Convergence of moving point inside of unit disk. Suppose Set D :
$$D=\left\{(x, y)\vert x^2 + y^2 \leq 1 \right\}$$
And Point $P(0, 0)$ on coordinate plane.
Define 'Movement' : 
For a point P, select any direction and move $\frac{1}{2^n}$ to straight (Sorry, my english is poor.) When 'Movement' Execute $n$-th times

For example : 
$1$st time, Suppose I selected positive-x-axis direction
Then, Point $P(0,0)$ moves to $P\left(\frac{1}{2},0\right)$
$2$nd time, Suppose I selected Positive-y-axis direction
Then, Point $P\left(\frac{1}{2}, 0\right)$ moves to $P\left(\frac{1}{2},\frac{1}{4} \right)$

If I take $n\longrightarrow\infty$, Arbitrary point $A(x, y)\in D$ can be expressed by 'Movement'?
I Know that perimeter of $x^2+y^2=1$ can be expressed by 'Movement' because $$\sum_{n=1}^\infty \frac{1}{2^n}=1$$
But how about inside of $x^2 + y^2 =1$?

Example

The point with $1, 2, 3, 4, \cdots$ is a point which moved by $n$-th 'Movement'
The angle(direction) is free.
the problem is : If $n \longrightarrow\infty$, arbitrary points in $D$ can be expressed by 'Movement'?

Short answer: Not only you can do this, it is even possible to move on one line connecting the origin and your point of interest. Maybe not surprisingly, this is related to the binary representation of a number.
Consider a point $p=re^{i\theta}=x+iy$, and let us limit ourselves to movement at angle $\theta$ only (i.e. on the line connecting the origin and $p$). Thus, this problem essentially boils down to the question if every number $-1\leq z\leq 1$ can be written as
$$z=\sum_{n=1}^{\infty}\dfrac{a_{n}}{2^{n}}=\sum_{n=1}^{\infty}\dfrac{\frac{1+a_{n}}{2}}{2^{n-1}}-\sum_{n=1}^{\infty}\dfrac{1}{2^{n}}=\sum_{n=1}^{\infty}\dfrac{b_{n}}{2^{n-1}}-1=\sum_{n=0}^{\infty}\dfrac{c_{n}}{2^{n}}-1$$
with $a_{n}\in\left\{-1,1\right\}$, $b_{n}=\frac{1+a_{n}}{2}\in\left\{0,1\right\}$ and $c_{n}=b_{n+1}\in\left\{0,1\right\}$. Let $q=z+1$, so $0\leq q\leq 2$ and the above condition is
$$q=\sum_{n=0}^{\infty}\dfrac{c_{n}}{2^{n}}$$
with $c_{n}\in\left\{0,1\right\}$. But this is just binary representation.

Example 1: Want to get back to the origin? No problem. Take $z=0$, thus $q=1$ and $c_{n}=\left(1,0,0,\dots\right)\Longrightarrow a_{n}=\left(1,-1,-1,\dots\right)$. In this case $\theta$ is arbitrary so let's walk on the $x$ axis. This means you should go
$$\left(0,0\right)\rightarrow\left(\frac{1}{2},0\right)\rightarrow\left(\frac{1}{2}-\frac{1}{4},0\right)\rightarrow\left(\frac{1}{2}-\frac{1}{4}-\frac{1}{8},0\right)\rightarrow\dots\rightarrow\left(\frac{1}{2}-\frac{1}{2}\sum_{n=1}^{\infty}\frac{1}{2^{n}},0\right)=\left(0,0\right)$$
Example 2: Want to go to $\left(\frac{1}{2},0\right)$? Take $\theta=0$ and $z=\frac{1}{2}$, thus $q=\frac{3}{2}$ and $c_{n}=\left(1,1,0,0,\dots\right)\Longrightarrow a_{n}=\left(1,1,-1,-1,\dots\right)$. This time your path is
$$\left(0,0\right)\rightarrow\left(\frac{1}{2},0\right)\rightarrow\left(\frac{1}{2}+\frac{1}{4},0\right)\rightarrow\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{8},0\right)\rightarrow\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{8}-\frac{1}{16},0\right)\rightarrow\dots\rightarrow\left(\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\sum_{n=1}^{\infty}\frac{1}{2^{n}},0\right)=\left(\frac{1}{2},0\right)$$
