# Thread: Intervals and Continuous Functions

1. ## Intervals and Continuous Functions

Hello,

I'm trying to prove that if $\displaystyle f(x)$ is continuous and if $\displaystyle f(a)=c,f(b)=d$ then $\displaystyle f \colon [a,b] \longrightarrow [c,d]$ or $\displaystyle f:]a,b[ \longrightarrow ]c,d[$ in brief the image of a closed interval is a closed interval if f is continuous and the same with an open interval. Since I'm doing this just for fun I'm wondering if the definition of a continuous function is enough to prove it?( The definition I'd use is : f is continuous at x if and only if $\displaystyle \forall \varepsilon \exists \delat>0, |f(x_1)-f(x_2)| \leq \varepsilon, \forall x_1,x_2 \in ]x- \delta,x + \delta[ \cap \mathbf{D}$ (with $\displaystyle f \colon \mathbf{D}\subset \mathbf{R} \longrightarrow \mathbf{R}$.
If you can give me the full proof I'll gladly look at it !

Thanks

2. Are you sure this is true?
What about $\displaystyle f(x)=sinx$ with domain $\displaystyle (- 2 \pi, 2 \pi)$? What about constant functions?

3. hoooo so I think I need f bijective then. Well thanks now I think I know what to do.

4. Originally Posted by sunmalus
hoooo so I think I need f bijective then. Well thanks now I think I know what to do.
It is a well known theorem that the continuous image of a connected set is connected. The non-trivial connected sets in the real numbers,$\displaystyle \mathbb{R}^1$, are simply the non-degenerate intervals.

That does what you are trying to do.