Decide on the validity of the following conjectures a) Continuous functions take bounded open intervals to bounded open intervals.

I can pick a counter example f(x) = tan(x) which is bounded -2 < x < 2 but f(x) is unbounded

b) Continuous functions take bounded open intervals to open sets. For this question, what is the difference between open intervals and open sets?

  • $\begingroup$ There's a simple example of a continuous function which maps an open interval to a closed interval... $\endgroup$ – aschepler Dec 9 '13 at 2:09
  • $\begingroup$ a flat line without two end points will do it right? $\endgroup$ – user114252 Dec 9 '13 at 2:10
  • $\begingroup$ Yes, $f(x)=C$ also works. I was thinking $f(x) = \sin x$: maps $(0,2\pi)$ to $[-1,1]$. $\endgroup$ – aschepler Dec 9 '13 at 2:13

Open sets are arbitrary unions and finite intersection of open intervals. For example $(0,1)\cup (2,5)$ is an open set, but it is not an interval.

  • $\begingroup$ Also the unbounded sets written $(-\infty,x)$ and $(y,\infty)$. $\endgroup$ – aschepler Dec 9 '13 at 2:08
  • $\begingroup$ is (0,1) U {5} an open set? $\endgroup$ – user114252 Dec 9 '13 at 2:08
  • $\begingroup$ And by the way, the statement is not true. $\endgroup$ – LASV Dec 9 '13 at 2:09
  • $\begingroup$ No. ${5}$ has no neighbourhood $N$ around it such that $N$ is completely contained in $(0,1)\cup \{5\}$. $\endgroup$ – LASV Dec 9 '13 at 2:12
  • $\begingroup$ @aschepler 4 These cases are taken care of by the arbitrary unions. $\endgroup$ – LASV Dec 9 '13 at 2:13

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