Prove that this set is open, hopefully just need help with the inequalities

Hi, I've been stuck on this problem for a while now and I'm scared it's just me failing with the inequalities, but it might be something more fundamentally wrong with my understanding of open sets.

From what I've learnt, a set S is open if for all z in S, there exists an r>0 such that D(z;r) is a subset of S.

My problem is here: I need to prove that

$\displaystyle S:= \left\{z \in \mathbb{C} : |z-1| < |z+i|\right\}$ is open.

My attempt at a proof so far gets here:

Let $\displaystyle z \in S$. Then $\displaystyle D\left\(z;r\right\) \subseteq S$ if, $\displaystyle \forall x \in \mathbb{C} \quad s.t \quad 0<|x|<=r, \quad z+x \in S$

So my aim now is to show that $\displaystyle z+x \in S$, equivalently, that $\displaystyle |z+x-1| < |z+x+i|$

I figure I start with $\displaystyle |z+x-1| \leq |z-1|+|x| \leq |z-1|+r$, but from there I don't know where to go.

I've been on this problem a while, and I think if I can figure this out, I should be okay with problems in the future.

Thankyou for reading, and I appreciate all help.

Re: Prove that this set is open, hopefully just need help with the inequalities

Quote:

Originally Posted by

**WonderingUser** Hi, I've been stuck on this problem for a while now and I'm scared it's just me failing with the inequalities, but it might be something more fundamentally wrong with my understanding of open sets.

From what I've learnt, a set S is open if for all z in S, there exists an r>0 such that D(z;r) is a subset of S.

My problem is here: I need to prove that

$\displaystyle S:= \left\{z \in \mathbb{C} : |z-1| < |z+i|\right\}$ is open.

My attempt at a proof so far gets here:

Let $\displaystyle z \in S$. Then $\displaystyle D\left\(z;r\right\) \subseteq S$ if, $\displaystyle \forall x \in \mathbb{C} \quad s.t \quad 0<|x|<=r, \quad z+x \in S$

So my aim now is to show that $\displaystyle z+x \in S$, equivalently, that $\displaystyle |z+x-1| < |z+x+i|$

I figure I start with $\displaystyle |z+x-1| \leq |z-1|+|x| \leq |z-1|+r$, but from there I don't know where to go.

I've been on this problem a while, and I think if I can figure this out, I should be okay with problems in the future.

Have you considered what set $\displaystyle S$ looks like?

It is the set of points which closer to $\displaystyle 1$ than to $\displaystyle -i$.

Can you prove that is a half-plane. Can you show that?

Are half-planes open?

Re: Prove that this set is open, hopefully just need help with the inequalities

METHOD 1: (The best) In general, if $\displaystyle f, g : X \rightarrow \mathbb{R}$ are continuous, then $\displaystyle ( f - g ) : X \rightarrow \mathbb{R}$ is continuous.

Thus in $\displaystyle X$, $\displaystyle ( f - g )^{-1}( (0, \infty) )$ is open, hence $\displaystyle \left\{ x \in X : (f-g)(x)>0 \right\}$ is open, hence $\displaystyle \left\{ x \in X : f(x) > g(x) \right\}$ is open.

This is the general way to handle this. Now just note that $\displaystyle f, g : \mathbb{C} \rightarrow \mathbb{R}$ by $\displaystyle g(z) = |(z-1)|$ and $\displaystyle f(z)=|(z+i)|$ are continuous, and you're done.

METHOD 2: Let $\displaystyle U = \mathbb{C}-\{-i\}$, which is open in $\displaystyle \mathbb{C}$.

Note that $\displaystyle {-i} \notin S$, so that $\displaystyle S = \left\{z \in U : |(z-1)/(z+i)| < 1 \right\} = f^{-1}(D(0;1))$, where $\displaystyle f : U \rightarrow \mathbb{R}$ is given by $\displaystyle f(z) = |(z-1)/(z+i)|$.

(There, $\displaystyle D(0;1)$ is the open interval $\displaystyle (-1, 1)$ in $\displaystyle \mathbb{R}$.) Since $\displaystyle f$ is a composition of continuous functions, $\displaystyle S$ is open in $\displaystyle U$, and hence $\displaystyle S$ is open in $\displaystyle \mathbb{C}$.

(That's because $\displaystyle S$ open in the relative topology implies $\displaystyle S = U \cap V$ for some $\displaystyle V$ open in $\displaystyle \mathbb{C}$. But since $\displaystyle U$ is open in $\displaystyle \mathbb{C}$, conclude that $\displaystyle S$ is open in $\displaystyle \mathbb{C}$).

METHOD 3: "Read" what S is. S is the set of z that are closer to 1 than they are to -i. That's clearly the upper part of the plane that's split by the perpendicular bisector from 1 to -i., which is graphed in the plane as y > -x. Thus $\displaystyle S = \left\{z \in \mathbb{C} : Im(z)+Re(z)>0 \right\}$. An open ball around any such point is would be $\displaystyle D(z;r), r = (Im(z)+Re(z))/\sqrt{2}$.

METHOD 4: The algebraic way to do method 3: Let $\displaystyle z=x+iy \in S$. Have:

$\displaystyle | z - 1 | < | z + i | \Leftrightarrow | z - 1 |^{2} < | z + i |^{2} \Leftrightarrow (x-1)^{2} + y^{2} < x^{2} + (y+1)^{2} \Leftrightarrow x^{2} - 2x + 1 + y^{2} < x^{2} + y^{2} + 2y + 1$

Simplifying that gives: $\displaystyle z=x+iy \in S \Leftrightarrow x+y>0.$

Now let $\displaystyle z_{0}=x_{0} + iy _{0} \in S$, let $\displaystyle r = (x_{0} + y_{0})/\sqrt{2}$ (note $\displaystyle r > 0$), and let $\displaystyle z_{1}=x_{1} + iy _{1} \in D(z_{0};r)$.

Then $\displaystyle | z_{1} - z_{0} |^{2} = (x_{1}-x_{0})^{2} + (y_{1}-y_{0})^{2} < r^{2} = (x_{0} + y_{0})^{2}/2 = 2 ( (x_{0} + y_{0}) / 2 )^{2}$. Then do the algebra to prove that $\displaystyle x_{1} + y_{1} > 0$.

I didn't work it out, but you get the idea. From what's already been said, it *must* be true that that suffices to prove $\displaystyle x_{1} + y_{1} > 0$.