Why does this (complex) $\varepsilon$-$\delta$ argument prove that the statement is true?

Question

I'm rather comfortable with $\varepsilon$-$N$ proofs in real analysis, but I'm taking a complex analysis course and (even in real analysis) the $\varepsilon$-$\delta$ proofs still confuse me a bit.

So we're told to prove that $\lim_{z\to z_0}\mathrm{Re}(z)=\mathrm{Rz}(z_0)$.

The proof goes as follows:

Let $z=x+iy$ and $z_0=x_0+iy_0$. Then $\mathrm{Re}(z) = x$, and $\mathrm{Re}(z_0)=x_0$, and $$\left|\mathrm{Re}(z)-\mathrm{Re}(z_0)\right|= \left|x-x_0\right|<\epsilon$$ If $\delta=\epsilon$, then $\left|\mathrm{Re}(z)-\mathrm{Re}(z_0)\right|<\epsilon$ when $\left|z-z_0\right|<\delta$, $\forall \epsilon > 0$.

To me, this doesn't seem to actually be proving anything. Is there a (hidden) assumption that $x\to z$ and $x_0\to z_0$?

Answer 1

Note that $$|z - z_0| = |(x-x_0) + i(y-y_0)| = \sqrt{(x-x_0)^2 + (y-y_0)^2} <\delta$$ implies that $$|x-x_0| =\sqrt{(x-x_0)^2} < \delta.$$ What this shows is that if we choose $\delta = \epsilon$ (for whatever $\epsilon>0$ was given to us) then $|z-z_0| < \delta$ implies that $|\text{Re}(z)-\text{Re}(z_0)| = |x-x_0|<\epsilon$.

Answer 2

You are asked to prove $\lim_{z→z_0}\Re z\ = \Re z_0$. With the notation $z =x+iy = x(z) + iy(z)$, $z_0 = x_0 + iy_0$ (i.e. $x(z) = \Re z$), this is $$\lim_{z→z_0}x(z)\ = x_0$$

We wish to show that:

given any $ε>0$, we can find $δ>0$ such that if $|z-z_0|<δ$, then $|x(z) - x_0| < ε $.

Note the similarity to the $ε-N$ definition:

given any $ε>0$, we can find $N>0$ such that if $n>N$, then $|x(n) - x_0| < ε $.

Conceptually very similar, where "$n$ is large enough" is replaced with "$z$ is close enough to $z_0$".

Anyway. Lets see where the definition takes us, and decide on what $δ$ to use later.

If $|z-z_0| = \sqrt{|x-x_0|^2 + |y-y_0|^2}<\delta$, we know that $$|x-x_0| = \sqrt{|x-x_0|^2} \leq \sqrt{|x-x_0|^2 + |y-y_0|^2} = |z-z_0| <δ $$

We want this last bound to be $ε$. So a good choice of $δ$ is…

Answer 3

The proof goes as follows:

No it doesn't. Not quite as you've quoted. Just as with $\varepsilon-N$ proofs, you are given an arbitrary $\varepsilon$, you pick a $\delta$ or $N$ based on this $\varepsilon$, and show that if the $\delta / N$ condition holds, then the $\varepsilon$ condition holds. So the proof actually is this:

For any $\varepsilon > 0$, let $\delta = \varepsilon$. If $| z - z_0 | < \delta$, then (as Alex G. has said) $$| x - x_0 | \le \sqrt{(x - x_0)^2 + (y - y_0)^2} = |z - z_0| < \delta = \varepsilon$$ So $\lim\limits_{z \to z_0} \text{Re}(z) = \text{Re}(z_0)$.