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

**WonderingUser** · September 9th 2012, 02:59 PM

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.

**Plato** · September 9th 2012, 03:37 PM

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
$S:= \left\{z \in \mathbb{C} : |z-1| < |z+i|\right\}$ 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?

**johnsomeone** · September 9th 2012, 10:05 PM

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 $x_{1} + y_{1} > 0$ .

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