# Mapping of open interval

Let $f$ be a continuous function whose domain contains an open interval $(a, b)$. What form can $f(a, b)$ have?

Assume that $(a, b)$ is bounded. Does anyone know examples for the different forms this might take?

• I think we could easily map this to just a single constant. E.g., $f(x)$ = $0$. Then $f(0,1)$ = {$0$}.

• Or, we can easily map it to another open interval. E.g., $f(x)$ = $x$. Then $f(0,1)$ = $(0, 1)$.

Are there other possibilities?

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Consider $\sin(0,10)$... – Henning Makholm Feb 16 '12 at 18:12
Think about where $\tan x$ sends the interval $(-\pi/2, \pi/2)$. – Dylan Moreland Feb 16 '12 at 18:12

It has to be an interval because of the intermediate value theorem. Its length could be $0$, in which case it's a closed interval $[c,c]=\{c\}$. It can be unbounded (think of the graph of the tangent function) or bounded (think of the cosine function); it can be open (think of any strictly increasing continuous function) or closed (think of the cosine function and a case where $a$ is much less than $b$) or half-open (e.g. $a=0$, $b=\pi$, $f=\sin$).