Hi,
Could someone help me in proving this statement:
Let I be an interval and assume that f: I--->R is an increasing function. Prove that if the image of f(I) is connected then f must be continuous.
Thanks a lot
So far, I've written down the definition of an interval as follows:
There exists a,b in S and c in S such that a < c < b. S
Let S1 = Intersection (-inf,c)
Let S2 = Intersection (c,inf).
Then I've assumed f is not confinuous at a point a and started an epsilon-delta arguement.
I think it has something to do with f(c) + epsilon/2, but I can't complete the proof...
Use the notation $\displaystyle f(c+)$ for the limit on the right at $\displaystyle x=c$.
Likewise for the limit on the left, $\displaystyle f(c-)$.
Monotonic functions are quasi-continuous, so both exist.
If $\displaystyle f$ is not continuous at $\displaystyle x=c$ then either $\displaystyle f(c-)<f(c)$ or $\displaystyle f(c)<f(c+)$.
If $\displaystyle f(c-)<f(c)$ consider $\displaystyle \left( { - \infty ,f(c - )} \right] \cup \left[ {f(c),\infty } \right)$.
Try thinking of it this way. Suppose that $\displaystyle f$ was not continuous then you can find some $\displaystyle x_0\in S$, and some $\displaystyle \varepsilon>0$ such that for all ]math]\delta>0[/tex] there exists some $\displaystyle y_\delta$ such that $\displaystyle |x-y_\delta|<\delta$ and $\displaystyle |f(x)-f(y_\delta)|\geqslant \varepsilon$. Try creating a contradiction with this.
Hint: Think about the IV property.