Thursday, March 24, 2016

real analysis - A continuous bijection f:mathbbRtomathbbR is an homeomorphism?




A continuous bijection f:RR is an homeomorphism. With the usual metric structure.




I always heard that this fact is true, but anyone shows to me a proof, and I can't prove it. I was tried using the definition of continuity but it is impossible to me conclude that. I tried using the fact that f bijective has an inverse f1 and the fact that ff1 and f1f are continuous identity, but I can't follow.


Answer



I think that I have the answer using Michael Greinecker hint. I'll first prove it, because I did not saw this theorem before.





Hint: A continuous injective function from the reals to the reals must be strictly increasing or strictly decreasing.




Proof: Let I=[a,b] be a closed interval with $af(y)byinjectivitycantbetheequality.Iff(a)

If image is all, I'll prove that f1=g is continuous. Is well known that if f is strictly increasing then g so is. I'll prove continuity in some aR. Let ϵ>0 then g(a)ϵ/2 and g(a)+ϵ/2 have preimages let g(b)=g(a)ϵ/2 and g(c)=g(a)+ϵ/2 then $b

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