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Let $z_0\in\mathbb{C}$ and $f(z)=|z-z_0|$. Show that $f$ is continuous on $\mathbb{C}$. I expect to see a proof using the triangle inequality.

Note a function $f$ is continuous on $\mathbb{C}$ if for every $\alpha \in \mathbb{C}$, then \begin{equation} \lim_{z \to \alpha} f(z) = f(\alpha) \end{equation} i.e. \begin{equation} \lim_{z \to \alpha} |z-z_0| = |\alpha-z_0| \end{equation}

Edit: I posted my attempted solution as an answer below. Thanks Winther and Mario Carneiro for their help!

EthanAlvaree
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  • How about using $||x|-|y||\le||x||+|!-!|y||=|x|+|y|$? You are right that the step you make at the end is not valid, although you can proceed (a little less elegantly) by case analysis if you also show that $f(\alpha)-f(z)<\epsilon$, since $f$ is real. – Mario Carneiro Jan 28 '15 at 14:52
  • Great suggestion, thank you! I revised the proof and posted it as a potential answer below. – EthanAlvaree Jan 28 '15 at 15:20

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In what follows, we will use the Reverse Triangle Inequality: for complex numbers $x$ and $y$, \begin{equation} ||x|-|y|| \le |x-y| \end{equation} Let $\epsilon>0$, and choose $\delta=\epsilon$. Then $|z-\alpha|<\delta$ implies $|z-\alpha|<\epsilon$, i.e. \begin{equation} |z-z_0+z_0-\alpha|=|z-\alpha|<\epsilon \end{equation} which, applying the Reverse Triangle Inequality, implies \begin{align*} ||z-z_0|-|\alpha-z_0|| &\le |\left(z-z_0\right)-\left(\alpha-z_0\right)| \\ &= |z-z_0+z_0-\alpha| \\ &=|z-\alpha|<\epsilon \end{align*} which shows \begin{equation} |f(z)-f(\alpha)|<\epsilon \end{equation} proving $\lim_{z \to \alpha} f(z) = f(\alpha)$.

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EthanAlvaree
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    Why is $|z-z_0| + |z_0-\alpha| \leq |z-\alpha|$? If you take $z_0=0$ then it says $|z| + |\alpha| \leq |z-\alpha|$ which is not correct, just take $z=\alpha$ to see this. – Winther Jan 28 '15 at 15:23
  • You're right. I have the triangle inequality reversed in that statement. Any way I can fix this? – EthanAlvaree Jan 28 '15 at 15:28
  • Can you show $||x|-|y||\le|x-y|$? I take my earlier statement back, I don't think there is an easier way to do this than just by case analysis on the real $|\cdot|$ function, i.e. show $|x|-|y|\le|x-y|$ and $|y|-|x|\le|x-y|$. – Mario Carneiro Jan 28 '15 at 15:31
  • Yes, you need the reversed triange inequality: $||x|-|y|| \leq |x-y|$. See for example this answer for proofs of several expressions related to the triangle inequality. – Winther Jan 28 '15 at 15:40
  • Edited my answer. Does it work now? Thanks all! – EthanAlvaree Jan 31 '15 at 22:52