# Analysis, Proof about, f being continuous, bijective

1. Feb 28, 2010

### Robert IL

Let f: R$$\rightarrow$$R be a non-decreasing function. Suppose that f maps Q to Q and f: Q$$\rightarrow$$Q bijection. Prove that f: R$$\rightarrow$$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$$\rightarrow$$Q is somehow "inside" R$$\rightarrow$$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.

Last edited: Feb 28, 2010
2. Feb 28, 2010

### 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.

3. Feb 28, 2010

### 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?

4. Feb 28, 2010

### 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?

5. Feb 28, 2010

### 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.

6. Feb 28, 2010

### 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.

7. Feb 28, 2010

### Robert IL

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

8. Feb 28, 2010

### 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).