# Connected set imply continuous boundary

Is it true that a connected and bounded set of $\mathbb{R}^2$ has a boundary that can be parametrized by a continuous mapping?

• Did you mean "convex"? – user228113 Jun 29 '15 at 10:32
• possible duplicate of Continuous boundary of a convex set – user99914 Jun 29 '15 at 10:35
• No. I mean conex – Student Jun 29 '15 at 10:48
• @Student: "conex" does not mean anything. Maybe "connected"? – Jack D'Aurizio Jun 29 '15 at 12:17
• Yes, I mean connected. Sorry about that stupid mistake of mine. In my language connected=conex. – Student Jun 30 '15 at 16:22

This is not true in general. For a simple counterexample, consider the following set in $\mathbb{R}^2$: $\mathcal{S} = \{(x,y) \in \mathbb{R}^2 : 1 \leq x^2 + y^2 \leq 4\}$. $\mathcal{S}$ is just an annulus centered at the origin whose larger radius is $2$ and whose smaller radius is $1$.
• It is bounded because it is, by construction, entirely contained in the disk $\Omega = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 4\}$.
• However, its boundary $\delta \mathcal{S}$ is not connected since it comprises of two disjoint circles centered at the origin with radii $2$ and $1$ respectively.
The last point means that any function $f : [0,1] \to \delta \mathcal{S}$ whose image is the whole of $\delta \mathcal{S}$ cannot be continuous since the interval $[0,1] \subseteq \mathbb{R}$ is connected while our image is not. This follows from the fact that the continuous image of a connected space must be connected.
So you cannot have a continuous parameterization of the boundary of our annulus $\mathcal{S}$ despite the fact that it is both bounded and connected.