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
$$|f(x)-f(p)| = |x-p| < \delta = \epsilon$$