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There exists a continuous function f:[0,1] onto [0,10], but there exists no continuous function g:[0,1] onto (0,10) . The answer is True. How??

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    $\begingroup$ The image of a compact set under a continuous map is compact. $\endgroup$ – Daniel Fischer Sep 16 '13 at 11:39
  • $\begingroup$ "Functions" is (in my view) a silly tag. Half the questions on m.se involve functions. $\endgroup$ – Gerry Myerson Sep 16 '13 at 11:42
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    $\begingroup$ An example of what? $\endgroup$ – DonAntonio Sep 16 '13 at 11:42
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    $\begingroup$ Hardik, I'm certain you can find an example of a continuous function from $[0,1]$ onto $[0,10]$. $\endgroup$ – Gerry Myerson Sep 16 '13 at 11:43
  • $\begingroup$ ok.... i'll try to find such an example. Thanks. $\endgroup$ – Dysfunctional Sep 16 '13 at 11:48
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Using the extreme value theorem, a continuous function $f$ from the closed interval $[0,1]$ onto (or even into) $(0,10)$ takes on a maximum $M$ and a minimum $m$ somewhere in $(0,10)$. But then any real in the interval $(0,m)$ is not taken on by $f$, so that $f$ is not onto. [Also $f$ cannot hit any real in $(M,10).$]

The extreme value theorem may be known by any calc 1 student, who might not know what "compactness" is.

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  • $\begingroup$ Thanks ..... The idea just didnt come upon my mind. $\endgroup$ – Dysfunctional Sep 16 '13 at 15:51
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    $\begingroup$ @Hardik Yes, I thought for a moment one could start off with $f(0)$ somewhere in $(0,10)$, then "wiggle it around a lot" so as to pick up all of $(0,10)$. But then I recalled the max/min guaranteed by extreme value theorem. $\endgroup$ – coffeemath Sep 16 '13 at 16:22
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For a function to be continuous the pre-image of an open set should be open. Here the pre-image of the open set $(0,10)$ is closed $[0,1]$. Hence the function is not continuous

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    $\begingroup$ $[0,1]$ is open in the domain of the function (which is $[0,1]$). That argument doesn't settle it. $\endgroup$ – Daniel Fischer Sep 16 '13 at 11:49
  • $\begingroup$ @Hardik In fact, it's one of definitions of a continuous function. To prove equivalence, you need to give us your definition of continuity. $\endgroup$ – TZakrevskiy Sep 16 '13 at 11:53

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