Analysis, Proof about, f being continuous, bijective

Robert IL

Let f: R[tex]\rightarrow[/tex]R be a non-decreasing function. Suppose that f maps Q to Q and f: Q[tex]\rightarrow[/tex]Q bijection. Prove that f: R[tex]\rightarrow[/tex]R is continuous, one to one and onto.

Hello everyone, I have been staring at this statement for a while now and I just don't understand it, hence I can't even begin to prove it. Can someone explain to me in different words what am I being asked to do. Is f the same same function and Q[tex]\rightarrow[/tex]Q is somehow "inside" R[tex]\rightarrow[/tex]R. I don't understand how f could be the same if R and Q don't have the same cardinality, or in other words-I am lost. Any type of help is greatly appreciated.

VeeEight

f maps Q to Q means that if q is in Q, f(q) is also in Q.
I am assuming (unless otherwise noted) that Q stands for the rationals.

Robert IL

Yes, Q=rationals, R=reals; but how and why is mapping of Q to Q relevant here, since I am later asked about map of function R to R?

VeeEight

It is relevant because it is a critical hypothesis for the problem. You are also given that f is non decreasing and that f restricted to Q is a bijection. Think about what the map f restricted to Q looks like. Where does f(q) go to?

Now, where does f map the irrational points to?

Robert IL

Since I=irrationals are uncountable and so are reals I guess we could create a bijection between the two (not sure if I am correct here), but I suppose I would be mapped to R.

some_dude

Ah, I deleted my previous post, what I said was incorrect!. But... let me try to be more helpful: try this outline of a proof (if you like):

1) show that the restriction of f to Q is continuous (treat Q as a metric space on its own and apply the epsilon delta definition).

2) show that there exists a unique extension of f:Q->Q to f:R->R such that f is continuous on R (proving this is fairly easy via contradiction).

3) show that f being non-decreasing on Q and continuous on Q implies it is this candidate function (step 2 isn't entirely necessary, but it should help provide a lot more intuition to make this last leap).

I'm quite drowsy, so if something sounds confusing tell me.

Robert IL

I'm sorry, but I'm not familiar with the term "unique extension" what do you mean by that?

some_dude

Oh no prob. An extension of a function f:Q->Q to a function g:R->R is g such that for every x in Q, f(x) = g(x).