Proving $|z-1|<|z-i|$ is an open set Consider this set:
$$|z-1|<|z-i|$$
Suppose $z=x+iy$, then:
$$\sqrt{(x-1)^2+y^2}<\sqrt{x^2+(y-1)^2}\implies$$
$$(x-1)^2+y^2<x^2+(y-1)^2\implies$$
$$x^2-2x+1+y^2<x^2+y^2-2y+1\implies$$
$$-2x<-2y\implies y<x$$
First of all, am I right?
Now, in order to prove that the set $O = \{(x,y); y<x\}$ is open, I need to pick a point $z\in O$, then construct an open ball of radius $r$ and prove it's entirely contained in $O$. I think that $r$ in this case must be the distance from this point $z$ to the line $y=x$. I have, then, to pick a point $w\in B(z, r)$ and prove that $w\in O$. How can I do that?
 A: If you draw the picture, you see that your set is just $\mathcal O =\left \{ (x,y):y<x \right \}$ so pick a point $(x_0,y_0)\in \mathcal O$ and observe that the ball centered at $(x_0,y_0)$ of radius $\frac{\vert x_0-y_0\vert }{2\sqrt{2}}$ lies entirely in $\mathcal O$. 
Or if you know that $f(z)=|z-1|-|z-i|$ is continuous, then the result is immediate, since $\mathcal O=f^{-1}(\left \{ y\in \mathbb R:y<0 \right \})$.
A: Note that the equation $|z - 1| < |z - i|$ says that that distance of $z$ from $1$ should be less than its distance from point $i$. If you plot the points $A = 1$ and $B = i$ on the argand diagram you see that $1$ lies on positive real axis and $i$ lies on upper imaginary axis each at a distance of $1$ from origin. First we need to find the points which lie at exactly equal distance from both points $A$ and $B$ and by basic geometry we know that these points form the perpendicular bisector of $AB$.
From the position of $AB$ is it clear that the perpendicular bisector passes through origin and is given by equation $y = x$. Now this line $y = x$ divides the whole complex plane into two parts: one part contains the point $A$ and other contains point $B$. We have to find points $z$ which are near to $A = 1$ compared to the point $B$. So the points $z$ lie in the part of the plane which contains $A$. Visually this portion lies below the line $y = x$ and given by the inequality $x > y$.
It is now almost obvious that the region below the line $y = x$ is an open set. But you require a formal / symbolic proof. You have also got the right idea but perhaps find it difficult to express it symbolically. Let $z = c$ be a specific point in the region below line $y = x$. Then we know that this point does not lie on the line $y = x$. Thus the distance $r$ of point $c$ from line $y = x$ is positive. Let's put $ c = a + ib$ where $b < a$. Then by coordinate geometry we know that $$r = \frac{a - b}{\sqrt{2}}$$ Now you have to consider the points $z$ which lie in a circle with center $c$ and radius $r$ i.e. points $z$ which satisfy $|z - c| < r$. We need to show that these points $z$ also lie below the line $y = x$.
Suppose $|z - c| = r'$ where $r' < r$. Then we can write $z - c = r'(\cos \theta + i\sin \theta)$ so that $$z = (a + r'\cos \theta) + i(b + r'\sin \theta) = x + iy\text{ (say)}$$ so that $$x = a + r'\cos \theta, y = b + r'\sin \theta$$ and then
\begin{align}
x - y &= a - b + r'(\cos\theta - \sin\theta)\notag\\
&= a - b + r'\sqrt{2}\cos(\theta + \pi/4)\notag\\
&\geq a - b - r'\sqrt{2}\notag\\
&= r\sqrt{2} - r'\sqrt{2}\notag\\
&= (r - r')\sqrt{2}\notag\\
&> 0 \text{ (because }r' < r)\notag
\end{align}
So the point $z = x + iy$ with $|z - c| = r' < r$ is such that $x > y$ and hence it lies below the line $y = x$. Since $c$ was an arbitrary point of the region $x > y$ it follows that every point in the region has a small ball type neighborhood all of whose points lie in the region $x > y$. It follows that the region given by $x > y$ is an open set.
Note that the above symbolic proof is totally unnecessary unless demanded by some insistent teacher/professor. One can use the the following obvious fact from geometry that if $\ell$ is a line and $P$ is a point not on $\ell$ and $Q$ is a point on $\ell$ such that $PQ$ is perpendicular to $\ell$ then all points inside the circle with center $P$ and radius $PQ$ lie on the same side of line $\ell$ as the point $P$. In other words the line $\ell$ divides the plane into two disjoint open regions.
