Showing that $\{z\in\mathbb{C}:|z-1|<|z+i|\}$ is an open set Got stuck on some homework (from H. A. Priestley, Complex Analysis). My topology ain't quite up to speed yet.
So, I want to show that $S=\{z\in\mathbb{C}:|z-1|<|z+i|\}$ is open. Geometrically it's the points above the line through $1-i$ and the origin. (Eh?)
So, essentially what I want to do is to prove that for every $z\in S$, there's a $r>0$ such that $D(z;r)\subset S$. In more concrete words, I want to figure out an $r$ such that the implication $|w-z|<r \Rightarrow |w-1|<|w+i|$ holds.
I started to fiddle around with the triangle inequality
$$|w-1|=|w-z+z-1|\leq |w-z|+|z-1|$$
and then figured I could set $0<r<|z+i|-|z-1|$, which then would yield
$$|w-1|\leq |z+i|$$
but that isn't quite it. I tried to use more terms in the "triangle inequality trick" but I can't seem to get it the way I want it. Help me oh math.stackexchange! (If you want, of course.)
 A: Regardless of the form of the set $S,$ this is actually fairly straightforward to prove, using basic topology results.
Consider the function $f:\Bbb C\to\Bbb R$ given by $$f(z)=|z+i|-|z-1|.$$ One can show that $f$ is continuous, which I leave to you, and that $$S=\{z\in\Bbb C:f(z)>0\},$$ which again I leave to you. Put another way, $$S=f^{-1}\bigl[(0,\infty)\bigr]:=\bigl\{z\in\Bbb C:f(z)\in(0,\infty)\bigr\}.$$ Since $f:\Bbb C\to\Bbb R$ is continuous and $(0,\infty)$ is open in $\Bbb R,$ what can we then say about $S$?
If you have no earthly idea what I'm talking about, let me know, and I'll try to clear things up for you.

Added: Let's explore this a bit further, using the familiar (but a bit tedious) definition of continuity:

If $z_0\in\Bbb C$ and $g:\Bbb C\to\Bbb C,$ we say that $g$ is continuous at $z_0$ if $$\forall\epsilon>0,\exists\delta>0:\bigl(|z-z_0|<\delta\bigr)\implies\bigl(|g(z)-g(z_0)|<\epsilon\bigr).$$ We say that such a $g$ is a continuous function if it is continuous at all $z_0\in E.$ Put another way: $$\forall z\in \Bbb C,\forall\epsilon>0,\exists\delta>0:\bigl(|w-z|<\delta\bigr)\implies\bigl(|g(w)-g(z)|<\epsilon\bigr).$$

Well, that's all pretty messy looking. How does it help? There are a few steps to take.
First, we prove that the function $f$ described above is continuous. (The proof is basically just several applications of the triangle inequality, and is a very good exercise. Let me know if you get stuck, or if you just want to bounce your thoughts off somebody.)
Next, we note/prove that $S$ is precisely the set of all $z\in\Bbb C$ for which $f(z)>0.$
Next, we want to use these fact to show that $S$ is open, but it may not seem clear how this is possible! The kicker is to translate continuity into terms of open sets. Well, since $f:\Bbb C\to\Bbb R$ is continuous, that means $$\forall z\in\Bbb C,\forall\epsilon>0,\exists\delta>0:\bigl(|w-z|<\delta\bigr)\implies\bigl(|f(w)-f(z)|<\epsilon\bigr),$$ which can instead be put as $$\forall z\in\Bbb C,\forall\epsilon>0,\exists r>0:\bigl(w\in D(z;r)\bigr)\implies\bigl(|f(w)-f(z)|<\epsilon\bigr).$$ (Do you see why?) From this, it follows (since $S\subseteq\Bbb C$ that $$\forall z\in S,\forall\epsilon>0,\exists r>0:\bigl(w\in D(z;r)\bigr)\implies\bigl(|f(w)-f(z)|<\epsilon\bigr).$$ Note moreover that for all $z\in S,$ we have $f(z)>0,$ so it follows that $$\forall z\in S,\exists r>0:\bigl(w\in D(z;r)\bigr)\implies\bigl(|f(w)-f(z)|<f(z)\bigr).$$
Now, you should be able to show that if $|f(w)-f(z)|<f(z),$ then $f(w)>0.$ Hence, we have $$\forall z\in S,\exists r>0:\bigl(w\in D(z;r)\bigr)\implies(f(w)>0).$$ Can you justify the above claims and take it from there?
A: Choose $r = \frac{|z+i|-|z-1|}2$. Then, if $|w-z|<r$,
$$
|w-1|<|z-1|+r= |z+i|-r<|w+i|.
$$
