Let $f:\mathbb{R^n}\to\mathbb{R^m}$ be a function such that the image of any closed bounded set is closed and bounded. Must $f$ be continuous?
Must $f$ be continuous?
3
$\begingroup$
real-analysis
general-topology
continuity
2 Answers
5
No.
For example take $f : \mathbb{R} \rightarrow \mathbb{R}$ defined by
$f(x) = 0$ if $x < 0$
$f(x) = 1$ if $x \geq 0$
-
0Wow, this is a case o$f$ two solutions independently and simultaneously being developed i$f$ I ever saw one :) I'll delete mine since yours is timestamped earlier. – 2012-11-01
3
Assume that the image by $f$ of any set is closed and bounded, for example because $f(\mathbb R^n)=\{a,b\}$ for some $a\ne b$ in $\mathbb R^m$. Such functions $f$ need not be continuous (example?).
(Once you will have exhibited such a function discontinuous at a point, say, you might try to find one which is discontinuous everywhere, since the idea is the same.)
-
0If $n=m=1$, try $f(x)=\sin(1/x)$ for $x\ne0$ and $f(0)=0$. – 2012-11-01