Boundary of a certain set in ${R}^2$ I've been studying topology on my own and came upon a curious question.
Let the function $f$ be a function from $(0,1)$ to a compact subset $K$ of ${R}^2$.
This function is injective and continuous.
Then, does the boundary of $f((0,1))$ have a subset which is not in $f((0,1))$ and is a continuous curve?
Thanks in advance! :)
 A: Sure. Let $f(x) = (x,\sin (1/x)).$ Then $f$ is bounded, hence its range is contained in a compact set, and it is continuous and injective. The boundary of $f((0,1))$ contains $\{0\}\times [-1,1],$ which is a continuous curve not in $f((0,1)).$
A: You can even have a continuous injective map $f: (0,1)\to R^2$ whose image is compact, hence, its boundary is entirely contained in $f((0,1))$. For instance, the image could the the figure 8. 
Edit. To get an explicit example, start with the map 
$$h(t)= (\cos(t- \frac{1}{2}\pi), \sin(t- \frac{1}{2}\pi)+1)= (x(t), y(t)), t\in [0, 2\pi),$$
 which is a continuous bijection to the unit circle $C_1$ centered at the point $(0,1)$. Now, extend $h$ to a map $g$ on the interval $(-2\pi, 0]$ by the formula
$$
g(t)= (x(-t), -y(-t)). 
$$
The image of this map is the unit circle $C_2$ centered at $(0,-1)$. 
Note that $f(0)=g(0)=(0,0)$. 
Then define a continuous bijection $f: (-2\pi, 2\pi)\to C_1\cup C_2$, where $f$ restricts to $h$ on the interval $[0,2\pi)$ and to $g$ on 
on the interval $(-2\pi,0]$. The image of this continuous bijection is, of course, compact (the "figure 8").  
