Prove $D(a;R_1,R_2)$ is a connected set I was solving problems from the start of my Complex Analysis course, and I found this one (the beggining of my course focuses a lot in topology):

Prove that $D(a; R_1,R_2)$ is a connected set.

The notation $D(a;R_1,R_2)$ refers to the open disk of center $a$ and with radius greater that $R_1$ but lower that $R_2$ (I guess, because this is the first time I see a disk notation with two different radius, tell me if I'm wrong) (Personally, I prefer the ball notation, but this is the one my course notes use). Also consider I'm working inside the metric space $(\mathbb{C},d)$, where $d$ is the euclidean metric function. To be said:
$$D(a;R_1,R_2)=\{z\in\mathbb{C} : |z-a|>R_1, |z-a|<R_2\}$$
Intuitively, I understand this set is connected, since it's like a "donut", but I have trouble proving it by using the definition of connected space, to be said, proving it isn't union of to open (or closed) sets. I'm not sure if the concept of connected by arcs is useful here, though I clearly see it in my mind I don't know how to write a consistent proof of it.
How can I proof that set is connected? Any help will be appreciated, thanks in advance.
 A: These things are often called "annuli" (in the singular, "annulus"), from the Latin for something that looks like a ring.  It makes intuitive sense that these things are arcwise connected because there is nothing stopping you from drawing a curve from any point in one of these things to any other without leaving the annulus.
If you want example formulas, I'll handle the case $a = 0$ because it's simple.  In polar form, $D(a,R_1,R_2) = \{r e^{i\theta}: R_1 < r < R_2\}$.  I'll just write $D$ for this set.
Fix any $z$ and $w$ in $D$ and choose positive $r_0, r_1$, and real $\theta_0$ and $\theta_1$ such that $z = r_0 e^{\theta_0}$ and $w = r_1 e^{\theta_1}$.  Note that because $z$ and $w$ are in $D$, $r_0$ and $r_1$ are both in the real interval $(R_1, R_2)$.
Consider the path $\gamma: [0,1] \to \mathbb{C}$ given by
$$
\gamma(t) = ((1-t) r_0 + t r_1) \exp(i((1-t) t_0 + t \theta_1)), \qquad t \in [0,1].
$$
Convince yourself that $\gamma$ is continuous and that $\gamma(0) = z$ and $\gamma(1) = w$, and it may help to think geometrically about what $\gamma$ does.  As $t$ goes from $0$ to $1$ it just moves the modulus and angle from that of $z$ to that of $w$ in a linear way; it'll look a bit like part of a spiral or something - maybe graph a few examples.
Fix $t$.  We have that
$$
|\gamma(t)| = (1-t) r_0 + t r_1
$$
Convince yourself via the fact that $r_0, r_1$ are both in $(R_1, R_2)$ that $|\gamma(t)| \in (R_1, R_2)$ also.  (In other language $|\gamma(t)|$ is a "convex combination" of $r_0$ and $r_1$ and $r_2$, so will lie in any convex set that contains $r_1$ and $r_2$.  But you don't need this language to verify the simple inequalities.)  So $|\gamma(t)| \in D$.
So $\gamma$ is an arc in $D$ connecting $z$ to $w$.  Since $z$ and $w$ were arbitrary, $D$ is arcwise connected.
The general case could be handled via a use of this case, or by just adding "$a+$" to a lot of the formulas.
