Does level set has a boundary?

Question

Let $g:\mathbb{R}^n \to \mathbb{R}$ be a continuous function and for some $c>0$, we think its $c$-level set $g^{-1}(c)$.

Then does $g^{-1}(c)$ has a boundary? For example, is it possible that the set $\{x\in \mathbb{R} | 0\le x \le 1\}$ could be a level set of some continuous function?

I heard that if $g$ is differentiable, then its level set should be always manifold and so it thus not have boundary.

But how about level set for continuous function?

Any comments will be appreciated!

Answer 1

Yes the segment $[0,1]$ can be a level set of a continuous function. But for the example I'm going to construct it will be more convenient to show that $[1,2]$ is a level set of a continuous function

Consider the following continuous "trapezoid" function $f:\mathbb{R}\rightarrow\mathbb{R}$

1. For $x<0$ we let $f(x)=0$.
2. For $0\leq x<1$ let $f(x)=x$
3. For $1\leq x\leq 2$ let $f(x)=1$
4. for $2< x\leq 3$ let $f(x)= 3-x$
5. for $3<x$ let $f(x)=0$

(draw it)

I leave it as an exercise for you to show that this function is continuous (note that it is far from being differentiable). And it is easy to see that $[1,2]=f^{-1}(1)$.