Is My Epsilon-Delta Proof Correct?

1. Jul 7, 2011

BrianMath

1. The problem statement, all variables and given/known data
Using the Epsilon-Delta Definition of a limit, show that

$$\lim_{z\to z_0} \overline{z} = \overline{z_0}$$

where $\overline{z}$ is the conjugate of z.

2. Relevant equations
$$|\overline{z}|=|z|$$
$$\overline{z_1}-\overline{z_2} = \overline{z_1-z_2}$$

3. The attempt at a solution
$$\forall \varepsilon > 0, \exists \delta > 0: 0 < |z-z_0| < \delta \implies |\overline{z} - \overline{z_0}| < \varepsilon$$
$$|\overline{z}-\overline{z_0}| = |\overline{z-z_0}| = |z-z_0|$$
Choose $\delta = \varepsilon$

$$|\overline{z-z_0}| = |z-z_0| < \delta = \varepsilon$$

I'm currently teaching myself Complex Variables using Churchill and Brown's text, but do not have access to any instructors to comment on my work (summer before university). This is the first time that I'm seeing $\varepsilon-\delta$ proofs, so I want to make sure that my proof is correct.

2. Jul 7, 2011

SammyS

Staff Emeritus
You've got the main part sort of backwards. What you've shown is what I call the scratch work that you do on the side before writing up the formal proof. It's how you decide what relationship δ should have to ε so that when z is within the 'δ - neighborhood' of z0, then your function of z, in this case $\bar{z}$, is within the 'ε - neighborhood' of the limit L, in this case $\bar{z}_0$.

Yes, you start with ε > 0 as you did. Then you set δ, in this case you said δ = ε.

Next show that for any z such that 0 < | z - z0 | < δ, then it follows (by algebra, etc.) that $\left|\bar{z}-\bar{z}_0\right|<\epsilon\,.$

3. Jul 7, 2011

BrianMath

Okay, thank you. I was wondering if I had the right order of everything.
So is this right, now?

Let $\varepsilon > 0$ and choose $\delta = \varepsilon$. Then
$$0 < |z-z_0| < \delta \implies |\overline{z}-\overline{z_0}| = |\overline{z-z_0}| = |z-z_0| < \delta = \varepsilon$$

Last edited: Jul 7, 2011
4. Jul 7, 2011

SammyS

Staff Emeritus
That's much better.

Probably should say $0<|z-z_0|<\delta \implies |\overline{z}-\overline{z_0}\,| = \dots$

You don't want (or need) z = z0.

5. Jul 7, 2011

BrianMath

Oops, oh yeah, I did forget the greater than 0 part.

Thank you very much for your help.