5

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.)

Tobias Shuxue Laoshi
  • 145

2 Answers2

5

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?

Cameron Buie
  • 102,994
  • wow, it is excellent – R.N Sep 01 '15 at 18:47
  • Hello, I have no earthly idea what you're talking about. – Tobias Shuxue Laoshi Sep 01 '15 at 18:51
  • @Tobias: What definition of continuity are you familiar with? – Cameron Buie Sep 01 '15 at 18:52
  • The one back from calculus, which isn't particularly helpful here. $lim_{x\rightarrow a}f(x)=f(a)$ – Tobias Shuxue Laoshi Sep 01 '15 at 18:54
  • Gotcha. What definition of $L=\lim_{x\to a}f(x)$ are you familiar with? – Cameron Buie Sep 01 '15 at 18:55
  • The good ole epsilon delta one. $\forall \epsilon > 0.\exists \delta. |x-a|<\delta\Rightarrow |f(x)-L|<\epsilon$. – Tobias Shuxue Laoshi Sep 01 '15 at 18:57
  • Excellent! I can work with that. – Cameron Buie Sep 01 '15 at 18:58
  • @Tobias: I have expanded my answer to give you an idea how we may use continuity to our advantage. See what you can do with it, and let me know if you have any questions. – Cameron Buie Sep 01 '15 at 19:25
  • @CameronBuie: All very nice. I get it, but I must be able to reproduce such an argument. Yep, all very nice. Pretty. Thanks! – Tobias Shuxue Laoshi Sep 01 '15 at 19:45
  • @Tobias: Following through with the proof will not only make you more familiar with methods that you'll be using more later. It will also show you how to choose $r$ and use triangle inequality to prove it (as you originally intended). – Cameron Buie Sep 01 '15 at 19:50
  • @CameronBuie: But I don't quite see how this quite formal approach can help me get to a short solution like Litho's? However, I'd probably be way better at reproducing an argument like this, than the "out of a hat" argument (albeit very pretty) that Litho used. – Tobias Shuxue Laoshi Sep 01 '15 at 19:54
  • @Tobias: To prove that $f$ is continuous, you'll be applying the same sort of triangle inequality approach you were already taking, and after you've gone as far as you can with those, a "natural choice" for $r$ (for $\delta$) should become self-evident. See if you can get that, at least, and let me know if you get stuck. I'll help you get "unstuck" as soon as I can. – Cameron Buie Sep 01 '15 at 20:00
  • Ok, I'll check the continuity a bit more. If I do get more stuck I at least have to spend a day or so to resolve it myself. You sure gave me a lot to work with. Thanks! – Tobias Shuxue Laoshi Sep 01 '15 at 20:08
3

Choose $r = \frac{|z+i|-|z-1|}2$. Then, if $|w-z|<r$, $$ |w-1|<|z-1|+r= |z+i|-r<|w+i|. $$

Litho
  • 4,396
  • Why $|z+i|-r<|w+i|$? – Ángel Mario Gallegos Sep 01 '15 at 19:00
  • @MarioG : Because by triangle inequality $|z+i|<|w+i|+|(z+i)-(w+i)|=|w+i|+|z-w|<|w+i|+r$. – Litho Sep 01 '15 at 19:02
  • Heck yeah! Nice, just what I was looking for. – Tobias Shuxue Laoshi Sep 01 '15 at 19:05
  • @Litho: How did you figure that out? Any particular strategy? – Tobias Shuxue Laoshi Sep 01 '15 at 19:07
  • @TobiasShuxueLaoshi : Well, you can look at it like this: we have the function $f(z)=|z+i|-|z-1|$. It consists of two summands such that, when the argument $z$ is changed by $a$, each summand changes by at most $|a|$. It means that $|f(z+a)-f(z)|\leq 2|a|$. So, if $f(z)>0$ and $a<r=f(z)/2$, then $f(z+a)>0$. Note, though, that it's easy to find an estimate on $r$ in this case only because the function $f$ is nice enough. Cameron Buie's answer contains a much more general method applicable for an arbitrary continuous $f$. – Litho Sep 02 '15 at 07:41
  • @Litho: Yes, after fully digesting it, I see the power of his method. – Tobias Shuxue Laoshi Sep 04 '15 at 14:00