0).$ Commented May 21, 2021 at 14:56
• @Thomas, Yes, you are right! Commented May 21, 2021 at 14:56
• To denote set removal use \backslash or \setminus, not -. Commented May 21, 2021 at 15:01
• @JohnInfinity Proving the contrapositive is logically equivalent to proving the statement posed. Is there a particular reason you are opposed to this method of proof? (This is a matter of logic and irrelevant to the equivalence of sequential continuity to continuity in metric spaces, which is what you stated in the question you wished to avoid.) Commented Jan 31, 2022 at 9:08

Consider $$D^+=\{(x,y)\ |\ y>f(x)\}$$ and $$D^-=\{(x,y)\ |\ y. Note that both those subsets are connected, even path connected when $$\text{im }f\subseteq[-1,1]$$ (regardless of whether $$f$$ is continuous or not). For example in $$D^+$$ we can connect any two points, say $$(x,y)$$ and $$(x',y')$$ by connecting first $$(x,y)$$ to $$(x,2)$$, then $$(x,2)$$ to $$(x',2)$$ and finally $$(x',2)$$ to $$(x',y')$$, all by straight lines. Analogously for $$D^-$$.
So the question is: can we connect $$D^+$$ with $$D^-$$ when $$f$$ is disconnected? Not by a path, because that's not necessarily true, e.g. if $$f(x)=\sin(1/x)$$, $$f(0)=0$$ then $$\mathbb{R}^2\backslash D$$ is not path connected.
Let $$\alpha\in\mathbb{R}$$ be a point of discontinuity of $$f$$. Thus we have a sequence $$x_n$$ such that $$x_n\to\alpha$$ but $$f(x_n)\not\to f(\alpha)$$. Since $$\text{im }f\subseteq[-1,1]$$ then $$f(x_n)$$ has a convergent subsequence, so WLOG we may assume that $$f(x_n)$$ converges to some $$\beta\neq f(\alpha)$$. And again WLOG we may assume that $$\beta>f(\alpha)$$, i.e. $$(\alpha,\beta)\in D^+$$.
Now $$(x_n,f(x_n)-1/n)\in D^-$$ and $$f(x_n)-1/n\to\beta$$. This shows that $$(\alpha,\beta)\in\overline{D^-}$$. But the closure of a connected subset is connected. Therefore $$\overline{D^-}$$, and thus $$D^-$$, is a subset of the connected component of $$(\alpha,\beta)$$. But so is $$D^+$$. Which finally means that $$D^+\cup D^-=\mathbb{R}^2\backslash D$$ is the connected component of $$(\alpha,\beta)$$, showing that the space is indeed connected.