$Let~~f:\mathbb R\rightarrow \mathbb R$ be a function. The graph of f, $\large \Gamma =\{(x,y) \in \mathbb R^2: y=f(x)\}$. Suppose $f$ is continous.
1) Prove that $g:\mathbb R\rightarrow \mathbb R^2, g(x) =(x,f(x)) $is continuous.
2)Prove that $\Gamma_f$ is path connected.
3) Prove that $\Gamma_f$ is closed in $\mathbb R^2$ . Hint: prove that it contains all of its limit points.
4) Suppose $\Gamma_f$ is closed and $f$ is bounded. Prove that $f$ is continuous.
I need a big hand. Please, help.

Question

loser66 · Answer

I have class now. Please leave the guidance. I 'll be back later.
Thanks in advance.

anonymous · Answer

I take it you know that f is continuous if and only if for any sequence $ x_n$  converging to$ x$ then$ f(x_n)$ converges to $ f(x)$

anonymous · Answer

Another fact that we need is a bounded sequence $z_n $ of reals has a convergent subsequence

anonymous · Answer

To prove that g is continuous, let $ x_n$ be a sequence converging to x, then
$$
f(x_n) \to f(x) \implies g(x_n)= (x_n, f(x_n)) \to ( x,f(x))=g(x)
$$
Hence g is continuous

anonymous · Answer

2) The image of a path connected set by a continuous map is itself path connected.

anonymous · Answer

3) Let $$ 
(x_n, f(x_n)) \in \Gamma_f \quad \text { that    converges to } 
(x,u)
$$
We need to show that u=f(x)

anonymous · Answer

Since f is continuous$$
f(x_n)\to f(x)
$$
but
$$
f(x_n)\to u
$$
then 
$$
u=f(x)
$$

anonymous · Answer

4) let $x_n$ be a sequence converging to x, we need to prove that
$ f(x_n) \to f(x) $. This can be done by proving that every 
 $x_{n_k} $subsequence of 
of $x_n$  has a further subsequence   $x_{n_{k_p}} $ such that
$$
f(x_{n_{k_p}} )\to f(x)
$$
I would let you finish it now.

loser66 · Answer

Thanks a lot. :)

anonymous · Answer

YW