A continuous bijection f:R→R 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 f−1 and the fact that ff−1 and f−1f 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)byinjectivitycan′tbetheequality.Iff(a)
If image is all, I'll prove that f−1=g is continuous. Is well known that if f is strictly increasing then g so is. I'll prove continuity in some a∈R. 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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