# show that p is not complete

• Nov 25th 2011, 05:45 PM
wopashui
Prove that D is connected
Let $\displaystyle D \subset R$, and let f: D-->R be continuous. Prove that D is connected if {$\displaystyle x,f(x): x \in D$}, the graph of f, is a connected subset of $\displaystyle R^2$
• Nov 25th 2011, 07:57 PM
Drexel28
Re: Prove that D is connected
Quote:

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

This is an if and only if. Let $\displaystyle \Gamma_f$ denote the graph, note then that $\displaystyle D=\pi_1(\Gamma_f)$ where $\displaystyle \pi_1:R^2\to R$ is the canonical projection onto the first coordinate--why does this tell us $\displaystyle D$'s connected. Conversely, if $\displaystyle D$ is connected then $\displaystyle \Gamma_f=h(D)$ where $\displaystyle h: D\to R^2: x \mapsto (x,f(x))$--why is $\displaystyle h$ continuous and why does this tell us that $\displaystyle \Gamma_f$ is connected?
• Nov 26th 2011, 08:35 AM
wopashui
Re: Prove that D is connected
Quote:

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

sorry, what is canonical projection ?