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So, the question is which one of the following is true?

a) There exists a continuous function from $[-\pi/2,\pi/2]$ onto $(0,1)$

b) There exists a continuous function from $[-\pi/2,\pi/2]$ onto R

c)There exist a continuos function from $[0,\pi]\cup[2\pi,3\pi]$ onto $[0,1]$

d)There exist a continuos function from $[-\pi/2,\pi/2]$ onto $[0,1/3]\cup[2/3,1]$

I know that a) and b) are not true because continuous functions map closed intervals to closed intervals. However, I don't understand how to know which one of c) or d) is true. How do I tackle c) and d) Thanks in advance!!

Natasha J
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  • You are contradicting yourself and b) is missing something. – copper.hat Aug 07 '21 at 03:56
  • How am I contradicting myself? I corrected b) – Natasha J Aug 07 '21 at 03:58
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    You have some misunderstanding: continuous functions does not necessarily take closed sets onto closed sets, such maps are called closed maps. One one hand, by the definition of continuity the preimage of any open set is open, and by taking complement it works for closed sets as well. On the other hand, what is true is that continuous maps take connected (resp. compact) sets onto connected (resp. compact) sets (take a look carefully at the item b and c). – rebo79 Aug 07 '21 at 03:59
  • Isn't the image of a closed interval under a continuos function is a closed interval? – Natasha J Aug 07 '21 at 04:03
  • For a closed and bounded interval, the answer is yes, because such sets are compact. – rebo79 Aug 07 '21 at 04:04
  • Why isn't (a) true? Doesn't $y=\cos(x)$ satisfy the conditions in (a)? – Aditya Aug 07 '21 at 04:04
  • In a) both are closed intervals. – copper.hat Aug 07 '21 at 04:04
  • @rebo79 so open intervals are mapped to open intervals under continuous function and closed and bounded are mapped to closed and bounded respectively? Am I getting it correct?Also is there any condition on boundedness of open intervals? And thank you so much for answering! – Natasha J Aug 07 '21 at 04:08
  • @NatashaJ the first statement you made is false, look at my answer: by the definition of continuity the statement is exactly the opposite one. Which is true is that intervals are mapped onto intervals under a continuous mapping, because intervales are connected. – rebo79 Aug 07 '21 at 04:16
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    I am sorry, it isn't clicking with me. So continuous function maps intervals onto intervals(irrespective of whether they are open , closed or semi closed, etc) because intervals are connected? – Natasha J Aug 07 '21 at 04:22
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    @NatashaJ yep, thats it – rebo79 Aug 07 '21 at 06:37

1 Answers1

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Your answers for a) and b) are correct (if you interpret closed interval as closed and bounded interval).

c) $f(x)=\sin (\frac x 2)$ for $0\leq x \leq \pi$ and $f(x)=\sin (\frac {x-2\pi} 2)$ for $2\pi \leq x \leq 3\pi$

d) DNE exist because continuous maps map intervals to intervals (by connectedness).

Kavi Rama Murthy
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  • Thank you so much! And you are right, I was interpreting closed interval as closed and bounded interval. – Natasha J Aug 07 '21 at 05:56