Let $f:[a,b]\to\mathbb R$ continuous. Prove that $G=${${(x,f(x): x\in [a,b]}$} (graph of $f$) is connected
Suppose $G$ is disconnected then $\exists A,B$ relatively open disjoint sets so that $A\neq \varnothing$, $B\neq \varnothing$, $G=A\cup B$. Let $A_1=${$ x\in [a,b] : (x,f(x))\in A$} and $B_1=${$x\in [a,b]: (x,f(x))\in B$}
I want to prove that $A_1$ and $B_1$ form a disconnection of $[a,b]$ which is a contradiction
$A_1\neq \varnothing$, $B_1\neq \varnothing$(because $A,B\neq \varnothing$), $A_1\cap B_1= \varnothing$(because $A\cap B= \varnothing$), $A_1\cup B_1=[a,b]$ ($A_1\cup B_1\subseteq [a,b]$ by construction; Let $x\in [a,b]$ then $(x,f(x))=G=A\cup B$ if $(x,f(x))\in A$ then $x\in A_1$ and if $(x,f(x))\in B$ then $x\in B_1$, hence $x\in [a,b]$)
The problem I have is that I get lost when I try to prove that $A_1, B_1$ are open
Let $x_0\in A_1$then $(x_0,f(x_0))\in A$: $A$ is open then $\exists B_r(x_0,f(x_0))\subseteq A$. I need to prove that $\exists V_\epsilon(x_0)\subseteq A_1$ but I don´t know hot to do this can you help me please? And I also would appreciate if you can tell me if this proof is correct , I would really appreciate it :)
I think you can take $V_\epsilon(x_0) = \{x \mid (x, f(x)) \in B_r(x_0, f(x_0)) \}$. You just have to show that this is open, using epsilon-delta definition because $f$ is continuous.