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Is My Epsilon-Delta Proof Correct?

  1. Jul 7, 2011 #1
    1. The problem statement, all variables and given/known data
    Using the Epsilon-Delta Definition of a limit, show that

    [tex]\lim_{z\to z_0} \overline{z} = \overline{z_0}[/tex]

    where [itex]\overline{z}[/itex] is the conjugate of z.

    2. Relevant equations
    [tex]|\overline{z}|=|z|[/tex]
    [tex]\overline{z_1}-\overline{z_2} = \overline{z_1-z_2}[/tex]


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

    [tex]|\overline{z-z_0}| = |z-z_0| < \delta = \varepsilon[/tex]


    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 [itex]\varepsilon-\delta[/itex] proofs, so I want to make sure that my proof is correct.
     
  2. jcsd
  3. Jul 7, 2011 #2

    SammyS

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    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 [itex]\bar{z}[/itex], is within the 'ε - neighborhood' of the limit L, in this case [itex]\bar{z}_0[/itex].

    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 [itex]\left|\bar{z}-\bar{z}_0\right|<\epsilon\,.[/itex]
     
  4. Jul 7, 2011 #3
    Okay, thank you. I was wondering if I had the right order of everything.
    So is this right, now?

    Let [itex]\varepsilon > 0[/itex] and choose [itex]\delta = \varepsilon[/itex]. Then
    [tex]0 < |z-z_0| < \delta \implies |\overline{z}-\overline{z_0}| = |\overline{z-z_0}| = |z-z_0| < \delta = \varepsilon[/tex]
     
    Last edited: Jul 7, 2011
  5. Jul 7, 2011 #4

    SammyS

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    That's much better.

    Probably should say [itex]0<|z-z_0|<\delta \implies |\overline{z}-\overline{z_0}\,| = \dots[/itex]

    You don't want (or need) z = z0.
     
  6. Jul 7, 2011 #5
    Oops, oh yeah, I did forget the greater than 0 part. :biggrin:

    Thank you very much for your help.
     
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