# IVT problem from Spivak

1. Aug 9, 2008

### mrb

I have the 4th edition of Spivak's Calculus. Problem 13(b) in Chapter 7 says:

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Suppose that f satisfies the conclusion of the Intermediate Value Theorem, and that f takes on each value only once. Prove that f is continuous.
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f(1) = 2
f(2) = 1
f(x) = x for all other x

And if you look at this on the interval [0, 3] then certainly for every c between f(0)=0 and f(3)=3 there is an x such that f(x)=c, so the IVT conclusion is satisfied. And f takes on each value only once. But f isn't continuous.

So isn't this a counterexample to what I'm supposed to prove?

2. Aug 9, 2008

### kidmode01

You're right, the converse of the intermediate value theorem is not true. Since your function is discontinuous, it works as a counterexample. There might be a misprint?

3. Aug 9, 2008

### HallsofIvy

Staff Emeritus
No, that's not a counter example. Suppose a= 0 and b= 1/4. Then the intermediate value theorem says that f must take on all values between f(a)= f(0)= 2 and f(b)= f(1/4)= 1/4 for some x between 0 and 1/4. But that is not true. For example, 1.5 is between 2 and 1/4 but f(x) is never equal to 1.5 between them. f does not "satisfy the intermediate value theorem".

4. Aug 9, 2008

### mrb

f(0) is not 2 but I see what you are saying.

I took it to mean that I was supposed to prove essentially this:

If for a function f and an interval [a, b], for any c between f(a) and f(b) there is some x in [a, b] such that f(x) = c and f takes on each value only once, then f is continuous on [a, b].

But apparently I was supposed to hold the hypothesis to be true for any interval, not just a specific one...

Thanks.