# State True or False

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??

• The image of a compact set under a continuous map is compact. – Daniel Fischer Sep 16 '13 at 11:39
• "Functions" is (in my view) a silly tag. Half the questions on m.se involve functions. – Gerry Myerson Sep 16 '13 at 11:42
• An example of what? – DonAntonio Sep 16 '13 at 11:42
• Hardik, I'm certain you can find an example of a continuous function from $[0,1]$ onto $[0,10]$. – Gerry Myerson Sep 16 '13 at 11:43
• ok.... i'll try to find such an example. Thanks. – Dysfunctional Sep 16 '13 at 11:48

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 $(M,10).$]
• @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. – coffeemath Sep 16 '13 at 16:22
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
• $[0,1]$ is open in the domain of the function (which is $[0,1]$). That argument doesn't settle it. – Daniel Fischer Sep 16 '13 at 11:49