Thread: continious function

A function f: A-> R is increasing if f(x)<= f(y) for every x, y in A such that x<=y.


suppose that f:[a,b] ->R is increasing and that for every Y in [f(a), f(b)] there is some x[0] in [a,b] sucht that f( x[0] ) = Y

How do i show that fis a continious function?

I have an idea.

The limit
And
So the left and right limits exists and are equal to the functional evaluation.

Because since the function is increasing it places the role of the least upper bound.

Thankz...
i want to show that the asuumption that f is increasing is essential by
finding an example of a function f:[-1,1] ->R which has the property that for every Y in [f(-1), f(1)] there is some x[0] in [-1,1] such that f( x[0] ) = Y , bu it is not continious and so fcan't be increasing.

Does any one know any examples with these values for a=-1 and b=1?

thankz

Originally Posted by dopi

Thankz...
i want to show that the asuumption that f is increasing is essential by
finding an example of a function f:[-1,1] ->R which has the property that for every Y in [f(-1), f(1)] there is some x[0] in [-1,1] such that f( x[0] ) = Y , bu it is not continious and so fcan't be increasing.

What you actually are asking is for the converse of the intermediate value theorem. You know that any continous function on a closed interval satisfies the Intermediate property. However you wish to now if a functions satisfies the Intermediate property then it is not necessarly continous.

Consider a semicircle
Now take any point not on endpoints of its extreme value and move it done "slightly" IVT still works but the function has replacable discontinuity.