If f is continuous at c and f(c)>1

[H2Pendragon]

Homework Statement

Hi everyone, this is probably an easy question but I'm having trouble on the wording of the proof.

Let f be continuous at x=c and f(c) > 1

Show that there exists an r > 0 such that $$\forall x \in B(c,r) \bigcap D : f(x) > 1$$

Homework Equations

$$\forall \epsilon > 0, \exists r > 0, \forall x \in B(c,r) \bigcap D \Rightarrow |f(x) - f(c)| < \epsilon$$

f(c) > 1

The Attempt at a Solution

I'd put something here but I don't really have any attempts. I originally thought of using the limit rule that lim f(x) = f(c) but that doesn't work since c could be an isolated point.

If I used a graph it just seems so obvious, but I have to use words to prove it. I tried also to use a contradiction, but using just what I put down in (2) didn't lead to any direct contradiction that I could see.

A nudge in the right direction would be helpful.

 

[matt grime]

If in doubt try an example.

Let's say for the sake of argument that f(c)=2. What choice of epsilon, and r, from 2. above will mean that f(x) > 1 in the ball around c?

 

[MathematicalPhysicist]

-e<f(x)-f(c)<e
f(x)>f(c)-e
f(c)=1+h^2 for some h real, do you understand how to pick e such that f(x)>1.

 

[H2Pendragon]

-e<f(x)-f(c)<e
f(x)>f(c)-e
f(c)=1+h^2 for some h real, do you understand how to pick e such that f(x)>1.

by this reasoning couldn't I just claim that f(c) = 1 + e?

 

[matt grime]

Yes, you can do that.