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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.

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    $\begingroup$ It might be easier to prove the contrapositive: If $\mathbb{R}^2-D$ is disconnected, then $f$ is continuous. $\endgroup$ Commented May 21, 2021 at 14:54
  • $\begingroup$ Also, discontinuous allows for $f(x_n)$ to not converge at all, not just that $f(x_n)$ converges to some $y\neq f(x_0).$ $\endgroup$ Commented May 21, 2021 at 14:56
  • $\begingroup$ @Thomas, Yes, you are right! $\endgroup$
    – Unknown
    Commented May 21, 2021 at 14:56
  • $\begingroup$ To denote set removal use \backslash or \setminus, not -. $\endgroup$
    – K.defaoite
    Commented May 21, 2021 at 15:01
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    $\begingroup$ @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.) $\endgroup$
    – Math1000
    Commented Jan 31, 2022 at 9:08

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Consider $D^+=\{(x,y)\ |\ y>f(x)\}$ and $D^-=\{(x,y)\ |\ y<f(x)\}$. 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.

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