f: [-1, 1]---R and the graph of f is compact, prove that f is continuous?

Any hint will be appreciated!!

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- Oct 11th 2007, 05:03 PMvioletsfadvanced calculus, help!
f: [-1, 1]---R and the graph of f is compact, prove that f is continuous?

Any hint will be appreciated!! - Oct 11th 2007, 05:43 PMtukeywilliams
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.

- Oct 11th 2007, 05:53 PMThePerfectHacker
- Oct 11th 2007, 06:02 PMtukeywilliams
I think I proved the right statement?

- Oct 11th 2007, 06:08 PMvioletsf
- Oct 11th 2007, 06:53 PMThePerfectHacker
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? - Oct 11th 2007, 06:59 PMtukeywilliams
$\displaystyle f $ is continuous on $\displaystyle [-1,1] $ if and only if $\displaystyle G(f) $ is compact. I proved the $\displaystyle \Leftarrow $ direction.

I defined $\displaystyle G(f) = \{x, f(x))| x \in [0,1] \} $ - Oct 11th 2007, 07:05 PMThePerfectHacker
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.