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Analysis, Proof about, f being continuous, bijective

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Robert IL
#1
Feb28-10, 07:33 PM
P: 4
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.
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VeeEight
#2
Feb28-10, 07:43 PM
P: 612
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
#3
Feb28-10, 07:45 PM
P: 4
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
#4
Feb28-10, 07:48 PM
P: 612
Analysis, Proof about, f being continuous, bijective

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
#5
Feb28-10, 07:56 PM
P: 4
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
#6
Feb28-10, 08:30 PM
P: 93
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
#7
Feb28-10, 08:38 PM
P: 4
I'm sorry, but I'm not familiar with the term "unique extension" what do you mean by that?
some_dude
#8
Feb28-10, 08:50 PM
P: 93
Quote Quote by Robert IL View Post
I'm sorry, but I'm not familiar with the term "unique extension" what do you mean by that?
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).


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