f: [-1, 1]---R and the graph of f is compact, prove that f is continuous?
Any hint will be appreciated!!
Assume that the graph is compact. Assume for contradiction that $\displaystyle f $ is discontinuous at $\displaystyle x_0 \in [-1,1] $. Then there exists an $\displaystyle \epsilon > 0 $ and a sequence $\displaystyle \{x_n \} \subset [-1,1] $ such that $\displaystyle \{x_n \} $ converges to $\displaystyle x_0 $. Look at the sequence $\displaystyle \{(x_n, f(x_n)) \} $. Since the graph is compact it has a subsequence $\displaystyle \{(x_{n_{k}}, f(x_{n_{k}})) \} $ which converges to a point $\displaystyle (x_1, f(x_1)) $. So $\displaystyle \{x_{n_{k}} \} $ converges to $\displaystyle x_1 $. But $\displaystyle \{f(x_{n_{k}}) \} $ converges to $\displaystyle f(x_0) $ which is a contradiction. This implies that $\displaystyle G(f) \to X $ is continuous and onto.
This is my understanding. By "graph" the poster means "image".
So the poster is asking: given a function mapping a compact set ([-1,1] is compact) into a compact set does it mean that the function is continous on the set?
That is what I did.
Now, I gave an example involving the Dirichelt (discontinous) function which shows it is false. You are saying you proved it, if so, then why do I have a conter-example?
I am not going to read your proof now, but I think I see out difference. You use $\displaystyle \{ (x,f(x)) \}$, i.e. as a point in $\displaystyle \mathbb{R}^2$, while I used $\displaystyle \{ f(x) \}$ as a point in $\displaystyle \mathbb{R}$. So maybe it is necessary for it to be continous.