A: Your set  consists of  the points in the plane that are closer to $1$ than $i$.
Consider the sequence $x_n=1/n$ you can see for every $n \in N$  it belongs to your set.
In fact every  natural number is in your set .Now through that sequence you can approximate zero which makes zero a limit point of your set (also belongs to the closure of the set)  but $0$  does not belong in your set .
So by definition if it was closed it should contain all of its limit points.Now this proves that it is not closed. So it can be neither open. So no idea on what to do maybe you can figure it out.
Another way to go is Consider the boundary of your  plane $y=x$  so now consider a point in your set $(a,b)$ what is its distance from  the line $y=x$ ? Now take a ball around $(a,b)$ with radius the half distance of the point $(a,b)$ from your line .It should be contained completely in your set. 
HINT:TO find the distance read this https://en.wikipedia.org/wiki/Distance_from_a_point_to_a_line 
A: It is an open set if $\forall z$ s.t. $|z-1|<|z+i|\; \exists \epsilon>0$ s.t. $|z-w|<0\implies |w-1|<|w+i| $.
By the triangle inequality, $|a|-|b|\leq|a-b|\; \forall a,b\in\mathbb{C}$. Set $a=|z-1|$ and $b=|w-1|$. Then clearly $|z-1|-|w-1|<\epsilon$. Likewise $|z+i|-|w+i|<\epsilon$.
So if $|z-1|<|z+i|$ and $|z-w|<\epsilon$ then $|w-1|<|w+i|$. I.e. if $z$ is in the set, you can find an open ball around $z$ which is also in the set.
A: I see many answers already, I hope I'm getting the point of what you want to clarify.
Take a point $z=(z_x,z_y)$ in $O$, i.e. $z_y<z_x$. Then $z_x=z_y+\delta$ for a positive number $\delta$.
You must prove that there's a ball $B(z,r)$ contained entirely in $O$. Since you can choose the radius, let's take it small, say $r=\delta/100$. If you take $w\in B(z,r)$ by definition
$$|w_x-z_x|=\sqrt{(w_x-z_x)^2}\leq \sqrt{(w_x-z_x)^2+(w_y-z_y)^2}<\frac{\delta}{100}.$$
Then
$$w_x\geq z_x-|w_x-z_x|>z_x-\frac{\delta}{100}=z_y+\frac{99}{100}\delta$$
where the first inequality is simply $z_x-w_x\leq|z_x-w_x|$.
Doing the same for the $y$ coordinates we obtain
$$|w_y-z_y|=\sqrt{(w_y-z_y)^2}\leq \sqrt{(w_x-z_x)^2+(w_y-z_y)^2}<\frac{\delta}{100}$$
and similarly (but this time reversing the role of $z$ and $w$!)
$$w_y\leq z_y+|w_y-z_y|<z_y+\frac{\delta}{100}.$$
In the end we obtain
$$w_y<z_y+\frac{\delta}{100}<z_y+\frac{99}{100}\delta <w_x$$
i.e. $w_y<w_x$.
A: The first part of your answer is correct. And by checking that the implications validly run backward, we have $|z-1|<|z-i|\iff Im (z)<Re(z).$
More simply: If $|z-i|-|z-1|=d>0$ then every $z'$ in the open ball $B(z,d/2)$ satisfies $|z'-i|-|z'-1|>0$  by the triangle inequality : $$|z'-1|=|(z-1)+(z'-z)|\leq |z-1|+|z'-z|<|z-1|+d/2<(|z-i|-d)+d/2=$$ $$=|z-i|-d/2.$$ while $$|z'-i|=|(z-i)+(z'-z)|\geq |z-i|-|z'-z|>|z-i|-d/2.$$
A: Proof by definition: every point has a ball around it contained in the set.
Assume $z$ is a point satisfying the inequality. The point is you can insert a real number in between,
$$ \exists \ \ c>0 \ \ \ \ (|z-1|<c<|z-i|). \ \ \ \  (*)$$
Now, $z$ is in the interior of a circle of radius $c$ centered at $1$, and therefore a smaller circle around $z$ is contained in it. Similarly, $z$ is in the exterior of the circle at $i$ with radius $c$, and again a (possibly smaller) circle around $z$ is still in the exterior.
Any point near $z$ in the circle with the smaller of the two radii above still satisfies the inequality (*), and hence contained in the set
$$|z-1|<|z-i|.$$
