Is $A=\{f(x)\in\mathbb{R} : ||x||=1\}$ an interval if $f:\mathbb{R}^2\mapsto \mathbb{R}$ is continous?

Question

So I have $f:\mathbb{R}^2\mapsto \mathbb{R}$ a continous function and the set $A=\{f(x)\in\mathbb{R} : ||x||=1\}$ I have to prove that A is an interval, but I don't have any idea on how to do it. What specific property of intervals can use to compare it with A and see that A holds this property?

Answer 1

Choose a continous surjective curve $\gamma:[0,1]\rightarrow S^1=\{x\in\mathbb{R^2}:\|x\|=1\}$, for example $t\mapsto (\cos(2\pi t),\sin(2\pi t)$). Then $f\circ\gamma:[0,1]\rightarrow\mathbb{R}$ is continous and hence by the intermediate value theorem it's image, which is $A$, is an intervall.

Answer 2

$A = f[S^1]$ where $S^1 = \{x \in \mathbb{R}^2: \|x\|=1\}$ is the unit sphere in the plane, which is compact and connected.

If $f$ is continuous, $A$ is thus a compact and connected subset of $\mathbb{R}$ hence of the form $[a,b]$ for some $a \le b$.