Consider the Cantor set $C$ in $[0,1]$. And a function $f: [0,1] \rightarrow [-1,1]$ defined by
$$f(x) = \frac{2(x-a)}{b-a}-1 \text{ if } x \in [a,b] \text{ where }(a,b) \text{ is a contiguous interval of } C, \quad f(x) = 0 \text{ otherwise }$$
Is the $G_{r}(f)$ is connected ?, where $G_{r}(f)=\{(x,f(x)): x \in [0,1]\}$. We know that $f$ is not continuous.
I supposed it doesn't. if we define $U= \{(x,0): x \in C \}$ and $V=\{ (x, f(x)): x \in [a,b], (a,b) \text{ contiguous interval }\}$
we can check $\overline{U} \cap V = U \cap \overline{V}= \emptyset$ and $G_{r}(f)=U \cup V$ then $G_{r}(f)$ is disconnected.
Could you check if I am right? Thanks for your help.
If $x$ is an endpoint of one of the deleted intervals, $\langle x,0\rangle$ is in your set $U$ but is not in $G_r(f)$. I think that you meant $U$ to be the set of $\langle x,0\rangle$ such that $x\in C$ and $f(x)=0$; if $E$ is the set of endpoints of the deleted intervals, and $C_0=C\setminus E$, this set would be $U=\{\langle x,0\rangle:x\in C_0\}=G_r(f)\setminus V$.
With that fixed, let’s take a look at your proposed separation.
Let $\langle x,y\rangle\in V$. Then there is a deleted interval $(a,b)$ such that $a\le x\le b$. If $a<x<b$, let $W=(a,b)\times\Bbb R$; if $x=a$, so that $y=-1$, let $W=\Bbb R\times(-2,0)$, and if $x=b$, so that $y=1$, let $W=\Bbb R\times(0,2)$. In each case $W\cap G_r(f)$ is a relatively open nbhd of $\langle x,y\rangle$ in $G_r(f)$, and $W\cap U=\varnothing$, so $\langle x,y\rangle\notin\operatorname{cl}U$, and therefore $V\cap\operatorname{cl}U=\varnothing$. (All closures are taken in $G_r(f)$.) So far so good.
Now let $\langle x,0\rangle\in U$, and suppose that $\langle x,0\rangle\notin\operatorname{cl}V$; then there is an $\epsilon\in(0,1)$ such that if $W=(x-\epsilon,x+\epsilon)\times(-\epsilon,\epsilon)$, then $W\cap V=\varnothing$. But this is impossible: there is a deleted interval $(a,b)$ such that $x-\epsilon<a<b<x+\epsilon$, and clearly
$$\left\langle\frac{a+b}2,0\right\rangle\in W\cap V\,.$$
Thus, not only is $U$ not disjoint from $\operatorname{cl}V$, but in fact $U\subseteq\operatorname{cl}V$, and hence $\operatorname{cl}V=G_r(f)$. And that means that if we could show that $V$ is connected, it would follow that $G_r(f)$ is also connected, so perhaps we should try that instead.
Unfortunately, $V$ is not connected: if $x\in C_0$, $W_0=(x-1,x)\times\Bbb R$, and $W_1=(x,x+1)\times\Bbb R$, then $W_0\cap V$ and $W_1\cap V$ are a separation of $V$. However, each of the line segments making up $V$ is connected and must therefore lie entirely in one piece of any separation of $V$, so we might reasonably suspect that this is the only kind of separation of $V$. Let’s see if we can use this idea to show that $G_r(f)$ is connected anyway.
Suppose that $G_r(f)=A\cup B$, where $A$ and $B$ are disjoint, relatively clopen subsets of $G_r(f)$. We may assume that $V\cap A\ne\varnothing$. If $V\subseteq A$, then
$$U\subseteq\operatorname{cl}V\subseteq\operatorname{cl}A=A$$
as well, so $A=G_r(f)$, and $A$ and $B$ do not form a separation of $G_r(f)$. Suppose, then, that $V\cap B\ne\varnothing$ as well. Let
$$T_A=\{x\in C:\langle x,-1\rangle,\langle x,1\rangle\in A\}$$
and
$$T_B=\{x\in C:\langle x,-1\rangle,\langle x,1\rangle\in B\}\,;$$
both of these sets are non-empty. (Why?) Without loss of generality there are $x_A\in T_A$ and $x_B\in T_B$ such that $x_A<x_B$. Let $x_0=\sup\big(T_A\cap[x_A,x_B]\big)$. There are three possibilities: $x_0\in T_A$, $x_0\in T_B$, or $x_0\in C_0$.
Conclude from this that $G_r(f)$ is connected.
Obviously this is not a neatly packaged argument such as you might find in a textbook. For the most part I’ve simply followed my own thought processes on encountering the problem in your question, in hopes that seeing how someone with more experience attacked the problem, including a false start, might be more illuminating than a sanitized version.