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Let $X = \{(a,b) \in \Bbb R^2 : a^2+b^2 =1 \}$ be the unit circle inside $\Bbb R^2$. Let $f : X \to \Bbb R$ be a continuous function. Then

  1. Image($f$) is connected.

  2. Image($f$) is compact.

  3. The given information is not sufficient to determine whether Image($f$) is bounded.

  4. $f$ is not injective.

My Attempt:

Since $X$ is compact and connected and $f$ is continuous then $f(X)$ is compact and connected. So options 1,2 are true and 3 is false. I have no knowledge how to handle last option. Please help me. Thanks in advance.

Mera bhai
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  • Your reasoning for 1 and 2 is sound. In regards to 3, what do you know about compact sets? Do you know any theorems about compact sets in $\Bbb{R}^n$? – Theo Bendit Jun 25 '21 at 16:23
  • That's good! Now, what about 4? Do you have a guess as to whether it's true or not? – Theo Bendit Jun 25 '21 at 16:28
  • Hint for 4: there are two ways of going from $(0,1)$ to $(0,-1)$ in $X$. – TonyK Jun 25 '21 at 16:36
  • @TonyK, By the hint, $(0,1) \neq (0,-1)$ but $f(0,1) = f(0,-1)$ so it is not injective. Am I right ? – Mera bhai Jun 25 '21 at 16:46
  • Please either remove the negative vote or tell me a reason behind it so that I can improve my knowledge. – Mera bhai Jun 25 '21 at 16:50
  • @Merabhai Regarding the negative vote, there's no way to figure out who it was (it wasn't me, and probably not Tony). I would guess that whoever it was thought you hadn't written down enough of your efforts? I totally disagree. Have a +1 from me. – Theo Bendit Jun 25 '21 at 16:56
  • @Theo Bendit, thanks, you are great and kind hearted. – Mera bhai Jun 25 '21 at 17:10
  • No, you have no reason to assume that $f(0,1)=f(0,-1)$. But if they are not equal, then by the Intermediate Value Theorem you can find two different points $(x,y)$ where $f(x,y)=\frac12(f(0,1)+f(0,-1))$, by going from $(0,1)$ to $(0,-1)$ along the two different paths. – TonyK Jun 25 '21 at 17:58

2 Answers2

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The number 4 is a particular case of the borsuk-ulam theorem, but I give another proof for this case.

Let be $f:S^1\to\Bbb{R}$, $x$ and $y$ two points in $S^1$ such that $f(x)$ is the minimum in $f(S^1)$ and $f(y)$ is the maximum in $f(S^1)$. If $f(x)=f(y)$ we got it, if not, we can consider the two arcs in $S^1$, connecting the points $x$ and $y$, let $R_1$, $R_2$ be the two arcs. We have: $$f(R_1)=[f(x),f(y)]=f(R_2),$$ therefore $f$ is not injective.

Sorry for my english.

Theo Bendit
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Gerson
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    Because the other user deleted their comment, have a link to the MathJax tutorial. It's basically the same as the mathematical bits of LaTeX, if you're familiar with it. feel free to check out how I've formatted your post by pressing "edit". If you play around with the code, you can see in real time how the code displays on the site. Good luck! – Theo Bendit Jun 25 '21 at 17:15
  • ah ok. Thank you very much – Gerson Jun 25 '21 at 23:56
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4 is true, and you can prove it by considering the functions $g(a,b)=f(a,b)-f(-a,-b)$ and $h(\theta)=(cos(\theta),sin(\theta))$. The latter maps any angle to a point on $X$.

Consider $g \circ h$. This is a continuous function from $[0,2\pi]$ to $\mathbb{R}$. By definition of $g$, it follows that $g \circ h(0)=-g \circ h(2\pi)$. By the Intermediate Value Theorem, there is a point $\theta\in [0,2\pi]$ where $g \circ h(\theta)=0$. At that point, by definition of our functions $g$ and $h$, we have $f(h(\theta))=f(-h(\theta))$. Since $h$ is nonzero on $[0,2\pi]$, this proves 4.

MathTrain
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