Real Analysis proof continuity

[kbrono]

Show that the function f(x)=x is continuous at every point p.

Here's what I think but not sure if i can make one assumption.

Let \epsilon>0 and let \delta=\epsilon such that for every x\in\Re |x-p|<\delta=\epsilon. Now x=f(x) and p=f(p) so we have |f(x)-f(p)|<\epsilon.




Or...

can i just say that |x-p| \leq |f(x)-f(p)|<\epsilon. ?


Thanks

 

[micromass]

Let \epsilon>0 and let \delta=\epsilon such that for every x\in\Re |x-p|<\delta=\epsilon. Now x=f(x) and p=f(p) so we have |f(x)-f(p)|<\epsilon.

This is correct!
I didn't really understand what your point was in your other idea...

 

[lanedance]

and you can say
|f(x)-f(p)| = |x-p| &lt; \delta = \epsilon