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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?

Robert Lewis
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Armando Rosas
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    Hint: Continuous functions map connected sets to connected sets. – MSDG Dec 10 '18 at 20:12
  • @MisterRiemann Ok, so, since all intervals in R are connected, if A is connected, then is an interval, and to prove that I have to see the preimage of A is connected? And thats true, since it's preimage is the circunference of radius 1? – Armando Rosas Dec 10 '18 at 20:15
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    Yes, you should prove that the set set $B={x \in \mathbb R^2 , \mid, \Vert x\Vert=1}$ is connected, and then so is $A=f(B)$. Be careful with your wording though. The set $A$ is indeed the image of the connected set $B$, but this does not mean that the whole preimage of $A$ is connected (the preimage could be contain elements which are not in $B$), but you don't even need that. Also check out this post for a proof of the fact that all connected sets on $\mathbb R$ are intervals. – MSDG Dec 10 '18 at 20:19
  • Can you precise what $f(x) \in(R)$ means? – mathcounterexamples.net Dec 10 '18 at 20:22
  • @mathcounterexamples.net made the correction, it was an error in the coding, it's the real numbers – Armando Rosas Dec 10 '18 at 20:25
  • This is about images, not pre-images. The circle $S^1$ is a connected space so any continuous image of $S^1$ is a connected space. However, the usual proof that a continuous image of a connected space is connected $does$ discuss pre-images. – DanielWainfleet Dec 11 '18 at 09:27

2 Answers2

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$A = f[S^1]$ 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 $a \le b$.

Henno Brandsma
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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.

Claire
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