# A continuous bijection $f:\mathbb{R}\to \mathbb{R}$ is an homeomorphism?

A continuous bijection $f:\mathbb{R}\to \mathbb{R}$ 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 $f^{-1}f$ are continuous identity, but I can't follow.

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Hint: A continuous injective function from the reals to the reals must be strictly increasing or strictly decreasing. – Michael Greinecker May 15 '12 at 21:55
@MichaelGreinecker I'll try now. – Gastón Burrull May 15 '12 at 22:03
@MichaelGreinecker Thanks for your answer. after hint, my question was much easier to prove, but Hint is not very hard to prove I think that I was proved correctly. – Gastón Burrull May 15 '12 at 22:44

Proof: Let $I=[a,b]$ be a closed interval with $a<b$ then without loss of generality I can asume $f(a)<f(b)$, I want to prove that $f|_{I}$ is strictly increasing. If not, there exists $x< y$ with $f(x)>f(y)$ by injectivity can't be the equality. If $f(a)<f(y)<f(x)$ intermediate value theorem give me $c\in(a,x)$ such that $f(c)=f(y)$ but it can't be by injectivity. If $f(y)<f(a)$ is the same. Hint follows from the fact that if $f$ is strictly increasing in any closed interval, obviously so is in reals. Then $f$ is strictly increasing. $\square$
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 $a\in\mathbb{R}$. Let $\epsilon>0$ then $g(a)-\epsilon/2$ and $g(a)+\epsilon/2$ have preimages let $g(b)=g(a)-\epsilon/2$ and $g(c)=g(a)+\epsilon/2$ then $b<a<c$ by hint, then we obtain $\delta=\min\{|b-a|,|c-a|\}$ then $g$ is continuous, therefore $f$ is a homeomorphism.