Previous questions established, for example, that a continuous bijection $f:(0,1)\to [0,1]$ does not exist, but the proof relied on continuity. Clearly, with continuity similar proofs can be invoked to show there is no bijection $f:(-\infty,\infty)\to[0,1]$.
Similarly, one can show there is no order-preserving bijection: Suppose there were, then there is an $x\in(-\infty,\infty)$ such that $f(x)=1$. But there exists some $y\in(-\infty,\infty)$ such that $y>x$, and for a bijection this requires $f(y)>f(x)=1$, which is a contradiction.
But what if the bijection need not be continuous? Is there a bijection, or can you prove there isn't?