Analysis, Proof about, f being continuous, bijectiveby Robert IL Tags: analysis, bijective, continuous, nondecreasing, topology 

#1
Feb2810, 07:33 PM

P: 4

Let f: R[tex]\rightarrow[/tex]R be a nondecreasing 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 wordsI am lost. Any type of help is greatly appreciated. 



#2
Feb2810, 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. 



#3
Feb2810, 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?




#4
Feb2810, 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? 



#5
Feb2810, 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.




#6
Feb2810, 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 nondecreasing 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
Feb2810, 08:38 PM

P: 4

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




#8
Feb2810, 08:50 PM

P: 93




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