Proving, using the definition of a limit that the function is continuous everywhere where $z,a \in C$

$$f(z) = \operatorname{Re}(z) $$

$f:S$ $\subset C \to C$

So I got to $|\operatorname{Re}(z) - \operatorname{Re}(a)| = 0.5|z - a + \bar z - \bar a|$ but I don't know how I can use this expression to show that if $|z - a| < \delta$ then $|\operatorname{Re}(z) - \operatorname{Re}(a)| < \varepsilon$

for any $\varepsilon > 0$

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    $\begingroup$ What you need to do is FIX an arbitrary $\epsilon$. Then FIND a $\delta$ that will work. Hint: Recall that $|x+iy| = \sqrt{x^2 + y^2}$ and so $\min(|x|, |y|) \leq |x+iy| \leq \max(|x|, |y|)$. $\endgroup$ – 4-ier Aug 22 '18 at 6:17
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    $\begingroup$ Hint for your approach: Remember the triangle inequality. $\endgroup$ – 4-ier Aug 22 '18 at 6:18
  • $\begingroup$ Thanks, forgot about the triangle inequality $\endgroup$ – Gordon Ramsey Aug 22 '18 at 6:29

Let $z=(x+iy)$ and $w=(a+ib)$

Note that your function is $$ f(x+iy)=x$$and $$ f(a+ib)=a$$

$$ |f(z)-f(w)|=|f(x+iy)-f(a+ib)|=|x-a|\le \sqrt {(x-a)^2+(y-b)^2}=|z-w|$$

For a given $\epsilon >0$ let $\delta = \epsilon $

If $$|z-w|<\delta$$ then $$ |f(z)-f(w)|\le |z-w|<\delta =\epsilon$$

Thus $ f $ is continuous.


It is easy to show that $z_n \to z$ implies $\overline{z}_n \to \overline{z}$. Hence, $z_n \to z$ implies $$\operatorname{Re} z_n = z_n + \overline{z}_n \ \to \ z + \overline{z} = \operatorname{Re} z.$$ This shows that $\operatorname{Re}(\cdot)$ is continuous.


Forgot about the Triangle Inequality $|\operatorname{Re}(z) - \operatorname{Re}(a)| = 0.5|z - a + \bar z - \bar a| \leq 0.5(|z - a| + |\bar z - \bar a|) = 0.5(2|z - a|) = |z-a|$ using triangle inequality

then pick $\delta = \varepsilon$

Therefore if $|z - a| < \delta$

$|\operatorname{Re}(z) - \operatorname{Re}(a)| < \delta = \varepsilon$

Thanks the guy in the comments

  • $\begingroup$ Something to keep in mind is that the real and imaginary parts are both smaller than the absolute value. So, $|Re(z-a)| \leq |z-a|$. This is proved in the way that you did it above. $\endgroup$ – 4-ier Aug 22 '18 at 6:34

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