Does there exist a continuous onto/surjective function from $[0, 1) \to \Bbb R$?
Finding difficult to site an example...
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2 Answers
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What about $f\colon[0,1)\to\mathbb{R}$ defined by $$f(x) = \frac{1}{1-x}\sin\frac{1}{1-x}?$$
answered Jul 8, 2016 at 16:36
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$\begingroup$ how did you construct the example?? $\endgroup$– User8976Commented Jul 8, 2016 at 16:37
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$\begingroup$ I was trying to think of something oscillating! (+1) Cool! $\endgroup$– snultyCommented Jul 8, 2016 at 16:37
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2$\begingroup$ @user8795 Examples of this sort are quite common: $\sin\frac{1}{x}$ and similar expressions are nice, as they will change sign increasingly faster near a point while still being bounded... it only remains to modulate with another expression to have their amplitude go up as well. $\endgroup$ Commented Jul 8, 2016 at 16:39
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1$\begingroup$ @user8795 I think it's kind of visual. You need a function nicely defined at zero, but somehow it needs to go to both $\pm \infty$ as it approaches $1$, hence you need something going to $\infty$ at $1$, like $\frac{1}{1-x}$, and something to make it oscillate, to cover all positives and negatives! Hence a $\sin$ $\endgroup$– snultyCommented Jul 8, 2016 at 16:40
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$\begingroup$ @ArticChar Not certain I agree with the edit for the punctuation, however... I liked the interrogation mark where it was. :) $\endgroup$ Commented Jul 8, 2016 at 16:40
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We can modify $f(x)=x\sin(x)$ to get our desired function. See that $f$ maps $[0,\infty)$ to $\mathbb{R}$ and $g(x)=\tan\left(\frac{\pi}{2}x\right)$ maps $[0,1)$ to $[0,\infty)$. Then we can choose $h=f\circ g$ to be the required continuous surjection.
