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

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    $\begingroup$ Did you mean "convex"? $\endgroup$ – user228113 Jun 29 '15 at 10:32
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    $\begingroup$ possible duplicate of Continuous boundary of a convex set $\endgroup$ – user99914 Jun 29 '15 at 10:35
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    $\begingroup$ No. I mean conex $\endgroup$ – Student Jun 29 '15 at 10:48
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    $\begingroup$ @Student: "conex" does not mean anything. Maybe "connected"? $\endgroup$ – Jack D'Aurizio Jun 29 '15 at 12:17
  • $\begingroup$ Yes, I mean connected. Sorry about that stupid mistake of mine. In my language connected=conex. $\endgroup$ – 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 $\Omega = \{(x,y) \in \mathbb{R}^2 : x^2 + y^2 \leq 4\}$.
  • Not only is it connected, it is even path connected. Check this link out for a simple proof of this.
  • 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.

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