Finding a choice for Epsilon for open/closed set proofs I'm studying the proofs for open/close sets by using the following definition:

I'm having problems to understand the proofs.
The proofs sounds pretty straightforward: just choose a value for Epsilon and you can show that the open disc D(w, Epsilon) is or is not contained in the set.
Now, most exercise have a pretty vague and random way to guess a workable Epsilon.
Is there a rule of thumb I can use to decide which one to use?
For example, sometimes I see $\epsilon = \text{Re}(z)/2$, other times $\epsilon = (|z| + 1) / 2$  and this puzzles me as I never know which value to use in order to prove it.
Thanks.
 A: I would say this is more of an art, and you get used to it the more you do. (I know that's horrible to hear - I'm only a third year mathmo, so I'm still in the early stages!) A basic principle is that you don't pick your $\epsilon$ at the start. You say "Let $\epsilon >0$." and then you work through with $\epsilon$, and then after manipulation, it will be become clear(er) what form $\epsilon$ should be. Sometimes, part way through the manipulation, you have to make an educated guess at the form (maybe write $\epsilon = \eta \|x\|,$ or something like that to make it easier to see).
Also, remember this: it doesn't matter how small $\epsilon$ is, just provided it is greater than $0$. So if you think that $\epsilon = \frac{1}{2} \|x\|$ will work, but maybe it needs to be $1/3$ not $1/2$, then just take $\epsilon = \frac{1}{10}\|x\|$ to be sure. Again, the specifics come with practice.
As you learn more methods, you don't have to use the $\epsilon$ notation every time, but instead you can deduce it with a lot less writing!
Anyway, hope this helps! :)

Comments Related Appendix (sounds important)
Let $R = \{ z \in \Bbb C \ | \ Re(z) > 0 \}$. This is open. Using the $\epsilon$ definition, pick $z \in R$ and write $z = x + i y$ with $x,y \in \Bbb R$. By definition of being in $R$, $x > 0$ and there is no constraint on $y$. Let $\epsilon > 0$. We need to show that $ |z - z'| < \epsilon \Rightarrow z' \in R$. We write $z' = z + (\eta + i\xi)$ where $|\eta + i\xi| < \epsilon$. We don't care about the imaginary part - there's no constraint on that - so our only constraint is $x + \eta >0$. Here's the important part. $x$ is given, so we just need $\eta > -x$.
$$|\eta + i\xi| < \epsilon \Rightarrow |\eta| < \epsilon,$$
so just take $\epsilon < x$. Then $|\eta| < \epsilon < x$, so $-x < \eta < x$, and we have our required bound. Note that this $\epsilon$ isn't necessarily small, but is only small if $x$ is small. To be on the safe side, I'd probably just take $\epsilon < \frac{1}{2}x$.
Now let $I = \{ z \in \Bbb C \ | \ Im(z) \ge 0 \}$. In your lecture notes, it will show that a set is closed if and only if its complement is open, and a set is open if and only if its complement is closed.
$$I^c = \{ z \in \Bbb C \ | \ Im(z) < 0 \},$$
which we know is open by analogous argument to above.
We can also show it directly. Just let $(z_n)$ be a sequence in $I$ converging to $z \in \Bbb C$ - we must show that $z \in I$; write $z_n = x_n + i y_n$ and $z = x + y$ as before. We know that $y_n \ge 0$ for all $n$, since $z_n \in I$. Since $z_n \to z$, we have that $y_n \to y$ for some $y \in \Bbb R$.[1] Given any $\epsilon > 0$, there exists $N \in \Bbb N$ such that $| y_n - y| < \epsilon$ for all $n \ge N$ (by definition). If $y > 0$, then just take $\epsilon = \frac{1}{2}y > 0$, and then $|y_N - y| < \frac{1}{2}y$, and so $y_N > 0$, which is a contradiction since $y_N \in I$.

Ok, that took quite a lot longer than expected - hopefully it is helpful though! :)
A: Keeping in mind that the set of all $w$ such that $|w-z| < \epsilon(z)$
is a disk with center at $z$, you need the entire disk to be inside
your set $A$ in order for the proof to work.
You need this to be true no matter which $z \in A$ someone might ask about.
That's basically all there is to it.
Choose an arbitrary $z \in A$.
If you can find the distance from $z$ to the closest point on the boundary
of $A$ as a function of $z$, then you can set $\epsilon(z)$ to that function,
and the entire disk of points $w$ such that $|w-z| < \epsilon(z)$
will be inside $A$.
If $A$ includes any points on its boundary, then this will not work,
because when you choose $z$ on the boundary of $A$ then the distance to
the boundary is zero, and we need $\epsilon(z) > 0$ in order to prove
that $A$ is open. But in this case $A$ is not open, so of course we
cannot find a suitable $\epsilon(z)$ for every $z \in A$.
It is often difficult to come up with a function that gives you the exact
distance to the boundary of a set $A$ from any interior point.
That is where you may need to be a little creative.
You just need to ensure that $\epsilon(z) > 0$
and that $\epsilon(z)$ is no greater than the distance to the boundary.
You can never go wrong by choosing a smaller positive value of $\epsilon(z)$.
