Property about bijection $f:\mathbb{R \backslash Z} \to \mathbb{R \backslash Z} $ Could someone help me with the following question?
Let $f:\mathbb{R \backslash Z} \to \mathbb{R \backslash Z} $ be a continuous bijection. It is true that for any integer $n$ there exists another integer $m$ such that $ f(]n,n+1[)=]m,m+1[$?
My attempt:
By using that $f$ is continuos we obtain that $f(]n,n+1[)$ must be a connected set of $\mathbb{R \backslash Z}$, that is, there eists an integer $m$ such that $f(]n,n+1[)$ is an open interval contained in $]m,m+1[$. I ve not been able to prove the contrary inclusion.
It seems to me that the proof is very simple but I have not found it.
Thanks.
 A: Since $f$ is a continous bijection, it has to be strictly monotone on an interval $]n,n+1[$, and therefore as you state, $f(]n,n+1[)$ is an open interval contained in some interval $]m,m+1[$. 
Assume for contradiction that $f(]n,n+1[) \ne ]m,m+1[$. Then, since $f$ is surjective, $]m,m+1[$ is covered by open intervals $f(]n_i,n_i+1[)$, and since $f$ is injective, these intervals are disjoint. So $]m,m+1[$ is the disjoint union of at least 2 open sets, which is a contradiction since it is connected.
Edit: Let me answer your questions in the post instead of the comments.
"Why can we claim that there exists other $n$ such that $f(]n;n+1[) \subset ]m,m+1[$?"
This is because we assumed (to reach a contradiction) that $f(]n;n+1[) \ne ]m,m+1[$. Let $y \in ]m,m+1[\setminus f(]n;n+1[)$. Since $f$ is surjective, there must exist some $x$ such that $f(x)=y$, and notice that $x \ne ]n,n+1[$.
Let $n^\prime \in \mathbb{Z}$ such that $x \in ]n^\prime, n^\prime + 1[$.  Now $f(]n^\prime, n^\prime + 1[)$ is connected, so it is contained in $]m,m+1[$.
"Are we using that the preimage of an open interval is an open interval?"
No, that would make the question really easy.
A: Edit: Sorry about earlier reply, apparently  misread the question. Original answer at the bottom.
Since $f$ is a bijection, it surjects onto the interval $]m,m+1[$.
Assume $f(]n,n+1[)=]a,b[$, and consider  the preimage of $b$, when $b \ne m+1$. It should lie in some other interval, which maps under $f$ to another open interval, say $]c,d[$. But then the intersection of these two intervals is nonempty, contradicting the injectivity.
Previous answer:
Note that $f$ maps integers to integers, hence both $f(n)$ and $f(n+1)$ are integers. If there is an integer between them, show that it violates bijectiveness  - you will have two numbers between $n$ and $n+1$ mapping to the same value. Then you may apply your argument of connectivity to finish the proof.
A: After a more careful reading I realize my answer is not really different than the one already posted by Richard Jensen. (For that matter, both answers already posted are valid and complete in my opinion, so mine is redundant, but I will leave it, as a different way to write the same thing.) 
So for the other direction, suppose toward a contradiction, that $f(]n,n+1[)$ is contained in some $]m,m+1[$, but not equal to it. Then there is some $j\not=n$ such that $f(]j,j+1[)$ is also contained in $]m,m+1[$. Let $K=\{k:f(]k,k+1[)\subseteq ]m,m+1[\}$. 
Then $|K|\ge2$ and 
this gives a representation of $]m,m+1[$ as the disjoint union of more than one open intervals, namely $]m,m+1[=\cup\{f(]k,k+1[):k\in K\}$. This is a contradiction, since each of these open intervals 
$f(]k,k+1[)$ is both open-and-closed in the relative topology of $]m,m+1[$, contradicting that $]m,m+1[$ is connected. 
Here is the same argument again, explained in a slightly different way. So, 
$f(]n,n+1[)$ is contained in some $]m,m+1[$, but not equal to it. Let $$J=\cup\{f(]j,j+1[):j\not=n,\ f(]j,j+1[)\subseteq ]m,m+1[\}$$ Then $f(]n,n+1[)$ and $J$ are non-empty disjoint open sets covering $]m,m+1[$, contradicting that $]m,m+1[$ is connected. 
We have used that $f$ is a bijection (in particular onto), so the interval $]m,m+1[$ must be covered by $\cup\{f(]l,l+1[):l\in\Bbb Z\}$. 
