Is there any continuous function defined on $(0, 1]$ whose range is $(0, 1)$?
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2$\begingroup$ f cannot be uniformly continuous, or it would extend to a continuous function $F:[0,1] \rightarrow (0,1) $. $\endgroup$– Passing ByDec 5, 2014 at 3:50
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2$\begingroup$ Why would the extension have (0,1) as range and not include its limit points possibly $\endgroup$– AmanDec 5, 2014 at 4:03
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$\begingroup$ @Aman: If f was extended continuously into $[0,1]$ the range would have to be compact, so could not be $(0,1)$ $\endgroup$– Passing ByDec 5, 2014 at 4:06
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1$\begingroup$ @Passing By: By the same argument, I could say "There is no uniformly continuous and surjective function $f: (0,1] \to (0,1]$, because if there were, it could be extended to a continuous surjective function $F: [0,1] \to (0,1]$, which is impossible". Which is obviously false. $\endgroup$– D. ThomineDec 5, 2014 at 12:42
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2$\begingroup$ @D.Thomine: Good point, but with a small tweak, Passing By has a very good point also. A hypothetical extension would have range $(0,1)$ or $[0,1)$ or $(0,1]$, none of which is compact. $\endgroup$– Jonas MeyerDec 5, 2014 at 14:19
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4 Answers
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**Answer without words! see the figure **
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8$\begingroup$ I see words in the figure ... $\endgroup$ Dec 6, 2014 at 16:46
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Something like $$x\mapsto \frac{1+(1-x)\sin(1/x)}{2} $$ will do.
answered Dec 5, 2014 at 3:59
hmakholm left over Monicahmakholm left over Monica
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Yes, there is. First send $(0,1]$ onto a spiral that winds inside a circle (e.g centered at $x=1/2$ and radius $1/2$) and then project it onto $(0,1)$.
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$\begingroup$ <img src="i.stack.imgur.com/ipfcP.png" alt="spiral in the plane"> $\endgroup$– MirkoJun 30, 2021 at 15:40
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$$f(x)=\frac12+\frac1\pi\arctan\left(\frac1x\cos\left(\frac1x\right)\right)$$
