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

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

    My attempt at a proof so far gets here:

    Let z \in S. Then D\left\(z;r\right\) \subseteq S if, \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 z+x \in S, equivalently, that |z+x-1| < |z+x+i|
    I figure I start with |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.
    Last edited by WonderingUser; September 9th 2012 at 03:05 PM.
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  2. #2
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    Re: Prove that this set is open, hopefully just need help with the inequalities

    Quote Originally Posted by WonderingUser View Post
    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
    S:= \left\{z \in \mathbb{C} : |z-1| < |z+i|\right\} is open.
    My attempt at a proof so far gets here:
    Let z \in S. Then D\left\(z;r\right\) \subseteq S if, \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 z+x \in S, equivalently, that |z+x-1| < |z+x+i|
    I figure I start with |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 S looks like?
    It is the set of points which closer to 1 than to -i.

    Can you prove that is a half-plane. Can you show that?
    Are half-planes open?
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  3. #3
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    Re: Prove that this set is open, hopefully just need help with the inequalities

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

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

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

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

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

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

    (That's because S open in the relative topology implies S = U \cap V for some V open in \mathbb{C}. But since U is open in \mathbb{C}, conclude that S is open in \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 S = \left\{z \in \mathbb{C} : Im(z)+Re(z)>0 \right\}. An open ball around any such point is would be D(z;r), r = (Im(z)+Re(z))/\sqrt{2}.

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

     | 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: z=x+iy \in S \Leftrightarrow x+y>0.

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

    Then  | 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 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 x_{1} + y_{1} > 0.
    Last edited by johnsomeone; September 10th 2012 at 12:10 AM.
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