# Thread: Variation on the Intermediate value theorem

1. ## Variation on the Intermediate value theorem

Suppose I have a collection of functions so that:
(a) each function maps Q into Q
(b) for each function f and each p in Q, there exists a q so that the limit of f(x) as x --> p = q.
Would the intermediate value theorem then hold?

2. Is Q here the set of rational numbers? If so, then no, the set of rational numbers is not connected and the intermediate value theorem, in a more abstract setting, essentially says that if f is continous then it maps a connected set to a connected set. (In the real numbers, a set is connected if and only if it is an interval. There are no connected subsets of the rational numbers.)

3. Thank you, HallsofIvy. Yes, Q is meant as the set of rational numbers, and in what follows R will be the set of real numbers. I should have incorporated my title into my question. What I meant to ask is whether a variation on the intermediate value theorem, in its most basic form, could hold or Q. That is, I was taking the original theorem in the form:

(1) Suppose that f : [a, b] → R is continuous and that u is a real number satisfying f(a) < u < f(b) or f(a) > u > f(b). Then for some c ∈ [a, b], f(c) = u.

This is of course false for Q, as you point out. However, I was thinking of a variation something like the following:

(2) Suppose that a and b are rational numbers, f is a function from [a,b] ∩ Q → Q such that the closure of f in R is continuous (that is, there exists a continuous function g: [a, b] → R such that f = g ∩ QXQ) and that q is a rational number satisfying f(a) < q < f(b) or f(a) > q > f(b). Then for some
p ∈ [a, b] ∩ Q, f(p) = q.

So, I wondered whether, although (1) is obviously false for Q, something like (2) could hold. That is, perhaps (2) will also not work, but I hope I have formulated (2) sufficiently to give an intuitive idea of what I am looking for. So: could something along these lines be formulated to hold?

4. No, that is still not true. If f(x) is continuous on [a, b] and q is between f(a) and f(b) there must exist a real number c such that f(c)= q but there is no reason to think that c is rational.

5. Again, thank you, but notice that in my construction (2) I did not make f continuous. Indeed, with my conditions there could not be an irrational c as you describe, since the functions were only defined on the rational numbers. Hence I do not see your objection as being applicable, and am not yet convinced that something similar to the theorem could not be postulated for the functions restricted in the way I have done.