# show that p is not complete

• Nov 25th 2011, 06:45 PM
wopashui
Prove that D is connected
Let $D \subset R$, and let f: D-->R be continuous. Prove that D is connected if { $x,f(x): x \in D$}, the graph of f, is a connected subset of $R^2$
• Nov 25th 2011, 08:57 PM
Drexel28
Re: Prove that D is connected
Quote:

Originally Posted by wopashui
Let $D \subset R$, and let f: D-->R be continuous. Prove that D is connected if { $x,f(x): x \in D$}, the graph of f, is a connected subset of $R^2$

This is an if and only if. Let $\Gamma_f$ denote the graph, note then that $D=\pi_1(\Gamma_f)$ where $\pi_1:R^2\to R$ is the canonical projection onto the first coordinate--why does this tell us $D$'s connected. Conversely, if $D$ is connected then $\Gamma_f=h(D)$ where $h: D\to R^2: x \mapsto (x,f(x))$--why is $h$ continuous and why does this tell us that $\Gamma_f$ is connected?
• Nov 26th 2011, 09:35 AM
wopashui
Re: Prove that D is connected
Quote:

Originally Posted by Drexel28
This is an if and only if. Let $\Gamma_f$ denote the graph, note then that $D=\pi_1(\Gamma_f)$ where $\pi_1:R^2\to R$ is the canonical projection onto the first coordinate--why does this tell us $D$'s connected. Conversely, if $D$ is connected then $\Gamma_f=h(D)$ where $h: D\to R^2: x \mapsto (x,f(x))$--why is $h$ continuous and why does this tell us that $\Gamma_f$ is connected?

sorry, what is canonical projection ?