Hi all
It's been a long time since I did anything like this, but I'll give it a try. I'm confident in the basic approach of the proof, but I feel like there might be something missing between the first and second step...
Let $\displaystyle (x_0,y_0)$ be a limit point of gr(f).
$\displaystyle \rightarrow y_0=\lim_{x\rightarrow x_0} f(x)$
$\displaystyle \rightarrow y_0=f(x_0)$ because f(x) is continuous.
$\displaystyle \rightarrow (x_0, y_0)\ \epsilon\ gr(f)$
$\displaystyle \rightarrow$ gr(f) is a closed subset of $\displaystyle \Re^2$
See this thread.
Suppose that $\displaystyle \left( {a,b} \right) \notin gr(f)$ that implies $\displaystyle f(a) \ne b$.
Let $\displaystyle \varepsilon = \left| {f(a) - b} \right| > 0$. From continuity we get $\displaystyle \left( {\exists \delta > 0} \right)\left[ {\,\delta < \frac{\varepsilon }{4}\& \,\left| {x - a} \right| < \delta \Rightarrow \left| {f(a) - f(x)} \right| < \frac{\varepsilon }{4}} \right]$
As a sort of lemma, any point in the ball centered at (a,b), $\displaystyle \left( {x,y} \right) \in B_\delta \left[ {(a,b)} \right]\quad \Rightarrow \quad \left| {a - x} \right| < \delta \,\& \,\left| {b - y} \right| < \delta $.
Can you show that? Hint: $\displaystyle \left| {x - a} \right| = \sqrt {\left( {x - a} \right)^2 } \le \sqrt {\left( {x - a} \right)^2 + \left( {y - b} \right)^2 } $.
So suppose some point of the graph is such that $\displaystyle \left( {c,f(c)} \right) \in B_\delta \left[ {(a,b)} \right]$.
Can you explain the following contradiction?
$\displaystyle \varepsilon = \left| {f(a) - b} \right| \le \left| {f(a) - f(c)} \right| + \left| {f(c) - b} \right| < \frac{\varepsilon }{4} + \frac{\varepsilon }{4} = \frac{\varepsilon }{2}$
That shows that the complement of the $\displaystyle gr(f)$ is open.